Solved Exa
Example 1. An iron ball of mass 3 kg is
released from a height of 125 m and falls
freely to the ground. Assuming that the
value of g is 10 m/s2, calculate
(i) time taken by the ball to reach the
ground
(ii) velocity of the ball on reaching the
ground
(iii) the height of the ball at half the time it
takes to reach the ground.​

Answers

Answer 1

According to the equations of motion, the time taken to reach the ground is 5 seconds.

Using;

s = ut + 1/2gt^2

s = distance

u = initial velocity

t = time taken

g = acceleration due to gravity

Note that u = 0 m/s since the object was dropped from a height

Substituting values;

125 = 1/2 × 10  × t^2

125 = 5t^2

t^2 = 125/5

t^2 = 25

t = 5 secs

Velocity on reaching the ground is obtained from

v = u + gt

Where u = 0 m/s

v = gt

v = 10 × 5

v = 50 m/s

At half the time it takes to reach the ground;

s =  ut + 1/2gt^2

Where u = 0 m/s

s =  1/2gt^2

s = 1/2  × 10  × (2.5)^2

s = 31.25 m

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Answer 2

Answer:

(i) time taken by the ball to reach the

ground is 5 sec.

(ii) velocity of the ball on reaching the

ground is 50 m/s.

(iii) the height of the ball at half the time it

takes to reach the ground is 31.25 m.

Step-by-step explanation:

Solution :

(i) time taken by the ball to reach the

ground

[tex]\longrightarrow{\sf{ \: \: s= ut + \dfrac{1}{2} a{(t)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: 125= 0 \times t + \dfrac{1}{2} \times 10 \times {(t)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: 125= 0 + \dfrac{10}{2} \times {(t)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: 125= 0 + 5\times {(t)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: 125= 5\times {(t)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = \dfrac{125}{5}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = \dfrac{ \cancel{125}}{\cancel{5}}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = 25}}[/tex]

[tex]\longrightarrow{\sf{ \: \: t = \sqrt{25} }}[/tex]

[tex]\longrightarrow \: \: {\sf{\underline{\underline{\red{ t = 5 \: sec}}}}}[/tex]

Hence, the ball taken 5 sec to reach the ground.

[tex]\begin{gathered}\end{gathered}[/tex]

(ii) velocity of the ball on reaching the

ground

[tex]\longrightarrow{\sf{ \: \: {v}^{2} - {u}^{2} = 2as}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {v}^{2} - {0}^{2} = 2 \times 10 \times 125}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {v}^{2} = 20 \times 125}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {v}^{2} = 2500}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {v} = \sqrt{2500} }}[/tex]

[tex]\longrightarrow{\sf{ \: \: \underline{\underline{ \red{{v} = 50 \: m/s }}}}}[/tex]

Hence, the velocity of ball is 50 m/s.

[tex]\begin{gathered}\end{gathered}[/tex]

(iii) the height of the ball at half the time it

takes to reach the ground.

[tex]\longrightarrow{\sf{ \: \: s= ut + \dfrac{1}{2} a{(t)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= 0 \times \dfrac{5}{2} + \dfrac{1}{2} \times 10 \times { \left( \dfrac{5}{2} \right)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= 0 + \dfrac{10}{2} \times { \left( \dfrac{5}{2} \times \dfrac{5}{2} \right)}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times { \left( \dfrac{5 \times 5}{2 \times 2} \right)}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times { \left( \dfrac{25}{4} \right)}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times \dfrac{25}{4}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10 \times 25}{2 \times 4}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= \dfrac{250}{8}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= \dfrac{\cancel{250}}{\cancel{8}}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {\underline{\underline{\red{s= 31.25 \: m}}}}}}[/tex]

Hence, the height of the ball to reach the ground is 31.25 m.

[tex]\underline{\rule{220pt}{3.5pt}}[/tex]


Related Questions

Use part one of the fundamental theorem of calculus to find the derivative of the function.
f(x) =
0 5 + sec(5t)dt
x
Hint:
0
x
5 + sec(5t)
dt = −
x
0
5 + sec(5t)
dt
f '(x) =

Answers

Using the theorem of calculus the derived derivative of the function found is f(x) = ∫₀ˣ 5 + sec(5t) dt is f'(x) = -x^5 + sec(5t).

Using the first part of the Fundamental Theorem of Calculus, we can find the derivative of the function f(x) = ∫[0, x] (5 + sec(5t)) dt.

Let F(x) be the antiderivative of the integrand 5 + sec(5t) with respect to t. By evaluating the integral at the upper limit x and subtracting the value at the lower limit 0, we obtain F(x) - F(0).

To find the derivative of f(x), we differentiate both sides of the equation with respect to x. Using the chain rule, we have:

f'(x) = (d/dx)(F(x) - F(0))

Since F(0) is a constant, its derivative is zero. Therefore, the equation simplifies to:

f'(x) = d/dx (F(x)) = F'(x)

The derivative of F(x) is the original integrand, 5 + sec(5t). Therefore, the derivative of the function f(x) is:

f'(x) = 5 + sec(5t)

Hence, the derivative of f(x) is 5 + sec(5t).

The derivative of the function f(x) = ∫₀ˣ 5 + sec(5t) dt is f'(x) = -x^5 + sec(5t).

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What is the resistance (in kΩ) of a 5.00 ✕ 10² Ω, a 2.00 kΩ, and 3.50 kΩ resistor connected in series?
in kohms

Answers

The resistance of the circuit is 6.00 kΩ when resistor connected in series.

Resistance: It is the opposition to the flow of electric current. It is a measure of how much the resistor opposes the flow of electricity through it. Series: In a series circuit, the current that flows through each of the components is the same and the voltage across the circuit is the sum of the individual voltage drops. When we connect multiple resistors in series, we connect them end to end to create a single path for current flow. The total resistance of the series circuit is equal to the sum of the individual resistances. In this problem, three resistors are connected in series: a 5.00 * 10² Ω, a 2.00 kΩ, and a 3.50 kΩ resistor. We need to find the total resistance of the circuit. First, we need to convert the 5.00 * 10² Ω into kΩ by dividing by 1000. \frac{5.00 * 10² Ω }{ 1000 }= 0.5 kΩ

Now we can add up the resistances in kΩ to find the total resistance in kΩ.R(total) = 0.5 kΩ + 2.00 kΩ + 3.50 kΩR

(total) = 6.00 kΩ .Therefore, the resistance of the circuit is 6.00 kΩ.

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Compare the density of a 1,000,000 kg iceberg to that of a 10 g ice cube taken from the
iceberg. Explain your answer.

ANYONE ASAPP ITS SO IMPORTANT PLSSS

Answers

Answer:

The density of the 1,000,000 kg iceberg = The density of the 10 g iceberg

Explanation:

The given quantities of iceberg that will be the basis of the comparison of the densities of the iceberg;

The mass of the iceberg = 1,000,000 kg

The mass of the ice cube taken from the iceberg = 10 g

The density of a substance, is the measure of the mass of the substance per volume of the substance, it is a constant for the substance

[tex]Density = \dfrac{Mass}{Volume}[/tex]

For the iceberg, the density of the iceberg is the property that indicates the volume occupied by a given mass of the iceberg

Because of density, we can estimate that the volume occupied by a 1,000 kg iceberg is 10 times the volume occupied by a 100 kg iceberg

We can write

The density of the 100 kg iceberg = 100 kg/(x m³)

The density of the 1,000 kg iceberg = 1,000 kg/(10 × x m³) = 100 kg/(x m³)

Therefore, the density of the 100 kg iceberg = The density of the 1,000 kg iceberg

Similarly,

The density of the 1,000,000 kg iceberg = The density of the 10 g iceberg because the 10 g iceberg was obtained from the 1,000,000 kg iceberg, and therefore, they have the same density.

Two charges, one with a charge of +10.0 x 10-6 C, the other with a charge of -3 x 10-6 C exert a force on each other with a magnitude of 1.7 Newtons on each other. Is this a repulsive or attractive force. What is the separation distance of these charges?

Answers

Answer:

b.

Explanation:

Answer:

Attractive force and r = 0.399 m

Explanation:

One charge is positive and the other charge is negative. Opposite charges attract, so there has to be a force that attracts between them.

q1 = 10.0 x 10^-6 C

q2 = -3 x 10^-6 C

F = 1.7 N

Plug those values into Coulomb's Law:

[tex]F = k\frac{q1q2}{r^{2} } \\1.7 = \frac{(9x10^{9})(10.0 x 10^{-6})(-3 x 10^{-6})}{r^{2} }[/tex]

Solve for r

r = 0.399 m

predict the ratio of the periods, t1/t2, of two masses m1 and m2 (=4m1) that oscillate in shm on springs that have the same spring constant k.Show the reasoning behind your prediction.

Answers

The ratio of the periods, t1/t2, for the two masses m1 and m2 in simple harmonic motion (SHM) on springs with the same spring constant k is 1:2.

In SHM, the period of oscillation (T) is given by the equation:

T = 2π√(m/k)

where T is the period, m is the mass, and k is the spring constant.

Let's calculate the periods for the two masses:

For mass m1:

T1 = 2π√(m1/k)

For mass m2 (which is 4 times m1):

T2 = 2π√(m2/k)

To find the ratio of the periods, we divide T1 by T2:

t1/t2 = (T1)/(T2)

Substituting the expressions for T1 and T2:

t1/t2 = (2π√(m1/k))/(2π√(m2/k))

Canceling out the common factors of 2π, we get:

t1/t2 = (√(m1/k))/(√(m2/k))

Since m2 = 4m1, we can substitute this value:

t1/t2 = (√(m1/k))/(√((4m1)/k))

Simplifying further:

t1/t2 = (√(m1/k))/(2√(m1/k))

The square root terms cancel out, resulting in:

t1/t2 = 1/2

The ratio of the periods, t1/t2, for the two masses m1 and m2 that oscillate in SHM on springs with the same spring constant k is 1:2.

This means that the period of oscillation for the larger mass (m2) is twice the period of the smaller mass (m1).

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Initial velocity vector vA has a magnitude of 3. 00 meters per second and points 20. 0o north of east, while final velocity vector vB has a magnitude of 6. 00 meters per second and points 40. 0o south of east. Find the magnitude and the direction of the change in velocity vector Δv (which is the vector subtraction of the two vectors: final velocity vector minus initial velocity vector).

Answers

We are givenInitial velocity vector vA has a magnitude of 3.00 meters per second and points 20.0o north of east, while final velocity vector vB has a magnitude of 6.00 meters per second and points 40.0o south of east. We need to find the magnitude and the direction of the change in velocity vector Δv (which is the vector subtraction of the two vectors:

final velocity vector minus initial velocity vector).Let's solve the given problem:From the above figure, the direction of Δv is at an angle θ to the east of south:

[tex]θ = θ2 - θ1= 40.0 - (-20.0)= 60.0o[/tex]

Magnitude of the Δv: Let's use the Pythagorean theorem to find the magnitude of Δv. We have:[tex]$$|\Delta \vec{v}| = \sqrt{|\vec{v}_B|^2+|\vec{v}_A|^2-2|\vec{v}_A||\vec{v}_B|\cos(\theta)}$$[/tex]  

Putting the given values in the above equation, we get

[tex]$$|\Delta \vec{v}| = \sqrt{(6.00)^2+(3.00)^2-2(6.00)(3.00)\cos(60.0)}$$$$|\Delta \vec{v}| = 3.10\ \text{m/s}$$[/tex]

So, the magnitude of the Δv is 3.10 m/s.Therefore, the magnitude and the direction of the change in velocity vector Δv is 3.10 m/s at an angle of 60.0o to the east of south.

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Find the Potential Difference across the 2 Ω resistor. Answer in units of V.
Image attached of circuit diagram, question needing help on is the second one in the picture. Thank you!!
(Please only answer if you know how to find the V I know what the current is already.)

Answers

The potential difference across the 2 Ω resistor is 2 V.

How to calculate the potential difference

The potential difference across the 2 Ω resistor is equal to the current flowing through it multiplied by the resistance of the resistor. The current flowing through the circuit is 1 A, and the resistance of the 2 Ω resistor is 2 Ω.

Therefore, the potential difference across the 2 Ω resistor is:

= 1 A * 2 Ω = 2 V.

V = I * R

V = 1 A * 2 Ω

V = 2 V

Therefore, the potential difference across the 2 Ω resistor is 2 V.

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A girl and a boy are riding on a merry-go-round that is turning at a constant rate. The girl is near the outer edge, and the boy is closer to the center. (1) For a given elapsed time interval, which rider has greater angular displacement?
(a) Both the girl and the boy have the same nonzero angular displacement.
(b) Both the girl and the boy have zero angular displacement.
(c) The boy has greater angular displacement.
(d) The girl has greater angular displacement.
(2) Who has greater linear speed?
(a) Both the girl and the boy have zero linear speed.
(b) The girl has greater linear speed.
(c) Both the girl and the boy have the same nonzero linear speed.
(d) The boy has greater linear speed.

Answers

(1) Both the girl and the boy have the same nonzero angular displacement.

Hence the correct option is A.

(2) The girl has greater linear speed.

Hence the correct option is B.

(1) For a given elapsed time interval, both the girl and the boy will have the same angular displacement. The angular displacement is determined by the angle swept out by the riders as the merry-go-round rotates. Since both riders are on the same merry-go-round and are moving with it at the same rate, they will both have the same angular displacement.

Therefore, Both the girl and the boy have the same nonzero angular displacement.

Hence the correct option is A.

(2) The linear speed of a rider depends on their distance from the center of the merry-go-round. The linear speed is given by the formula:

v = ω * r

Where:

v is the linear speed

ω is the angular speed (which is constant for the merry-go-round)

r is the distance from the center of the merry-go-round

Since the girl is near the outer edge of the merry-go-round, she has a greater distance from the center compared to the boy. As a result, the girl will have a greater linear speed than the boy.

Therefore, The girl has greater linear speed.

Hence the correct option is B.

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what is the minimum thickness of a soap bubble film with refractive index 1.50 that results in a constructive interference in the reflected light if this film is illuminated by a beam of light of wavelength 580 nm?

Answers

When light waves fall on a thin film of oil or soap bubble, they reflect back from both the top and bottom surfaces. Therefore, the minimum thickness of the soap bubble film is 386.7 nm.

The light waves interfere with one another and either enhance or cancel each other out, depending on their relative phase at the time of reflection. When two light waves reinforce each other, the interference is constructive. In this question, we have to find the minimum thickness of a soap bubble film with refractive index 1.50 that results in constructive interference in reflected light if this film is illuminated by a beam of light of wavelength 580 nm.So, we know that the condition for constructive interference is given by2t=nλ/1.5where, t is the thickness of the soap film, λ is the wavelength of the light and n is the order of interference. To find the minimum thickness, we need to consider the first order of interference, i.e., n=1.Substituting the values in the above equation, we get2t= (1 × 580 × 10-9) /1.5= 0.3867 × 10-6 m= 386.7 nm.

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Which human activity causes the most erosion?
А
building a bridge over a river

B
cutting down trees for lumber

С
building a dam in a stream

D
planting crops in a field

Answers

Answer:

B

Explanation:

calculate the magnetic field strength (t) needed on the loop to create a maximum torque of 320 n⋅m if the loop is carrying 21 a.

Answers

The magnetic field strength needed on the square loop to create a maximum torque of 320 N·m is approximately 43.24 N/A.

To calculate the magnetic field strength needed to create a maximum torque on a square loop, we can use the formula:

Torque (T) = N × B × A × sinθ

Where:

T = Torque

N = Number of turns in the loop

B = Magnetic field strength

A = Area of the loop

θ = Angle between the magnetic field and the plane of the loop

In this case, we are given:

N = 185 turns

A = (20.0 cm)² = 0.04 m² (since the loop is square)

T = 320 N·m

θ = 90 degrees (since the torque is maximum)

Rearranging the formula, we can solve for B:

B = T / (N × A × sinθ)

Substituting the given values:

B = 320 N·m / (185 × 0.04 m² × sin(90°))

B = 320 N·m / (7.4 m²)

B ≈ 43.24 N/A

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The question is -

Calculate the magnetic field strength needed on a 185-turn square loop 20.0 cm on a side to create a maximum torque of 320 N·m, if the loop is carrying 21 A.

T?

what matches ????????????????

Answers

Answer:

1st: Radiation

2nd: Conduction

3rd: Convection

Explanation:

I actually learned this before in school. Yay

which, if either, of the forces pictured as acting upon the rod in the diagram will produce a torque about an axis perpendicular to the plane of the diagram at the left end of the rod?

Answers

The force F2 will produce a torque about an axis perpendicular to the plane of the diagram at the left end of the rod.

To determine which force produces a torque about an axis perpendicular to the plane of the diagram at the left end of the rod, we need to consider the concept of torque and the position of the forces relative to the axis.

Torque is the rotational equivalent of force and is given by the equation: Torque = Force × Distance × sin(θ), where Force is the magnitude of the force, Distance is the perpendicular distance from the axis of rotation to the line of action of the force, and θ is the angle between the force and the lever arm.

In the given diagram, we have two forces acting on the rod, F1 and F2. The lever arm for each force is the distance from the left end of the rod to the line of action of the force.

For force, F1, the line of action passes through the left end of the rod. Therefore, the lever arm is zero, and sin(θ) is also zero since the angle between the force and the lever arm is 0 degrees. Consequently, the torque produced by force F1 is zero.

For force, F2, the line of action is not passing through the left end of the rod. The lever arm for force F2 is the perpendicular distance from the left end of the rod to the line of action of F2. Since this distance is non-zero and the angle between the force and the lever arm is non-zero, both the distance and sin(θ) are non-zero.

Therefore, the torque produced by force F2 is non-zero and will produce a torque about an axis perpendicular to the plane of the diagram at the left end of the rod.

Out of the two forces pictured, the only force F2 will produce a torque about an axis perpendicular to the plane of the diagram at the left end of the rod. Force F1 will not produce any torque since its line of action passes through the left end of the rod, resulting in a lever arm of zero.

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A charge of +5.0 x 10-6 C is situated 0.2 meters away from another isolated charge of -3.0 x 10-6 C. What is the magnitude of the electric force that these charges exert on each other? Is this a repulsive or attractive force?

Answers

Answer:

since the charges are of different nature it's a attractive force

Explanation:

magnitude of force=

9*10^9*5*10^-6*3*10^-6/0.04

= 3.375N answer

A train is moving with the velocity 10 m/s. It attains an acceleration of 4 m/s² after 5 seconds. Find the distance covered by the train in that time. ​

Answers

The train covers a distance of 100 meters in the given time.

To find the distance covered by the train in the given time, we can use the equations of motion.

S = ut + (1/2)at²

The equation S = ut + (1/2)at² is derived from the basic equations of motion. The first term (ut) represents the distance covered in the initial velocity u multiplied by time t. The second term (1/2)at² represents the distance covered due to the acceleration a during time t.

The initial velocity (u) of the train is 10 m/s, and the acceleration (a) is 4 m/s². We are given that this acceleration is attained after 5 seconds, so the time (t) is also 5 seconds. We need to find the distance covered (S).

Substituting the given values:

S = (10 m/s)(5 s) + (1/2)(4 m/s²)(5 s)²

S = 50 m + (1/2)(4 m/s²)(25 s²)

S = 50 m + 50 m

S = 100 m

It's important to note that the given problem assumes a constant acceleration throughout the entire time interval.

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the electric potential at a distance of 6 m from a certain point charge is 240 v relative to infinity. what is the electric potential (relative to infinity) at a distance of 2 m from the same charge?

Answers

Given that the electric potential at a distance of 6 m from a certain point charge is 240 V relative to infinity. The electric potential (relative to infinity) at a distance of 2 m from the same charge is 80 V.

Electric potential (relative to infinity) at a distance of 2 m from the same charge can be calculated as follows: By using the formula of electric potential, the electric potential at any point of space due to a point charge is given by;

V = kq / r

where,V = Electric potential due to point charge

q = Point charge

k = Coulomb's constant = 9 × 10^9 Nm^2C^-2

r = Distance between the charge and point where electric potential is to be calculated. Hence,Electric potential at a distance of 6 m from point charge q,V = kq / r1 = 9 × 10^9 × q / 6 ............(1)

Electric potential at a distance of 2 m from point charge q,

V = kq / r2 = 9 × 10^9 × q / 2 ............(2)

Divide equation (1) by equation (2), we get,

240 / V = 6 / 2V = 240 / (6 / 2)

By solving the above equation, we get

V = 80 V

Therefore, the electric potential (relative to infinity) at a distance of 2 m from the same charge is 80 V.

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A common way to measure the distance to lightning is to start counting, one count per second, as soon as you see the flash. Stop counting when you hear the thunder and divide by five to get the distance in miles. Use this information to estimate the speed of sound in m/s. Show your work below. This will require some conversions.

Answers

Answer:

d = 1.07 mile

Explanation:

The rationale for this method is that the speed of light is much greater than the speed of sound, the definition of speed in uniform motion is  

v = d / t  

d = v t  

the speed of sound is worth  

v = 343 m / s  

Therefore, the speed of sound must be multiplied by time to do this, all the units must be in the same system, as the distance in miles is requested  

v = 343 m/s (1mile/1609 m) (3600s/1 h) = 343 (2.24) = 767.4 mile/h  

v = 343 m / s (1 mile / 1609 m) = 0.213, mile/ s  

If the measured time is t = 5s we multiply it by the speed  

we substitute  

d = 0.213 5

d = 1.07 mile

If you want to calculate the speed, this method in general is not widely used, since you must know the distance where the lightning occurred, which is relatively complicated.

a soccor ball is dropped from a height of h1 = 3.05 m above the ground. after it bounces, it only reaches a height h2 = 2.12 m above the ground. the soccor ball has mass m = 0.115 kg.
Part (a) What is the magnitude of the impulse , in kilogram meters per second, the soccer ball experienced during the bounce?
Part (b) If the soccer ball was in contact with the ground for , what was the magnitude of the constant force acting on it, in Newtons?
Part (c) How much energy, in joules, did the soccer ball transfer to the environment during the bounce

Answers

(a) The magnitude of the impulse experienced by the soccer ball during the bounce is 0.6923 kg·m/s, (b) the magnitude of the constant force acting on the soccer ball during the bounce is 13.846 N and (c) the soccer ball transferred approximately 12.026 joules of energy to the environment during the bounce.

What is energy and how is it measured?

Energy is a fundamental concept in physics that refers to the ability of a system to do work or cause a change. It is a scalar quantity and is associated with various forms, such as kinetic energy, potential energy, thermal energy, and others.

The SI unit of energy is the joule (J).

Part (a):

The magnitude of the impulse (J) experienced by the soccer ball during the bounce can be calculated using the equation:

J = Δp,

where Δp is the change in momentum.

The change in momentum is given by:

Δp = m * Δv,

where m is the mass of the soccer ball and Δv is the change in velocity.

The initial velocity of the soccer ball is zero as it is dropped from rest. The final velocity can be calculated using the equation for final velocity in free fall:

v² = u² + 2gh,

where v is the final velocity, u is the initial velocity (zero in this case), g is the acceleration due to gravity, and h is the height.

Calculating the final velocity:

v²  = 0² + 2 * 9.8 m/s² * 2.12 m,

v ≈ 6.02 m/s.

Substituting the values into the equation for change in momentum:

Δp = m * (v - u),

Δp = 0.115 kg * (6.02 m/s - 0 m/s).

Calculating:

Δp ≈ 0.6923 kg·m/s.

Therefore, the magnitude of the impulse experienced by the soccer ball during the bounce is approximately 0.6923 kg·m/s.

Part (b):

The magnitude of the constant force (F) acting on the soccer ball can be calculated using the equation:

F = Δp / Δt,

where Δp is the change in momentum and Δt is the time interval.

Given that the soccer ball was in contact with the ground for Δt = 0.05 s, we can substitute the values into the equation:

F = 0.6923 kg·m/s / 0.05 s.

Calculating:

F = 13.846 N.

Therefore, the magnitude of the constant force acting on the soccer ball during the bounce is 13.846 N.

Part (c):

The energy transferred to the environment during the bounce can be calculated as the work done by the force of the ball on the ground.

The work done is given by:

W = F * d,

where F is the magnitude of the force and d is the distance over which the force acts.

In this case, the force acts over the distance between the initial and final heights, which is h₁ - h₂.

Substituting the values:

W = 13.846 N * (3.05 m - 2.12 m).

Calculating:

W ≈ 12.026 J.

Therefore, the soccer ball transferred approximately 12.026 joules of energy to the environment during the bounce.

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A uniform, solid disk with mass m and radius R is pivoted about a horizontal axis through its center. A small object of the same mass m is glued to the rim of the disk.
If the disk is released from rest with the small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.
(Express your answer in terms of the variables m, R, and appropriate constants.)

Answers

The angular speed ω[tex]_{final}[/tex]when the small object is directly below the axis is 0. This means that the system comes to rest in that position.

To solve this problem, we can use the principle of conservation of angular momentum. When the small object is directly below the axis, the total angular momentum of the system remains constant.

The angular momentum L of an object is given by the formula:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular speed.

The moment of inertia of a solid disk rotating about an axis through its center is given by:

I_disk = (1/2) × m × R²

where m is the mass of the disk and R is the radius.

Similarly, the moment of inertia of a point mass m located at the rim of the disk is given by:

I_object = m × R²

The total moment of inertia of the system, when the small object is glued to the rim, is the sum of the moments of inertia of the disk and the object:

[tex]I_{total}[/tex] = I_disk + I_object

[tex]I_{total}[/tex]= (1/2) × m × R² + m × R²

[tex]I_{total}[/tex]= (3/2) × m × R²

Now, at the initial position, the angular momentum of the system is given by:

I[tex]_{total}[/tex] × ω[tex]_{initial}[/tex]= L[tex]_{initial}[/tex]

Since the disk is released from rest, ω_initial is 0.

When the small object is directly below the axis, the moment of inertia becomes:

I[tex]_{final}[/tex]= I[tex]_{disk}[/tex]

The angular momentum at this position is:

L[tex]_{final}[/tex]= I[tex]_{final}[/tex] × ω[tex]_{final}[/tex]

Since angular momentum is conserved, we can equate the initial and final angular momentum:

L[tex]_{initial}[/tex] = L[tex]_{initial}[/tex]

I[tex]_{total}[/tex] × ω[tex]_{initial}[/tex]= I[tex]_{final}[/tex] × ω[tex]_{final}[/tex]

Substituting the expressions for the moments of inertia and simplifying:

[(3/2) × m × R²] * 0 = (1/2) × m × R² × ω_final

Simplifying further:

0 = (1/2) * ω[tex]_{final}[/tex]

Therefore, we find that the angular speed ω_final when the small object is directly below the axis is 0. This means that the system comes to rest in that position.

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Learning Goal: To understand the forces between a bar magnet and 1. a stationary charge, 2. a moving charge, and 3. a ferromagnetic object. A bar magnet oriented along the y axis can rotate about an axis parallel to the z axis. Its north pole initially points along j^.

Answers

Solution :

As the charge is stationary, hence

[tex]$F_m= qvB \sin \theta$[/tex]

[tex]$F_m=0$[/tex]

Hence, no torque at all.

When the charge is moving in positive x direction and the field will be in the negative y direction outside the bar, then :

[tex]$F = q(V \hat i \times B(- \hat j))$[/tex]

    [tex]$= -qV B (\hat i \times \hat j)$[/tex]

    [tex]$=qVB(- \hat k)$[/tex]

Hence, the force have direction [tex]$(- \hat k)$[/tex].

When instead of charge, an iron nail is used, then there will be induced magnetic field in the soft iron. The nature of the pole induced will be opposite near tot he bar. That is the north pole will be induced near the south pole and vice versa. That is why whichever be the pole of magnet closest to iron will be attracted by iron.

The energy used to move against the magnetic force is stored as (pick one: potential or kinetic)

Answers

I would pick potential

Please help, I'm taking a test mlnkhjbgvfgcfgvhb


What is the motion of the particles in this kind of wave?

A) The particles will move up and down over large areas.

B) The particles will move up and down over small areas.

C) The particles will move side to side over small areas.

D) The particles will move side to side over large areas.

Answers

Answer:

I think its A

Explanation:

Not 100 percent sure tho but they do go up and down in big movements.

What is a metallic bond?
Explain in like a simple way please

Answers

Metallic bond, force that holds atoms together in a metallic substance. ... The atoms that the electrons leave behind become positive ions, and the interaction between such ions and valence electrons gives rise to the cohesive or binding force that holds the metallic crystal together.
Metallic bond, force that holds atoms together in a metallic substance. ... The atoms that the electrons leave behind become positive ions, and the interaction between such ions and valence electrons gives rise to the cohesive or binding force that holds the metallic crystal together.

A metallic bond is the sharing of many detached electrons between many positive ions, where the electrons act as a "glue" giving the substance a definite structure. It is unlike covalent or ionic bonding. ... The electrons and the positive ions in the metal have a strong attractive force between them.

Station 1: Sierra is running in a school track
meet. She will need extra energy to complete
the race and her body systems need to work
together to help her get it. Cellular respiration
is the way our bodies get energy out of the food
we eat. Her body needs to make sure her cells
get enough oxygen for cellular respiration to
occur but removes the carbon dioxide that is
built up in this same process. Which body systems will work together to
maintain the energy Sierra needs?​

Answers

Answer:

The body systems that work together to maintain the energy Sierra needs are;

The digestive system, the respiratory system, and the circulatory system

Explanation:

Cellular respiration in the body cells require oxygen to produce energy which are used by the muscles and other body cells. Carbon dioxide is also produced and is the build up of carbon dioxide has to be removed from the body as the by product of cellular respiration which is toxic at the cell level

Therefore, the body systems that work together to maintain the energy Sierra needs are;

The digestive system; Takes in the energy containing food and brakes them into chemicals that are transported to the cells for cellular respiration

The respiratory system; Takes in oxygen and removes carbon dioxide from the blood from and to the atmosphere

The circulatory system; Supplies food and oxygen from the digestive and respiratory system to the cells and transports produced carbon dioxide from the cells to the lungs from where it is passed out of the body by th respiratory system.

Carter is going camping outside he wants to bring a frying pan that will heat up and cool down quickly Which frying pan should carter use that will heat up and cool down the fastest


A copper
B iron
C glass
D steel

Plz help

Answers

It’s a because the science backs it up as the science shows

The specific heat of the copper pan is the lowest among the other given pans. Therefore, a copper pan should carter used that will heat up and cool down the fastest. Therefore, option (A) is correct.

What is the specific heat?

Specific heat of a substance can be described as the heat energy required to raise the temperature of one unit mass of a material by 1 °C. The S.I. unit of the specific heat capacity of a material is J/g°C.

The thermal capacity of a substance is a physical characteristic of a substance so it depends upon the nature of the material.

The mathematical expression of specific heat can be represented as :

Q = m×C× ΔT        Where C is the specific heat.

The specific heat is an intensive characteristic of a material and does not depend upon the shape or size of the quantity.

As given the values of the specific heat of metals of the frying pans, the highest specific heat means that the pan will take more heat to increase just one-degree temperature. As copper has the lowest specific heat so it can easily heat up in comparison to other pans.

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For a mass hanging from a spring, the maximum displacement the spring is stretched or compressed from its equilibrium position is its

Answers

Answer: The maximum displacement the spring is stretched or compressed from its equilibrium position is its AMPLITUDE.

Explanation:

In simple harmonic motion, the restoring force which pulls the oscillating body back towards its rest position is proportional in magnitude to the displacement of the body from the rest position.

The simple harmonic motion in terms of MASS AND SPRING, simple pendulum and loaded test tube is the motion or movement of a particle in a to and fro movement along a straight line under the influence of force.

Mass and spring: This means when a string of suspended mass, M, with initial level of the spring is at rest, the spring will start moving upward and downward due to the imbalance of the suspended mass.

The maximum displacement as the spring is stretched or compressed from its equilibrium position is its AMPLITUDE. This is measured in units of meter.

the power dissipated in a series rcl circuit is 65.0 w, and the current is 0.530 a. the circuit is at resonance. determine the voltage of the generator.

Answers

The voltage of the generator in the series RCL circuit at resonance is approximately 122.64 volts.

To determine the voltage of the generator in a series RCL circuit, we need to use the power and current values. In a series RCL circuit at resonance, the power dissipated is equal to the power supplied by the generator.

In this case:

Power dissipated (P) = 65.0 W

Current (I) = 0.530 A

The formula for power in an electrical circuit is:

P = VI

Where:

P is the power in watts

V is the voltage in volts

I is the current in amperes

Rearranging the formula to solve for voltage (V), we get:

V = P / I

Substituting the given values:

V = 65.0 W / 0.530 A

V ≈ 122.64 volts

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what is the velocity of something that traveled 6 meters in .96 seconds

Answers

Answer: the average velocity should be 13.422

Explanation:

tell me if i'm wrong please because i want to know if my calculations are reliable

An ammeter measures that the current in a simple circuit is 0.22 amps. The circuit is connected to a 55V battery. What is the resistance in the circuit?

Answers

Answer: The resistance in the circuit is 250 ohms

Explanation:

According to Ohm's law:

[tex]V=IR[/tex]

where V = voltage  = 55 V

I = current in Amperes = 0.22 A

R = Resistance = ?

Putting in the values we get:

[tex]55V=0.22A\times R[/tex]

[tex]R=250ohm[/tex]

Thus the resistance in the circuit is 250 ohms

Scientists at the Hopkins Memorial Forest in western Massachusetts have been collecting meteorological and environmental data in the forest data for more than 100 years. In the past few years, sulfate content in water samples from Birch Brook has averaged 7. 48 mg/L with a standard deviation of 1. 60 mg/L

Answers

Hopkins Memorial Forest in western Massachusetts has been collecting meteorological and environmental data for more than 100 years. Scientists at the forest have observed that in the past few years, the sulfate content in water samples collected from Birch Brook has averaged 7.48 mg/L with a standard deviation of 1.60 mg/L.

Meteorological and environmental data are crucial for understanding the state of the natural environment. Hopkins Memorial Forest in western Massachusetts has been collecting this data for over a century. The data collected in the forest can provide valuable insights into how environmental factors such as climate change, pollution, and other factors affect the local ecosystem.They have found that the sulfate content in these samples has averaged 7.48 mg/L with a standard deviation of 1.60 mg/L.

This information is useful for understanding how sulfate levels in the water are changing over time, and whether this could have any implications for the local ecosystem.The scientists at Hopkins Memorial Forest in western Massachusetts have a unique dataset that can provide valuable insights into how environmental factors affect ecosystems over time. By continuing to collect meteorological and environmental data, they will be able to gain a better understanding of how the environment is changing and how these changes could affect the local ecosystem in the future.

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