Answer:
y = 6 10⁻² m
Explanation:
This is a diffraction exercise which is described by the expression
a sin θ = m λ
we can use trigonometry to find the distance from the inside of the screen to the dark point (y)
tan θ = y / L
angles are very small in diffraction experiments, so we can approximate
tan θ = sin θ /cos θ = sin θ
sin θ = y / L
substituting
a (y / L) = m λ
Let's find the points where the intensity becomes zero
y = m L λ / a
λ₁ = 4.0 10⁻⁷ m
m = 1
y = 1 2.0 4.0 10⁻⁷/ 4.0 10⁻⁵
y = 2 10⁻² m
m = 2
y = 4 10⁻² m
λ₂ = 6.0 10⁻⁷ m
m = 1
y = 1 2.0 6.0 10⁻⁷ / 4.0 10⁻⁵
y = 3 10⁻² m
m = 2
y = 6 10⁻² m
if we want a point where the two colors are dark, we set the two expressions equal
y₁ = y₂
m₁ L λ₁ / a = m₂ L λ₂ / a
m₁/m₂ = λ₂/λ₁
m₁ / m₂ = 6 10⁻⁷ / 4 10⁻⁷
m₁ / m₂ = 1.5
since the quantities m must be integers, the smallest relation that the relation fulfills is
m₁ = 3
m₂ = 2
the distance for this destructive interference is
y = 3 2 4.0 10⁻⁷ / 4.0 10⁻⁵
y = 6 10⁻² m
this is the first point where the minimum of the two wavelengths coincide
Aluminum wire with a diameter of 0.8650 mm is wound onto a spool. The wire is insulated, but you have access to both ends. The resistivity of aluminum at 20.0 °C is 2.65 x 10^-8 Ω-m. You measure the resistance of the wire at that temperature, and it is 2.48 Ω. What is the length of the wire?
a. 8.10 x 10^4 m
b. 22.0 m
c. 5.68 m
d. 0.111 m
e. 55.0 m
Answer:
e. 55.0 m
Explanation:
Given;
diameter of the aluminum wire, d = 0.865 mm
radius of the wire, r = d/2 = 0.4325 mm = 0.4325 x 10⁻³ m
resistivity of the wire, ρ = 2.65 x 10⁻⁸ Ω-m
resistance of the wire, R = 2.48 Ω
The resistance of a wire is given by;
[tex]R = \frac{\rho \ L}{A} \\\\[/tex]
where;
L is length of the wire
A is area of the wire = πr² = π(0.4325 x 10⁻³ )² = 5.877 x 10⁻⁷ m²
Substitute the givens and solve for L,
[tex]L = \frac{RA}{\rho} \\\\L = \frac{(2.48)(5.877*10^{-7})}{2.65*10^{-8}}\\\\L = 55.0 \ m[/tex]
Therefore, the length of the wire is 55.0 m
The the figure shows a famous roller coaster ride. You can ignore friction. If the roller coaster leaves Point Q from rest, what is its speed at the top of 25 m peak(Point s
Answer:
22 m/s
Explanation:
PEf +KEf =PE0 +KE0 →PE0 −PEf =KEf
−mgΔy= 1 mv2 →v= −2gΔy = −2(9.8 m/s2)(−25 m)=22 m/s
Measurements of the radioactivity of a certain isotope tell you that the decay rate decreases from 8255 decays per minute to 3110 decays per minute over a period of 4.50 days.
What is the half-life (T1/2) of this isotope?
Answer:half-life (T1/2) of this isotope =
Explanation:
The number of nuclei of any radioactive substance at a given time is expressed by
Nt = N0e⁻kt
Nt=decay of material at a time t, =3110 decays per minute
N=decays at t=0, 8255 decays per minute
k=constant
Nt=N0e−kt
3110= 8255 e⁻k(4.50)
3110/ 8255=e−k(4.50)
0.3767 = e−k(4.50)
In 0.3767 = -k (4.50)
0.976=-4.5k
k=0.976/4.5
=0.2159
Also we know that t 1/2= time that it takes half the original material to decay.it is related to the rate constant by
T₁/₂=ln 2 / k
Therefore half-life (T1/2) of this isotope
T₁/₂=ln 2/0.2159
T₁/₂=3.12 days
a 1000kg car uses a breaking force of 10,000N to stop in two second. What impulse acts on the car?
Answer:
5,000
Explanation:
Vf = Vi + a * t
Consider a block sliding down a ramp whose motion is opposed by frictional forces. The total energy of this system is modeled by the equation:
Etotal = 1/2mv^2 + mgh + Ff(f is underscore)d
Which part of the equation represents the amount of energy converted to thermal energy?
A. mg
B. Ffd
C. mgh
D. 1/2 mv^2
Answer:
Do you have the answers for the unit 6 test?
Explanation:
PLZ
How safe is Wind Energy? Use in your own words.
Answer:
wind energy is safe
Explanation:
because it just is
Answer:
Wind energy isn't perfect, but better than the alternatives. There is a risk of falling off a turbine in maitenence and there are instances where birds fly into the blades, however there are far more casualties as a result of coal and other fossil fuel extraction.
Explanation:
You are a guest magician in a circus. One of your tricks is to place a football on an inclined plane without the football rolling over. How can you achieve this?
Determine if the weight of the football is needed for you to prevent the football from rolling over. The coefficient of static friction of the plane is 0.067.
Answer:
You are a guest magician in a circus. One of your tricks is to place a football on an inclined plane without the football rolling over is explained below in details.
Explanation:
spinning ball halts after traveling some range due to friction energy act different direction of movement of the ball. you can observe in the figure.
Let any rolling ball of mass (m ) is traveling with velocity v ,
common effect on ball (N) = mg
because of motion, friction energy develops on the contact exterior and begins to resist the movement of the rolling ball.
hence,
fr = uN = umg act on communicating exterior, so, after any time due to friction energy rolling ball gets to rest.
SI unit differ from one country to another . true or false
Answer:
false ..........false
Answer:
FALSE
Explanation:
If the knee has coordinates of (1.2,2.7) and the ankle has coordinates of (1.5,2.1), with the shank weighing 3.2kg, what is the moment of inertia of the shank about the proximal end point? (%k=0.528)
Answer:
The value is [tex]I = 0.48 \ kg m^2[/tex]
Explanation:
From the question we are told that
The coordinate of the knee is [tex]knee = (1.2,2.7)[/tex]
The coordinate of the ankle is [tex]ankle = (1.5,2.1)[/tex]
The mass of the shank is [tex]m = 3.2 \ kg[/tex]
Generally the length between the knee and the ankle is mathematically represented as
[tex]D = \sqrt{( 1.5 - 1.2)^2+ [ 2.1- 2.7]^2 }[/tex]
=> [tex]D =0.67 \ m[/tex]
Generally the moment of inertia of the shank about the proximal end point is mathematically represented as
[tex]I= m * \frac{D^2}{3}[/tex]
=> [tex]I = 3.2 * \frac{0.67^2}{3}[/tex]
=> [tex]I = 0.48 \ kg m^2[/tex]
PRACTICE
z
1.
What happens to the force of gravity between two objects if the distance between them is doubled?
Answer:
If the mass of one of the objects is doubled, then the force of gravity between them is doubled. ... Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces
Explanation:
Two speeding lead bullets, one of mass 15.0 g moving to the right at 295 m/s and one of mass 7.75 g moving to the left at 375 m/s, collide head-on, and all the material sticks together. Both bullets are originally at temperature 30.0°C. Assume the change in kinetic energy of the system appears entirely as increased internal energy. We would like to determine the temperature and phase of the bullets after the collision. (Lead has a specific heat of 128 J/(kg K), a melting point of 327.3°C, and a latent heat of fusion of 2.45 104 J/kg.)
Answer:
The final temperature of the bullets is 327.3 ºC.
Explanation:
Let suppose that a phase change does not occur during collision and collided bullets stop at the end. We represent the phenomenon by the First Law of Thermodynamics:
[tex]K_{A, o} + K_{B, o}-K_{A}-K_{B}+U_{A,o} + U_{B,o}-U_{A}-U_{B} = 0[/tex] (1)
Where:
[tex]K_{A,o}[/tex], [tex]K_{A}[/tex] - Initial and final translational kinetic energies of the 15-g bullet, measured in joules.
[tex]K_{B,o}[/tex], [tex]K_{B}[/tex] - Initial and final translational kinetic energies of the 7.75-g bullet, measured in joules.
[tex]U_{A,o}[/tex], [tex]U_{A}[/tex] - Initial and final internal energies of the 15-g bullet, measured in joules.
[tex]U_{B,o}[/tex], [tex]U_{B}[/tex] - Initial and final internal energies of the 7.75-g bullet, measured in joules.
By definitions of translational kinetic energy and sensible heat we expand and simplify the equation above:
[tex]\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T) = 0[/tex] (2)
Where:
[tex]m_{A}[/tex], [tex]m_{B}[/tex] - Masses of the 15-g and 7.75-g bullets, measured in kilograms.
[tex]v_{A,o}[/tex], [tex]v_{A}[/tex] - Initial and final speeds of the 15-g bullet, measured in meters per second.
[tex]v_{B,o}[/tex], [tex]v_{B}[/tex] - Initial and final speeds of the 7.75-g bullet, measured in meters per second.
[tex]c[/tex] - Specific heat of lead, measured in joules per kilogram-Celsius degree.
[tex]T_{o}[/tex], [tex]T[/tex] - Initial and final temperatures of the bullets, measured in Celsius degree.
Now we clear the final temperature of the bullets:
[tex](m_{A}+m_{B})\cdot c \cdot (T-T_{o}) = \frac{1}{2}\cdot [m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})][/tex]
[tex]T-T_{o} = \frac{m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})}{(m_{A}+m_{B})\cdot c}[/tex]
[tex]T= T_{o}+\frac{m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})}{(m_{A}+m_{B})\cdot c}[/tex] (3)
If we know that [tex]T_{o} = 30\,^{\circ}C[/tex], [tex]m_{A} = 15\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.75\times 10^{-3}\,kg[/tex], [tex]v_{A,o} = 295\,\frac{m}{s}[/tex], [tex]v_{B,o} = 375\,\frac{m}{s}[/tex], [tex]v_{A} = 0\,\frac{m}{s}[/tex], [tex]v_{B} = 0\,\frac{m}{s}[/tex] and [tex]c = 128\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then the final temperature of the collided bullets is:
[tex]T = 30\,^{\circ}C+\frac{(15\times 10^{-3}\,kg)\cdot \left[\left(295\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]+(7.75\times 10^{-3}\,kg)\cdot \left[\left(375\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{(15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(128\,\frac{J}{kg\cdot ^{\circ}C} \right)}[/tex]
[tex]T = 852.534\,^{\circ}C[/tex]
Given that found temperature is greater than melting point, then we conclude that supposition was false. If we add the component of latent heat of fussion, then the resulting equation is:
[tex]\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T)-U = 0[/tex] (4)
[tex]U=\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T)[/tex]
If we know that [tex]T_{o} = 30\,^{\circ}C[/tex], [tex]T = 327.3\,^{\circ}C[/tex], [tex]m_{A} = 15\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.75\times 10^{-3}\,kg[/tex], [tex]v_{A,o} = 295\,\frac{m}{s}[/tex], [tex]v_{B,o} = 375\,\frac{m}{s}[/tex], [tex]v_{A} = 0\,\frac{m}{s}[/tex], [tex]v_{B} = 0\,\frac{m}{s}[/tex] and [tex]c = 128\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then latent heat received by the bullets during impact is:
[tex]U =\frac{1}{2}\cdot (15\times 10^{-3}\,kg)\cdot \left[\left(295\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] + \frac{1}{2}\cdot (7.75\times 10^{-3}\,kg)\cdot \left[\left(375\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]+(15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(128\,\frac{J}{kg\cdot ^{\circ}C} \right) \cdot (30\,^{\circ}C-327.3\,^{\circ}C)[/tex][tex]U = 331.872\,J[/tex]
The maximum possible latent heat ([tex]U_{max}[/tex]), measured in joules, that both bullets can receive during collision is:
[tex]U_{max} = (m_{A}+m_{B})\cdot L_{f}[/tex] (5)
Where [tex]L_{f}[/tex] is the latent heat of fusion of lead, measured in joules per kilogram.
If we know that [tex]m_{A} = 15\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.75\times 10^{-3}\,kg[/tex] and [tex]L_{f} = 2.45\times 10^{4}\,\frac{J}{kg}[/tex], then the maximum possible latent heat is:
[tex]U_{max} = (15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(2.45\times 10^{4}\,\frac{J}{kg} \right)[/tex]
[tex]U_{max} = 557.375\,J[/tex]
Given that [tex]U < U_{max}[/tex], the final temperature of the bullets is 327.3 ºC.
Which of the following statements describe a perfectly in elastic collision
A) two billiard balls bounce off each other
B) a wad of chewing gum is thrown and sticks to a moving truck
C)Two ice skaters hit each other and fall over in opposite directions
D) a car crashes into a tree and rebounds in the other direction
Answer:
B
Explanation:
A radio station can be heard if the receiver is tuned to a frequency of 6x10^5 Hz. What is the wavelength of the radio waves
5 * 10² m
Explanation:We are Given:
Frequency of the wave = 6 * 10⁵ Hz
Wavelength of the wave:
We know the relation:
c = λν
[where λ (Lambda) is the wavelength ,ν (nu) is the frequency and c is the speed of light ]
3 * 10⁸ m/s= (6 * 10⁵ )* λ [replacing known values]
λ = [tex]\frac{3 * 10^{8}}{6 * 10^{5}}[/tex] [dividing both sides by 6 * 10⁵]
λ = 1/2 * 10³
λ = 1/2 * 10 * 10² [10³ can be rewritten as 10 * 10²]
λ = 5 * 10² m
Therefore, the wavelength of the wave is 5 * 10² m
A 10.0 kg ball weighs 98.0 N in air and weighs 65.0 N when submerged in water. The volume of the ball is:_________.A) 0.00245 m3. B) 0.00337 m3. C) 0.00457 m3. D) 0.00766 m3 E) 0.00980 m3
Answer: B) 0.00337 m3.
Explanation:
Given data:
Mass of the ball = 10kg
Weight of the ball in air = 98N
Weight of the ball in water = 65N
Solution:
To get the Volume of the ball when submerged in water, we divide the weight of the ball in water with the difference in apparent weight by 9.8m/s^2.
= 98 - 65 / 9.8
= 33 / 9.8
= 3.37kg
The volume of the ball is 3.37kg
The density of water is 1kg per Liter.
So 3.37 kg of water would have a volume of 3.37 Liters.
Therefore the ball would have a volume of 3.37 Liters (or 0.00337 cubic meters).
If the speed of an object in uniform circular motion is constant and the radial distance is doubled, by what factor does the magnitude of the radial acceleration decrease?
Answer: Half
Explanation:
Given
The object is in uniform circular motion with constant speed.
Radial acceleration of the object is given by
[tex]\Rightarrow a_r=\dfrac{v^2}{r}[/tex]
Where, [tex]a_r=\text{Radial acceleration}[/tex]
[tex]v=\text{speed}\\r=\text{distance from the axis of rotation}[/tex]
If the radial distance [tex]r[/tex] is doubled, i.e. [tex]2r[/tex]
Radial acceleration is
[tex]\Rightarrow a'_r=\dfrac{v^2}{2r}\\\\\Rightarrow a'_r=\dfrac{a_r}{2}[/tex]
Radial acceleration becomes half of the initial value.
Learn more: https://brainly.in/question/21217248
Heidi (39 kg) is walking her tiny chihuahua, Chaxi (5.60 kg), on the sidewalk. To encourage Chaxi along, Heidi pulls forward with a force of 9.55 N. Identify the correct reaction force in response to Heidi’s action force.
a. The friction is less than 660 N since the beam can be moved at a constant velocity.
b. There is no friction acting on the beam since it is accelerating.
c. The friction is equal to 660 N since the beam is not accelerating.
d. The friction is greater than 660 N since the beam is not in equilibrium.
Answer:
The correct reaction force in response to Heidi's action force is:
c. The friction is equal to 660 N since the beam is not accelerating.
Explanation:
Heidi's action force does not affect the beam. Since friction resists the sliding or rolling of one solid object over another, there is no friction acting on the beam, in this respect. The reaction force is what makes the dog to move because it acts on it. According to Newton's Third Law of Motion, forces always come in action-reaction pairs. This Third Law states that for every action force, there is an equal and opposite reaction force. This means that the dog exerts some force on Heidi, as he pulls it "forward with a force of 9.55 N."
A potential difference of 5v is used to produce a current of 4A for 4 hours through a heating coil. What is the heat produced.
a 80j
b 4.8kj
c 20j
d 4800kj
Answer:
Explanation:
Heat generated by the coil is expressed as;
Heat generated = Power ×Time
Heat generated = IVt
I is the current = 4A
V is the voltage= 5V
t is the time = 4hrs
Convert to secs
t = 4(3600) = 14400secs
Substitute into the formula
Heat generated = 4×5×14400
Heat generated = 20×14400
Heat generated = 288000Joules
Heat generated =288KJ
A peregrine falcon dives at pigeon . the falcon start downward from rest with free-fall acceleration . if the pigeon is 76.0 m below the initial position of the falcon , how long the falcon take to reach the pigeon? Assume that the pigeon remains at rest .
Answer:
3.94 s
Explanation:
Take down to be positive. Given:
Δy = 76.0 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
76.0 m = (0 m/s) t + ½ (9.8 m/s²) t²
t = 3.94 s
hi! hina can you help me with those 2 questions
Which of the following is the main idea of Thomas Paine's "Common Sense"?
A simple pendulum has a period T on Earth. If it were used on Planet X and found to have a period of only 71% of T, what would be the free fall acceleration due to gravity on this planet X?
Answer:
gₓ = 1.4 g = 13.8 m/s²
Explanation:
The time period of simple pendulum on earth is given as:
T = 2π√(L/g)
where,
L = Length of pendulum
g = acceleration due to gravity on the surface of earth = 9.8 m/s²
T = Time Period on Earth
but, on the other planets, it becomes:
Tₓ = 2π√(L/gₓ)
where,
Tₓ = Time Period on Surface of Other Planet
gₓ = acceleration due to gravity on other planet = ?
It is given that:
Tₓ = 71% of T
Tₓ = 0.71 T
using values:
2π√(L/gₓ) = (0.71) 2π√(L/g)
√(1/gₓ) = (0.71)√(1/g)
squaring both sides:
1/gₓ = 0.71/g
gₓ = g/0.71
gₓ = 1.4 g = 13.8 m/s²
Bob rides his bike with a constant speed of 10 miles per hour. How long will he take to travel a distance of 15 miles?
[tex]{\underline{\pink{\textsf{\textbf{ Answer : }}}}}[/tex]
➡ 150hrs.
[tex]{\underline{\pink{\textsf{\textbf{Explanation : }}}}}[/tex]
➡ Time = distance × speed
➡ Time = 15*10
➡ Time = 150hrs ans.
A large pizza is cut into 8 even slices. A person orders 4 large pizzas from a restaurant. How many total slices of pizza did the person order?
Answer:
32 slicesExplanation:
Step one:
given data
we are told that 1 large pizza can be cut into 8 even slices
Required
we want to find how many slices are there in 4 large pizzas
Step two:
so if 1 pizza has 8 slices
4 pizza will have x
cross multiply we have
x= 8*4
x=32 slices
Find the weight of an object with mass 80 kg on the moon ( g = 1.6 m/s^2)
Answer:
80kg = 133 Newtons I'm pretty sure this is right.
The weight or gravitational force of an object with mass 80 kg on the moon is 130 Newtons.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
Weight of the object is force applied due to gravity, So force = mass.gravity .
on the moon gravity is 1/6 of that on the earth.
so weight or force = 80*(9.8/6)
= 130 Newtons
The weight or gravitational force of an object with mass 80 kg on the moon is 130 Newtons.
To learn more about force refer to the link:
brainly.com/question/13191643
#SPJ6
5.
State the number of significant digits in each measurement and
express the value in scientific notation.
a. 110 m
f. 20.40 m/s
b. 0.2°C
g0.71 ns
c. 0.090 11 kg
h. 0.06 kg
d. 52.5 X 10 - 3
7040 m
600.0 N
וורן
e.
Answer:
Tanong mo po sa guro mo welcome pasensya need ko po ng point
Which current is produced in homes
Answer:
answer is C on edge 2021
Explanation:
(iii) Calculate the distance travelled by the car in part Q.
Use the equation
distance travelled = average speed x time
(2)
distance travelled = ....... m
Answer:
distance travelled = 3000m
Explanation:
distance travelled = average speed x time
=30m/s*100s
=3000m
Find an expression for the kinetic energy of the car at the top of the loop.Express the kinetic energy in terms of m, g, h, and R.
Answer:
K.E₂ = mg(h - 2R)
Explanation:
The diagram of the car at the top of the loop is given below. Considering the initial position of the car and the final position as the top of the loop. We apply law of conservation of energy:
K.E₁ + P.E₁ = K.E₂ + P.E₂
where,
K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0 (car initially at rest)
P.E₁ = Initial Potential Energy = mgh
K.E₂ = Final Kinetic Energy at the top of the loop = ?
P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)
Therefore,
0 + mgh = K.E₂ + mg(2R)
K.E₂ = mg(h - 2R)
The kinetic energy of the car is expressed as K.E = mg(h - 2R). The energy of the object by the integrity of its movement is understood as kinetic energy.
What is kinetic energy?The energy of the body by the virtue of its motion is known as the kinetic energy of the body. It is defined as the product of half of mass and square of velocity.
The following data is observed from the figure;
K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0
P.E₁ = Initial Potential Energy = mgh
K.E₂ = Final Kinetic Energy at the top of the loop =?
P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)
According to the law of conservation of energy, total energy is defined as the sum of kinetic energy and potential energy.
Total energy = kinetic energy+potential energy
On applying the law of conservation of energy for top and bottom of the loop;
K.E₁ + P.E₁ = K.E₂ + P.E₂
0 + mgh = K.E₂ + mg(2R)
K.E₂ = mg(h - 2R)
Hence the kinetic energy of the car is expressed as K.E = mg(h - 2R).
To learn more about kinetic energy refer to the link;
brainly.com/question/999862
A car company wants to ensure its newest model can stop in less than 450 ft when traveling at 60 mph. If we assume constant deceleration, find the value of acceleration (in ft/sec2) that accomplishes this.
Answer:
The value of acceleration that accomplishes this is 8.61 ft/s² .
Explanation:
Given;
maximum distance to be traveled by the car when the brake is applied, d = 450 ft
initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s
final velocity of the car when it stops, v = 0
Apply the following kinematic equation to solve for the deceleration of the car.
v² = u² + 2as
0 = 88.02² + (2 x 450)a
-900a = 7747.5204
a = -7747.5204 / 900
a = -8.61 ft/s²
|a| = 8.61 ft/s²
Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .
Two horizontal forces, 230 N and 120 N, are exerted in opposite direction on a crate. What is the horizontal acceleration of the crate if the mass of the crate is 20 kg?
5.5 m/s2
20 m/s2
17.5 m/s2
72 m/s2
Answer:
a = 5.5 [m/s²]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the force is equal to the product of mass by acceleration.
ΣF =m*a
where:
F = Force [N] (units of Newtons)
m = mass = 20 [kg]
a = acceleration [m/s²]
Let's take the force of 230 [N] as positive, in this way the other force will be negative, by pointing in the opposite direction.
[tex]230 - 120 = 20*a\\110 = 20*a\\a=5.5 [m/s^{2} ][/tex]