An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be moving backward just after releasing the ball? Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

Answer: Kinectic Energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity.

Explanation: If an object with a mass of 10 kg (m=10 kg) is moving at a velocity of 5 meters per second (v=5 m/s), the kinetic energy is equal to 125 Joules, or (1/2* 10 kg) * 5 m/s^2.

Answer:

Explanation:

Moment of inertia I = M k² , where M is mass and k is radius of gyration .

Putting the given values in the equation

5000 = 200 x k²

k² = 25

k = 5 cm .

Radius of gyration is 5 cm .

Answer:

The answer is "0.92 c"

Explanation:

[tex]v_1\ (earth) = 0.62 \ c \\\\v_2\ ( enterprise ) = -0.30[/tex]

so,

[tex]v_2 \ (earth) = 0.62 \ c - (-0.30 \ c) \\\\[/tex]

               [tex]= 0.62 \ c +0.30 \ c\\\\= 0.92 \ c[/tex]

Answer:

xplanation:

Angle of reflection is measured between the incident ray and the angle which it makes with the normal at the point where incident ray strikes the mirror surface.

Further on reflection, it makes the same angle i.e. angle of reflection is equal to angle of reflection.

Hence, as angle of incidence is 55∘ angle of reflection too is 55∘ and the angle between the incident ray and the reflected ray is 55∘+55∘=110∘

Answer:

the magnitude of the magnetic force on the wire is 0.2298 N

Explanation:

Given the data in the question;

we know that, the magnitude of magnetic force is given as;

|F[tex]_{mg}^>[/tex] | = I([tex]B^>[/tex] × [tex]L^>[/tex] )

given that

I = 2.6 A

[tex]B^>[/tex] = 0.17

[tex]L^>[/tex] = 0.52

so we substitute

|F[tex]_{mg}^>[/tex] | = 2.6( 0.17i" × 0.52j" )

|F[tex]_{mg}^>[/tex] | = 0.2298 N

Therefore, the magnitude of the magnetic force on the wire is 0.2298 N

Answer: 550 cm

Explanation:

Original equation: 1/f= 1/do + 1/di.

F=50.0 cm, and do=55.0.

Since we don't have di, we'll have to subtract do to the other side, making the equation: 1/f - 1/do= 1/di.

Doing the math, 1/f - 1/do is 0.0018181818

Then to get di by itself, you multiply both sides by di. Then you divide by 0.0018181818 to get di by itself. You then get: di= 1/0.0018181818

At that point, you just divide 1 by 0.0018181818, which will give you 550 cm

There could be simpler way, but that is just what I did to get the answer. Answer was right on Acellus

1: Ming

2: Chloe

3: 40m

Answer:

The cart mark (a) has the most potential energy and the cart marked (b) has the most kinetic energy

Answer:

8 kV

Explanation:

Here is the complete question

Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500 μF capacitors and an 800−V charging source?

Solution

Since the capacitors are initially connected in parallel, the same voltage of 800 V is applied to each capacitor. The charge on each capacitor Q = CV where C = capacitance = 500 μF and V = voltage = 800 V

So, Q = CV

= 500 × 10⁻⁶ F × 800 V

= 400000 × 10⁻⁶ C

= 0.4 C

Now, when the capacitors are connected in series and the voltage disconnected, the voltage across is capacitor is gotten from Q = CV

V = Q/C

= 0.4 C/500 × 10⁻⁶ F

= 0.0008 × 10⁶ V

= 800 V

The total voltage obtained across the ten capacitors is thus V' = 10V (the voltages are summed up since the capacitors are in series)

= 10 × 800 V

= 8000 V

= 8 kV

Answer:

increasing the masses of the objects and decreasing the distance between the objects

Explanation:

Answer:

Potential and kinetic

Explanation:

At 30°C, the rms speed of a sample of oxygen with a molar mass of 16 g/mol is approximately 482.34 m/s.

The root mean square (rms) speed of a gas molecule is a measure of the average speed of the gas particles in a sample. It can be calculated using the formula:

vrms = √(3kT/m)

Where:

vrms is the rms speed

k is the Boltzmann constant (1.38 x 10^-23 J/K)

T is the temperature in Kelvin

m is the molar mass of the gas in kilograms

To calculate the rms speed of oxygen at 30°C (303 Kelvin) with a molar mass of 16 g/mol, we need to convert the molar mass to kilograms by dividing it by 1000:

m = 16 g/mol = 0.016 kg/mol

Substituting the values into the formula, we have:

vrms = √((3 * 1.38 x 10^-23 J/K * 303 K) / (0.016 kg/mol))

Calculating this expression yields the rms speed of the oxygen sample:

vrms ≈ 482.34 m/s

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Answer:18 watts

Explanation:i just got this question trust me

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Answer:

11.5 m/s²

Explanation:

The centripetal acceleration, a = rω² where r = radius of cylinder = 10.7 cm = 0.107 m and ω = angular speed = 2πN where N = number of revolutions per second = 1.65 rev/s

So, a = rω²

a = r(2πN)²

a = 4π²rN²

substituting the values of the variables into the equation, we have

a = 4π²rN²

a = 4π²(0.107 m)(1.65 rev/s)²

a = 4π²(0.107 m)(2.7225 rev²/s)²

a = 4π² × 0.2913075 mrev²/s)²

a = 11.5 m/s²

The force 1 is tension force.

To find the correct statement among all the options, we need to know more about friction, tension, normal force and weight.

Thus, we can conclude that the correct option is (B).

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Answer:

sorry I am not confident you the answer

Answer:

just search it up you'll get ur answer

Answer:

  k y -b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]

give us some initial conditions

1) friction force fr = 1N when v = 4m / s

2) an initial displacement of x = 0.08 m for t=0 s

Explanation:

In this exercise, you are asked to state the problem you are posing. We are going to find the equation of motion for this exercise. Let's start with Newton's second law

Let's set a reference system with the y-axis in a vertical and positive direction upwards.

We have four forces: an external downward force, negative in sign, the but that goes down and is negative, the Hook force that goes up and is positive and the friction force that opposes the movement, in this case it goes down being negative

let's write Newton's second law

          F_e -F -fr - W = m a

where

          F_e = -kDy = - k y

          fr = - b v = -b dy / dt

          W = mg

we substitute for the specific case, that is, using the signs

          k y  -b [tex]\frac{dy}{dt}[/tex] - m g - F = m [tex]\frac{d^2y}{dt^2}[/tex]

In the initial condition of the problem, before starting the movement, the friction force is zero and the acceleration is also zero

         k y - m g - F = 0

from this equation you can find the spring constant, y= 9m and F=2 N

It is not clear if when the movement starts this external force becomes zero, but since it balances the weight we can eliminate the two forces that have the same magnitude and opposite direction, so the equation remains

              k y - b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]

give us some initial conditions

1) friction force fr = 1N when v = 4m / s

2) an initial displacement of x = 0.08 m for t=0 s

therefore, to initiate the movement, a small external force F 'is applied that moves the system to a new equilibrium position and this small force F' is made zero, thus initiating an oscillatory movement, described by the equation.

             k y -b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]

This is a differential equation of the second degree, therefore it needs two initial conditions for its complete solution

The initial amount of displacement corresponds to the amplitude of movement A = 0.08 m

Chris used a non-plane mirror to check out a box resting on a shelf, the focal length of the mirror is mathematically given as

f=8.38cm

Question Parameter(s):

The image of the box was located 15 cm behind the mirror

and the box was placed 19 cm from the mirror.

Generally, the equation for the focal length is mathematically given as

1/f=1/u+1/v

Therefore

1/f=1/15+1/19

f=8.3823529cm

In conclusion,  the focal length of the mirror

f=8.3823529cm

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Answer:

1.72 m

Explanation:

Potential energy = mgh, where m is mass, g is acceleration due to gravity (9.8), and h is height

76 = (3.5)(9.8)h

76=44.1h

h=1.72335600907 ≈1.72 m

Answer:

:r

Explanation:r

Tobnbv346468this Ishmael

Answer:

a. The wooden cylinder b. the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.

Explanation:

a. Who reaches the bottom first

The kinetic energy of the objects is given by

K = 1/2mv² + 1/2Iω² where m = mass of object, v = velocity of object, I = moment of inertia and ω = angular velocity = v/r where r = radius of object

For the wooden cylinder, I = mr²/2 where m = mass of wooden cylinder and r = radius of wooden cylinder and v = velocity of wooden cylinder

So, its kinetic energy, K = 1/2mv² + 1/2(mr²/2)(v/r)²

K = 1/2mv² + 1/4mv²

K = 3mv²/4

For the steel hoop, I' = mr'² where m' = mass of steel hoop and r' = radius of steel hoop and v' = velocity of steel hoop

So, its kinetic energy, K' = 1/2m'v'² + 1/2(m'r'²)(v'/r')²

K' = 1/2m'v'² + 1/2m'v'²

K' = m'v'²

Since both kinetic energies are the same, since the drop from the same height,

K = K'

3mv²/4 = m'v'²

v²/v'² = 4m/3m'

v²/v'² = 4/3(m/m')

v/v' = √[4/3(m/m')]

Since the hoop weighs more than the cylinder m/m' < 1 and 4/3(m/m') < 4/3 ⇒ √ [4/3(m/m')] < √4/3 ⇒ v/v' < 1.16 ⇒ v'/v > 1/1.16 ⇒ v'/v > 0.866. Since 0.866 < 1, it implies v' < v.

Since v' = speed of steel hoop < v = speed of wooden cylinder, the wooden cylinder reaches the bottom first.

b. Why

Since the kinetic energy, K = translational + rotational

We find the translational kinetic energy of each object.

For the wooden cylinder,

K = K₀ + 1/2Iω² where K₀ = translational kinetic energy of wooden cylinder

K - 1/2Iω² = K₀

3/4mv² - 1/2(mr²/2)(v/r)² = K₀

3/4mv² - 1/4mv² = K₀

K₀ = 1/2mv²

For the steel hoop,

K' = K₁ + 1/2I'ω'² where K₁ = translational kinetic energy of steel hoop

K' - 1/2I'ω'² = K₁

m'v'² - 1/2(m'r'²)(v'/r')² = K₁

m'v'² - 1/2m'v'² = K₁

K₁ = 1/2m'v'²

So, K₀/K₁ =  1/2mv²÷1/2m'v'² = mv²/m'v'² = (m/m')(v²/v'²) = (m/m')4/3(m/m') = 4/3(m/m')².

Since (m/m') < 1 ⇒  (m/m')² < 1 ⇒ 4/3(m/m')² < 4/3 ⇒ K₀/K₁  < 1.33 ⇒ K₀ > K₁

So, the kinetic energy of the wooden cylinder is greater than that of the steel hoop.

So, the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.

a. The wooden cylinder b. the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.

The energy of the body due to its movement in a particular direction under the influence of a force like a free-falling body due to gravitaional force is called  Kinetic energy.

The kinetic energy of the objects is given by

[tex]K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}Iw^2[/tex]

where

m = mass of object,

v = velocity of object,

I = moment of inertia and

ω = angular velocity = v/r where r = radius of object

For the wooden cylinder, I = mr²/2 where m = mass of wooden cylinder and r = radius of wooden cylinder and v = velocity of wooden cylinder

So, its kinetic energy,

[tex]K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}(\dfrac{mr^2}{2})\dfrac{v}{r}^2[/tex]

[tex]K = \dfrac{3mv^2}{4}[/tex]

For the steel hoop,

I' = mr'²

where

m' = mass of steel hoop and

r' = radius of steel hoop and

v' = velocity of steel hoop

So, its kinetic energy,

[tex]K' = \dfrac{1}{2}m'v'^2 + \dfrac{1}{2}(m'r'^2)\dfrac{v'}{r'}^2[/tex]

[tex]K' = \dfrac{1}{2}m'v'^2 + \dfrac{1}{2}m'v'^2[/tex]

K' = m'v'²

Since both kinetic energies are the same, since the drop from the same height,

K = K'

[tex]\dfrac{3mv^2}{4 }= m'v'^2[/tex]

[tex]\dfrac{v^2}{v'^2} =\dfrac{ 4m}{3m'}[/tex]

[tex]\dfrac{v^2}{v'^2} = \dfrac{4}{3}(\dfrac{m}{m'})[/tex]

[tex]\dfrac{v}{v'} = \sqrt{[\dfrac{4}{3}(\dfrac{m}{m'})][/tex]

Since the hoop weighs more than the cylinder m/m' < 1 and 4/3(m/m') < 4/3 ⇒ √ [4/3(m/m')] < √4/3 ⇒ v/v' < 1.16 ⇒ v'/v > 1/1.16 ⇒ v'/v > 0.866. Since 0.866 < 1, it implies v' < v.

Since v' = speed of steel hoop < v = speed of wooden cylinder, the wooden cylinder reaches the bottom first.

(b) Since the kinetic energy, K = translational + rotational

We find the translational kinetic energy of each object.

For the wooden cylinder,

[tex]K = K_o + \dfrac{1}{2}Iw^2[/tex]

where

K₀ = translational kinetic energy of wooden cylinder

[tex]K - \dfrac{1}{2}Iw^2 = K_o[/tex]

[tex]\dfrac{3}{4}mv^2 - \dfrac{1}{2}(\dfrac{mr^2}{2})(\dfrac{v}{r})^2 = K_a[/tex]

[tex]\dfrac{3}{4}mv^2 - \dfrac{1}{4}mv^2 = K_o[/tex]

[tex]K_o = \dfrac{1}{2}mv^2[/tex]

For the steel hoop,

[tex]K' = K_1 + \dfrac{1}{2}I'w'^2[/tex]

where

K₁ = translational kinetic energy of steel hoop

[tex]K' - \dfrac{1}{2}I'w'^2 = K_1[/tex]

[tex]m'v'^2 - \dfrac{1}{2}(m'r'^2)(\dfrac{v'}{r'})^2 = K_1[/tex]

[tex]m'v'^2 - \dfrac{1}{2}m'v'^2 = K_1[/tex]

[tex]K_1= \dfrac{1}{2}m'v'^2[/tex]

So, K₀/K₁ =  1/2mv²÷1/2m'v'² = mv²/m'v'² = (m/m')(v²/v'²) = (m/m')4/3(m/m') = 4/3(m/m')².

Since (m/m') < 1 ⇒  (m/m')² < 1 ⇒ 4/3(m/m')² < 4/3 ⇒ K₀/K₁  < 1.33 ⇒ K₀ > K₁

So, the kinetic energy of the wooden cylinder is greater than that of the steel hoop.

So, the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.

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Answer:

e. Combustion

Explanation:

In Combustion reaction, a substance reacts with oxygen from the air and resultant product is that it releases carbon dioxide and water.

Here,

2C2H6 is the substance that reacted with 7O2 (Oxygen) to release 4CO2 (Carbon Dioxide) and 6H2O (Water).

Answer:

Conservation of Energy: Like all things, Stirling Engines follow the conservation of energy principle (all the energy input is accounted for in the output in one form or another). ... The hot one supplies all of the energy QH, while the cold one removes energy QC (a necessary part of the cycle).

Explanation:

Answer: Yes

Explanation: All the energy input is accounted for in the output in one form or another

Answer:

The equation of motion is [tex]x(t)=-[/tex][tex]\frac{1}{3} cos4\sqrt{6t}[/tex]

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is [tex]32ft/s^2[/tex]

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet [tex]x=\frac{4}{12} =\frac{1}{3}feet[/tex]

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

[tex]W=kx[/tex] ⇒ [tex]k=\frac{W}{x}[/tex]

The spring constant , [tex]k=\frac{24}{1/3}[/tex]

                            = 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

    [tex]m\frac{d^2x}{dt} +kx=0[/tex]

    [tex]\frac{3}{4} \frac{d^2x}{dt} +72x=0[/tex]

  [tex]\frac{d^2x}{dt} +96x=0[/tex]

Auxiliary equation is, [tex]m^2+96=0[/tex]

                                 [tex]m=\sqrt{-96}[/tex]

                               =[tex]\frac{+}{} i4\sqrt{6}[/tex]

Thus , the solution is [tex]x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}[/tex]

                                 [tex]x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2[/tex]  [tex]4\sqrt{6}[/tex] [tex]cos4\sqrt{6t}[/tex]

The mass is released from the rest x'(0) = 0

                    [tex]=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2[/tex] [tex]4\sqrt{6}[/tex] [tex]cos4\sqrt{6(0)}[/tex] =0

                                                    [tex]c_2[/tex] [tex]4\sqrt{6} =0[/tex]

                                     [tex]c_2=0[/tex]

Therefore , [tex]x(t)=c_1[/tex] [tex]cos 4\sqrt{6t}[/tex]

Since , the mass is released from the rest from 4 inches

                    [tex]x(0)= -4[/tex] inches

[tex]c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}[/tex] feet

   [tex]c_1=-\frac{1}{3}[/tex] feet

Therefore , the equation of motion is  [tex]-\frac{1}{3} cos4\sqrt{6t}[/tex]

Answer:

Wind deposits sand into a small mound. So the answer is Deposition