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Ghella [55]
2 years ago
10

A ball of mass 0.4 kg is initially at rest on the ground. It is kicked and leaves the kicker's foot with a speed of 5.0 m/s in a

direction 60° above the horizontal. The magnitude of the impulse imparted by the ball to the foot is most nearly:. ||J|| = 1. ||J|| = 1Ns.
2. ||J|| =(2/√3) Ns.
3. ||J|| = 2Ns.
4. ||J|| = 4Ns.
5. ||J|| = √3 Ns.
Physics
1 answer:
yawa3891 [41]2 years ago
4 0

Answer:

the answer the correct  is 3

Explanation:

Let's use the relationship between momentum and momentum

         I = Δp

         I = m v_{f} - m v₀

     

Let's calculate

         I = 0.4 5.0 - 0

         I = 2.0 N s

By Newton's law of action and reaction the force on the ball is equal to the force that the ball exerts on the foot, therefore the impulse on the foot of equal magnitude, but in the opposite direction

        I = 2.0 Ns with 60°

When reviewing the answer the correct  is 3

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An open-topped freight car with mass 24,000 kg is coasting without friction along a level track. It is raining very hard, and th
skelet666 [1.2K]

Answer:

(a) v = 3..6 m/s

(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.

Explanation:

from the question we have the following:

mass of the car (Mc) = 24,000 kg

initial velocity of the car (u) = 4 m/s

mass of water (Mw) = 3000 kg

final velocity of the car (v) = ?

(a) we can calculate the final momentum of the car by applying the conservation of momentum where

initial momentum = final momentum

Mc x U = (Mc + Mw) x V

24000 x 4 = (24000 + 3000) x v

96,000 = 27000v

v =3.6 m/s

(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.

7 0
2 years ago
A merry-go-round with a a radius of R = 1.63 m and moment of inertia I = 196 kg-m2 is spinning with an initial angular speed of
kondor19780726 [428]

Answer:

1) L = 299.88 kg-m²/s

2) L = 613.2 kg-m²/s

3) L = 499.758 kg-m²/s

4) ω₁ = 0.769 rad/s

5) Fc = 70.3686 N

6) v = 1.2535 m/s

7) ω₀ = 1.53 rad/s

Explanation:

Given

R = 1.63 m

I₀ = 196 kg-m²

ω₀ = 1.53 rad/s

m = 73 kg

v = 4.2 m/s

1) What is the magnitude of the initial angular momentum of the merry-go-round?

We use the equation

L = I₀*ω₀ = 196 kg-m²*1.53 rad/s = 299.88 kg-m²/s

2) What is the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round?

We use the equation

L = m*v*Rp = 73 kg*4.2 m/s*2.00 m = 613.2 kg-m²/s

3) What is the magnitude of the angular momentum of the person just before she jumps on to the merry-go-round?

We use the equation

L = m*v*R = 73 kg*4.2 m/s*1.63 m = 499.758 kg-m²/s

4) What is the angular speed of the merry-go-round after the person jumps on?

We can apply The Principle of Conservation of Angular Momentum

L in = L fin

⇒ I₀*ω₀ = I₁*ω₁

where

I₁ = I₀ + m*R²

⇒  I₀*ω₀ = (I₀ + m*R²)*ω₁

Now, we can get ω₁

⇒  ω₁ = I₀*ω₀ / (I₀ + m*R²)

⇒  ω₁ = 196 kg-m²*1.53 rad/s / (196 kg-m² + 73 kg*(1.63 m)²)

⇒  ω₁ = 0.769 rad/s

5) Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?

We have to get the centripetal force as follows

Fc = m*ω²*R  

⇒  Fc = 73 kg*(0.769 rad/s)²*1.63 m = 70.3686 N

6) Once the person gets half way around, they decide to simply let go of the merry-go-round to exit the ride.

What is the linear velocity of the person right as they leave the merry-go-round?

we can use the equation

v = ω₁*R = 0.769 rad/s*1.63 m = 1.2535 m/s

7) What is the angular speed of the merry-go-round after the person lets go?

ω₀ = 1.53 rad/s

It comes back to its initial angular speed

8 0
2 years ago
The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th
Rasek [7]

Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

<em>A freight train has a mass of </em>1.83\times 10^{7}\,kg<em>. The wheels of the locomotive push back on the tracks with a constant net force of </em>7.50\times 10^{5}\,N<em>, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?</em>

If locomotive have a constant net force (F), measured in newtons, then acceleration (a), measured in meters per square second, must be constant and can be found by the following expression:

a = \frac{F}{m} (1)

Where m is the mass of the freight train, measured in kilograms.

If we know that F = 7.50\times 10^{5}\,N and m = 1.83\times 10^{7}\,kg, then the acceleration experimented by the train is:

a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}

a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}

Now, the time taken to accelerate the freight train from rest (t), measured in seconds, is determined by the following formula:

t = \frac{v-v_{o}}{a} (2)

Where:

v - Final speed of the train, measured in meters per second.

v_{o} - Initial speed of the train, measured in meters per second.

If we know that a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}, v_{o} = 0\,\frac{m}{s} and v = 22.222\,\frac{m}{s}, the time taken by the freight train is:

t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s}  }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }

t = 542.265\,s

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

6 0
2 years ago
160 students sit in an auditorium listening to a physics lecture. Because they are thinking hard, each is using 125 W of metabol
anastassius [24]

Answer:

minimum power should be used to operate the air conditioner is 4000 W

Explanation:

given data

students  n = 160

power p = 125 W

COP = 5.0

to find out

what minimum power should be used

solution

we know the COP formula that is given below

COP = students × power  / minimum power

minimum power = n × p / COP

put all value

minimum power = n × p / COP

minimum power = 160 × 125 / 5

minimum power = 4000 W

minimum power should be used to operate the air conditioner is 4000 W

8 0
2 years ago
An unspecified force causes a 0.20-kg object to accelerate at 0.40 m/s2. If 0.30 kg is added to the 0.20-kg object and the force
Naddik [55]

Answer:

a = 0.16

Explanation:

given,

mass of the object 1 = 0.2 kg

mass of the object 2 = 0.3 kg

acceleration when force is on 0.2 kg = 0.4 m/s²

acceleration when both mass are combine = ?

F = m a

F = 0.2 × 0.4

F = 0.08 N

force acting is same and total mass  = 0.2 + 0.3 = 0.5 Kg

F = m a

a = \dfrac{F}{m}

a = \dfrac{0.08}{0.5}

a = 0.16 m/s²

the acceleration  acting when both the body is attached is a = 0.16

4 0
2 years ago
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