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nekit [7.7K]
3 years ago
13

A 75.0 kg man sits on a massless cart that is on a horizontal surface. The cart is initially stationary and it can move without

friction or air resistance. The man throws a 5.00 kg stone in the positive direction, applying a force to it so that it has acceleration +3.50 m/s2 as measured by a nearby observer on the ground. What is the man's acceleration during the throw, as seen by the same observer? Be careful to use correct signs.
Physics
1 answer:
Anna11 [10]3 years ago
6 0

Answer:0.233 m/s^2

Explanation:

Given

Mass of man m=75 kg

mass of stone m_0=5 kg

acceleration of stone a=3.5 m/s^2

To provide a acceleration of 3.5 m/s^2[/tex] he must exert a force of

F=m_0\times a=5\times 3.5=17.5 N

same Force is exerted by stone on cart

17.5=m\times a

a=\frac{17.5}{75}=0.233 m/s^2

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A truck pushes a pile of dirt horizontally on a frictionless road with a net force of 20\, \text N20N20, start text, N, end text
meriva

Answer:

300 Nm ; 300 J

Explanation:

Given that:

Force (F) = 20 N

Distance (d) = 15 m

The kinetic energy (Workdone) = Force * Distance

Kinetic Energy = 20N * 15m

Kinetic Energy = 300Nm

K. E = 1/2

4 0
3 years ago
Read 2 more answers
Tim and Rick both can run at speed Vr and walk at speed Vw, with Vr > Vw.
miss Akunina [59]

Answer:

Δt =  \frac{2D}{Vw+Vr} - \frac{D}{2Vr} - \frac{D}{2Vw}

Explanation:

Hi there!

Using the equation of speed for the whole trip, we can obtain the time each one needed to cover the distance D.

The speed (v) is calculated by dividing the traveled distance (d) over the time needed to cover that distance (t):

v = d/t

Rick traveled half of the distance at Vr and the other half at Vw. Then, when v = Vr, the distance traveled was D/2 and the time is unknown, Δt1:

Vr = D/ (2 · Δt1)

For the other half of the trip the expression of velocity will be:

Vw = D/(2 · Δt2)

The total time traveled is the sum of both Δt:

Δt(total) = Δt1 + Δt2

Then, solving the first equation for Δt1:

Vr = D/ (2 · Δt1)

Δt1 = D/(2 · Vr)

In the same way for the second equation:

Δt2 = D/(2 · Vw)

Δt + Δt2 = D/(2 · Vr) + D/(2 · Vw)

Δt(total) = D/2 · (1/Vr + 1/Vw)

The time needed by Rick to complete the trip was:

Δt(total) = D/2 · (1/Vr + 1/Vw)

Now let´s calculate the time it took Tim to do the trip:

Tim walks half of the time, then his speed could be expressed as follows:

Vw = 2d1/Δt  Where d1 is the traveled distance.

Solving for d1:

Vw · Δt/2 = d1

He then ran half of the time:

Vr = 2d2/Δt

Solving for d2:

Vr · Δt/2 = d2

Since d1 + d2 = D, then:

Vw · Δt/2 +  Vr · Δt/2 = D

Solving for Δt:

Δt (Vw/2 + Vr/2) = D

Δt = D / (Vw/2 + Vr/2)

Δt = D/ ((Vw + Vr)/2)

Δt = 2D / (Vw + Vr)

The time needed by Tim to complete the trip was:

Δt = 2D / (Vw + Vr)

Let´s find the diference between the time done by Tim and the one done by Rick:

Δt(tim) - Δt(rick)

2D / (Vw + Vr) - (D/2 · (1/Vr + 1/Vw))

\frac{2D}{Vw+Vr} - \frac{D}{2Vr} - \frac{D}{2Vw} = Δt

Let´s check the result. If Vr = Vw:

Δt = 2D/2Vr - D/2Vr - D/2Vr

Δt = D/Vr - D/Vr = 0

This makes sense because if both move with the same velocity all the time both will do the trip in the same time.

8 0
3 years ago
Current is directly proportional to resistance.<br> a. True<br> b. False
ikadub [295]

IF voltage remains constant, then current is
inversely proportional to resistance.

The correct response is "b).", signifying "false" as the choice.

4 0
3 years ago
2. A test reveals that 150 J of work is required to lift an object 3 m at a
Nuetrik [128]

Answer:

50N

Explanation:

W=Fd

150=F(3)

50N=F

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