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Sav [38]
3 years ago
11

A tow truck drags a stalled car along a road. The chain makes an angle of 30???? with the road and the tension in the chain is 1

500 N. How much work is done by the truck in pulling the car 1 km?
Physics
1 answer:
My name is Ann [436]3 years ago
3 0

Answer: work = 1,305kJ

Explanation:

angle= 30°

force= 1,500N

distance= 1,000m

The formula for work is : Work= force x distance, however there is an angle of 30° between the direction of force applied and the direction of motion, therefore force must be decomposed to its value on the horizontal axis which is the direction of motion by using the cosine of the very angle.

W= F×cos(α)×D

W= 1,500×cos (30)×1,000

W= 1,305kJ ( kilojoules)

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timama [110]

Answer:

A

Explanation:

The acceleration of an object is directly proportional to its net force.

a =  \frac{f}{m}

4 0
3 years ago
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A coin with a diameter 3.00 cm rolls up a 30.0 inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s
sweet [91]

This question is in complete.The question is

A coin with a diameter 3.00 cm rolls up a 30.0° inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s and rolls in a straight line without slipping. If the moment of inertia of the coin is(1/2) MR² , how far will the coin roll up the inclined plane (length along the ramp)? Hint: Conservation of mechanical energy.

Answer:

distance=0.124 m

Explanation:

mgh=mglSin\alpha =(1/2)Iw_{i}^{2}+(1/2)mv^{2}\\   v=wR\\Solve for L\\L=((1/2)(1/2)0.015^{2}*60^{2}+(1/2)(60*0.015^{2} ))/9.8Sin30\\   L=0.124m

6 0
3 years ago
A father is pulling his child on a sled in the snow. According to Newton’s Third Law, the force the father exerts on the sled is
viktelen [127]
That is because there are other forces like the friction forces that apply differently on both of them. The frictional forces applied to the sled are smaller than they are on the father, for example, so it's possible for him to pull it.
5 0
3 years ago
. Water is flowing at 12m/s in a horizontal pipe under a pressure of 600kpa radius 2cm. a. What is the speed of the water on the
lana66690 [7]

Answer:

Outlet Velocity = 192 m/s

Outlet Pressure = 510 kPa

Explanation:

Givens:

Inlet Velocity, V₁ = 12 m/s

Inlet Pressure, P₁ = 600 kPa = 600,000 Pa

Inlet Radius r₁ = 0.5 cm

Outlet Velocity, V₂ = not given (we are asked to find this)

Outlet Pressure,  P₂ = not given (we are asked to find this)

Outlet Radius, r₂ = 0.5 cm

From these, we can find the following:

Inlet Area, A₁ = π (r₁)² = π(2)² = 4π cm²

Outlet Area, A₂ = π (r₂)² = π(0.5)² = 0.25π cm²

<u>Part A :</u>

Assuming that water is incompressible, we can reason that within the same given time, the amount of volume of water entering the inlet must equal the volume of water exiting the outlet. Hence by the continuity equation (i.e. conservation of mass)

Inlet Volume flow rate = Outlet Volume flow rate

(recall that Volume flow rate in a pipe is given by Velocity x Cross Section Area), Hence the equation becomes

V₁ x A₁ = V₂ x A₂  (substituting the values that we know from above)

12 x 4π = V₂ x 0.25π  (we don't have to change all to SI units because the conversion factors on the left will cancel out the conversion factors on the right).

V₂ = (12 x 4π) / (0.25π)

V₂ = 192 m/s  (Answer)

<u>Part B:</u>

For Part B, if we assume a closed ideal system (control volume method), we can simply apply the energy equation (i.e Bernoulli's equation)

P₁ + (1/2)ρV₁ + ρgh₁ = P₂ + (1/2)ρV₂ + ρgh₂

Because the pipe is horizontal, there is no difference between h₁ and h₂, hence we can neglet this term:

P₁ + (1/2)ρV₁ = P₂ + (1/2)ρV₂  (rearranging)

P₂ = P₁ + (1/2)ρV₁ - (1/2)ρV₂

= P₁  + (1/2)ρ (V₁-V₂)

Assuming that the density of water is approx, ρ = 1000 kg/m³

P₂ = 600,000  + (1/2)(1000) (12-192)

= 600,000  + ( -90,000)

= 510,000 Pa

= 510 kPa (Answer)

8 0
3 years ago
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