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aliya0001 [1]
2 years ago
10

A wagon of dog treats (combined mass 55 kg) is rolling at 2.1 m/s. A dog with mass 21 kg dives into the wagon, colliding with ju

st enough momentum to make both stop. If the collision between the dog and the wagon lasts 0.1 s, what is the magnitude of the average force that will be exerted on the dog by the collision with the wagon
Physics
1 answer:
Lunna [17]2 years ago
5 0

Answer:

Explanation:

An impulse results in a change of momentum

If the wagon and dog both stop, they must have had equal and opposite momentums

FΔt = mΔv

F = mΔv/Δt = m(v₁ - v₀)/(t₁ - t₀)

v₁ = t₀ = 0

F = m(v₀)/t₁

F = 55(2.1)/0.1 = 1155 N

We could have also figured the dog's initial velocity and used the dog's mass in the equation as well. Result would be identical.

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A 1170-kg car is held in place by a light cable on a very smooth (frictionless) ramp, as shown in the figure (Figure 1) . The ca
liq [111]
Refer to the diagram shown below.

The mass of the car is 1170 kg, therefore its weight is
W = (1170 kg)*(9.8 m/s²) = 11466 N

The component of the weight acting down the incline is
F = W sin(25°) = (11466 N)*sin(25°) = 4845.7 N

The normal reaction from the inclined plane is
N = W cos(25°) = (11466 N) cos(25°) = 1039.2 N

T =  tension in the cable, acting at 31° above the surface of the ramp.

The Free Body Diagram on the right shows all the forces (friction is ignored)
and they FDB is sufficient for determining the value of T which establishes equilibrium.

4 0
3 years ago
A 2.00-kg box is suspended from the end of a light vertical rope. A time-dependent force is applied to the upper end of the rope
Alex

Answer:

T = 27.92 N

Explanation:

For this exercise let's use Newton's second law

      T - W = m a

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      W = mg

The acceleration can be found by derivatives

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     a = 2 + 0.6 t

We replace

      T - mg = m (2 + 0.6t)

      T = m (g + 2 + 0.6 t)               (1)

Let's look for the time for the speed of 15 m / s

       15 = 2 t + 0.6 t²

       0.6 t² + 2 t - 15 = 0

We solve the second degree equation

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        t = [-2 ±√40] / 1.3 = [-2 ± 6.325] / 1.2

We take the positive time

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       T = 2.00 (9.8 + 2 + 0. 6  3.6)

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3 years ago
Room temperature water is placed in an Erlenmyer flask and heated to the boiling point. After the flask is removed from the heat
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Explanation:

When water is boiled in the flask . Some portion of it is evaporated out . Now when cork is placed on it and is placed in the ice box . It cools down , by which the pressure inside decreases .

Due to decrease of pressure , the boiling point of water also decreases . Now it can boil at lower temperature . Thus it starts boiling at lower temperature even , when placed in the ice box .

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4 years ago
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Answer:

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