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sergij07 [2.7K]
3 years ago
8

Suppose two wagons, one with twice as much mass as the other, fly apart when a compressed spring that joins them is released. Th

e heavier wagon rolls ____ as fast as the lighter wagon.
Physics
1 answer:
sp2606 [1]3 years ago
7 0

Answer:

The heavier wagon rolls 1/2 as fast as the lighter wagon.

Explanation:

When the compressed spring that joins them is released then the force acts on both wagons will be of equal magnitude but in the opposite direction. However as the mass of one wagon is twice that of other, so the acceleration  will become half of the heavier wagon in comparison with lighter one.

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A lorry of mass 4000 kg is travelling at a speed of 4 m/s.
bulgar [2K]

Answer:

KE = 32,000J

v = 8m/s

Explanation:

KE = .5*m*v²

KE = .5*4000kg*(4m/s)²

KE = 32,000J

32,000J = .5*1000kg*v²

v² = 64

v = 8m/s

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3 years ago
52. How did people spend Life in the<br>hunting age-​
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Answer:

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A motorcycle is stopped at a traffic light. When the light turns green, the motorcycle accelerates to a speed of 91 km/h over a
zimovet [89]

Given :

Initial speed , u = 0 m/s .

Final speed , v = 91 km/h = 25.28 m/s .

To Find :

a) Average acceleration .

b ) Assuming the motorcycle maintained a constant acceleration, how far is it from the traffic light after 3.3 s .

Solution :

a )

We know ,by equation of motion :

v^2-u^2=2as\\\\a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{25.28^2-0^2}{2\times 47}\ m/s^2\\\\a=6.8\ m/s^2

b)

Also , by equation of motion :

s=ut+\dfrac{at^2}{2}\\\\s=0+\dfrac{6.8\times (3.3)^2}{2}\ m\\\\s=37.02\ m

Hence , this is the required solution .

6 0
3 years ago
A bowling ball with a momentum of 18kg-m/s strikes a stationary bowling pin. After the collision, the ball has a momentum of 13k
Veronika [31]

Answer:

14.98\ \text{kg m/s}

45.26^{\circ}

Explanation:

P_1 = Initial momentum of the pin = 13 kg m/s

P_i = Initial momentum of the ball = 18 kg m/s

P_2 = Momentum of the ball after hit

55^{\circ} = Angle ball makes with the horizontal after hitting the pin

\theta = Angle the pin makes with the horizotal after getting hit by the ball

Momentum in the x direction

P_i=P_1\cos55^{\circ}+P_2\cos\theta\\\Rightarrow P_2\cos\theta=P_i-P_1\cos55^{\circ}\\\Rightarrow P_2\cos\theta=18-13\cos55^{\circ}\\\Rightarrow P_2\cos\theta=10.54\ \text{kg m/s}

Momentum in the y direction

P_1\sin55=P_2\sin\theta\\\Rightarrow P_2\sin\theta=13\sin55^{\circ}\\\Rightarrow P_2\sin\theta=10.64\ \text{kg m/s}

(P_2\cos\theta)^2+(P_2\sin\theta)^2=P_2^2\\\Rightarrow P_2=\sqrt{10.54^2+10.64^2}\\\Rightarrow P_2=14.98\ \text{kg m/s}

The pin's resultant velocity is 14.98\ \text{kg m/s}

P_2\sin\theta=10.64\\\Rightarrow \theta=sin^{-1}\dfrac{10.64}{14.98}\\\Rightarrow \theta=45.26^{\circ}

The pin's resultant direction is 45.26^{\circ} below the horizontal or to the right.

4 0
3 years ago
A 7 ft tall person is walking away from a 20 ft tall lamppost at a rate of 5 ft/sec. Assume the scenario can be modeled with rig
aleksandr82 [10.1K]

Answer:

The rate of change of the shadow length of a person is 2.692 ft/s

Solution:

As per the question:

Height of a person, H = 20 ft

Height of a person, h = 7 ft

Rate = 5 ft/s

Now,

From Fig.1:

b = person's distance from the lamp post

a = shadow length

Also, from the similarity of the triangles, we can write:

\frac{a + b}{20} = \frac{a}{7}

a = \frac{7}{13}b

Differentiating the above eqn w.r.t t:

\frac{da}{dt} = \frac{7}{13}.\frac{db}{dt}

Now, we know that:

Rate = \frac{db}{dt} = 5\ ft/s

Thus

\frac{da}{dt} = \frac{7}{13}.\times 5 = 2.692\ ft/s

5 0
3 years ago
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