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faust18 [17]
2 years ago
5

(b)

Physics
1 answer:
m_a_m_a [10]2 years ago
4 0

Answer:

Driving force increases, friction forces increase, the driving force is bigger than friction 12.

Explanation:

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A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?
stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, m_1 = 300\ g = 0.3\ kg

Initial speed of the cart, u_1=1.2\ m/s

Mass of the larger cart, m_2 = 2\ kg

Initial speed of the larger cart, u_2=0

After the collision,

Final speed of the smaller cart, v_1=-0.81\ m/s (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let v_2 is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

m_1u_1+m_2u_2-m_1v_1=m_2v_2

v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}

v_2=0.301\ m/s

So, the speed of the large cart after collision is 0.301 m/s.

4 0
3 years ago
How does changing the lengthy but not the height of an inclined plane affect the work done to lift a load? PLZ HELP ME NOW'
Serhud [2]

Answer: Work Done would remain same.

Let us assume that the velocity is constant while taking the load up the inclined plane. Then, the kinetic energy would remain the same. This is because kinetic energy is dependent on velocity (K.E.=\frac{1}{2}mv^2). If that is constant, the kinetic energy would remain same. The potential energy is dependent on the height(P.E.=mgh). If the height is changed, then potential energy varies. In the question, it is mentioned that without changing the height, the length of the inclined plane is changed. Therefore, the potential energy would be same as before.

We know, work done is equal to potential energy plus kinetic energy. Since there is no change in any of these, the required work done would not change.


4 0
3 years ago
An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at 12.8 m/s2 . At t1 the rocket e
Simora [160]

Answer:4.39 s

Explanation:

Given

initial velocity u=0

acceleration a=12.8 m/s^2

velocity acquired by sled in t_1 time

v=0+at

v=12.8t_1

distance traveled by sled in t_1 s

v^2-u^2=2as

(12.8t_1)^2-0=2\times 12.8\times s_1

s_1=6.4\cdot t_1^2

distance traveled in t_2 time with velocity v=12.8t_1

s_2=v\times t_2

s_2=12.8\times t_1\times t_2

s_2=12.8\cdot t_1\cdot t_2

s_1+s_2=5.37\times 10^3

6.4t_1^2+12.8t_1t_2=5370----1

t_1+t_2=97.7 s

t_2=97.7-t_1

substitute the value of t_2 in 1

we get

6.4t_1^2-1250.56t_1+5370=0

thus t_1=\frac{1250.56-1194.33}{12.8}=4.39 s

t_1=4.39 s

5 0
3 years ago
Electromagnetic wav​
krek1111 [17]

Answer:

I dont get what your asking sorry boo

Explanation:

6 0
2 years ago
Read 2 more answers
What is a penalty stroke in field hockey?
Paha777 [63]

Answer: D

Explanation: Because the orginal Penalty stroke is Today's field hockey topic is the exciting penalty stroke! What is a stroke? A penalty stroke occurs when an intentional foul happens in the penalty circle (the circle around the goal), the foul prevents a scoring of a opportunity.

4 0
2 years ago
Read 2 more answers
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