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Crazy boy [7]
3 years ago
9

A constant force is exerted for a short time interval on a cart that is initially at rest on an air track. This force gives the

cart a certain final speed. Suppose we repeat the experiment but, instead of starting from rest, the cart is already moving with constant speed in the direction of the force at the moment we begin to apply the force.
After we exert the same constant force for the same short time interval, the increase in the cart's speed:

A. is equal to two times its initial speed.
B. is equal to the square of its initial speed.
C. is equal to four times its initial speed.
D. is the same as when it started from rest.
E. cannot be determined from the information provided.
Physics
1 answer:
Ilya [14]3 years ago
5 0

Answer:

d) is the same as when it started from rest

Explanation:

using equation of motion

v = u + at

second law of momentum defines

F = ma

a = F /m

the equation becomes

v = u + (F/m)t

from hear

since v is directly proportional to the force and the force remain the same, the increase in the cart speed will also remain the same.

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Two long, straight wires are parallel and 20 cm apart. One carries a current of 2.2 A, the other a current of 5.9 A. (a) If the
Leya [2.2K]

Answer: 1.298\times 10^{-5}\ N/m

Explanation:

Given

Current in the first wire I_1=2.2\ A

Current in the second wire I_2=5.9\ A

wires are 20\ cm apart

Force per unit length between the current-carrying wires is

\Rightarrow \dfrac{F}{l}=\dfrac{\mu_oI_1I_2}{2\pi r}

Force exerted  by the wires is the same

Put the values

\Rightarrow \frac{F}{l}=f=\dfrac{4\pi \times 10^{-7}\times 2.2\times 5.9}{2\pi \times 0.2}=1.298\times 10^{-5}\ N/m

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4 0
3 years ago
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.

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3 years ago
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Answer:

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Answer:

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