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yKpoI14uk [10]
3 years ago
12

A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T

he same cart again traveling at 0.3 m/s collides with a different stationary object. This time the cart is at rest after the collision. In which collision is the impulse on the cart greater?
A. The first collision.

B. Cannot be determined without knowing the rebound speed of the first collision.

C. The second collision.

D. Cannot be determined without knowing the mass of the cart.

E. The impulses are the same.
Physics
1 answer:
Nady [450]3 years ago
6 0
The first collision because a greater amount of momentum must be taken and used in order to push the cart back, giving it a greater mass and impulse
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Answer:

Explanation:

There will be loss of potential energy due to loss of height and gain of kinetic energy .

loss of height = R - R cos 14 ,    R is radius of hemisphere .

R ( 1 - cos 12 )

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h = .286 m

loss of potential energy

= mgh

= m x 9.8 x .286

= 2.8 m

gain of kinetic energy

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v² = 2 g h

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v = 7.40 m /s

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3 years ago
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krek1111 [17]

Answer:

Explanation:

The path length difference = extra distance traveled

The destructive interference condition is:

\Delta d = (m+1/2)\lambda

where m =0,1, 2,3........

So, ←

\Delta d = (m+1/2)\lamb da9/tex]so [tex]\Delta d = \frac{\lambda}{2}

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4 0
2 years ago
How much support force does a table exert on a book that weighs 15 N when the book is placed on the table?
mestny [16]

Answer:

15 N

Explanation:

According to Newton's third law of motion, to every action, there is an equal and opposite reaction. This reaction is equal in magnitude to the force acting but in an opposite direction.

Now, if the book weighs 15 N, an opposite equal force will be: N = -15 N

But the magnitude of this will be the absolute value which is 15N.

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2 years ago
What is the current in a 160V circuit if the resistance is 2Ω?<br> V=<br> I=<br> R=
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Explanation:

v = IR

v= 160 R = 2

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2 years ago
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Karolina [17]

Answer:

The average induced emf in the coil is 0.0286 V

Explanation:

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initial magnetic field, B₁ = 0.53 T

final magnetic field, B₂ = 0.24 T

time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

where;

A is area of the coil = πr²

r is the radius of the wire coil = 0.112m / 2 = 0.056 m

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E = -0.00985(B₂-B₁)/t

E = 0.00985(B₁-B₂)/t

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E = 0.00985 (0.29)/ 0.1

E = 0.0286 V

Therefore, the average induced emf in the coil is 0.0286 V

3 0
3 years ago
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