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alukav5142 [94]
2 years ago
8

A 3.0-kg cart is rolling across a frictionless, horizontal track toward a 1.3-kg cart that is held initially at rest. The carts

are loaded with strong magnets that cause them to attract one another. Thus, the speed of each cart increases. At a certain instant before the carts collide, the first cart's velocity is +4.6 m/s, and the second cart's velocity is −1.9 m/s.
(Indicate the direction with the sign of your answer.)

(a) What is the total momentum of the system of the two carts at this instant?
kg · m/s

(b) What was the velocity of the first cart when the second cart was still at rest?
m/s
Physics
1 answer:
Karolina [17]2 years ago
6 0

Answer:

Explanation:

a ) Momentum of first cart = mass x velocity

= 3 x 4.6 =+13.8 kg m /s

Momentum of second cart = 1.3 x - 1.9 = - 2.47 kg m /s

Total momentum = 13.8 - 2.47

= +11.33 kg m /s

b )

Let the velocity of first cart be v at the moment when second cart was at rest

total momentum = 3 x v + 0 = 3 v

Applying conservation of momentum law

3 v  = +11.33

v = +3.77 m /s

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2 years ago
An aluminum cylinder with a radius of 2.7 cm and a height of 67 cm is used as one leg of a workbench. The workbench pushes down
soldier1979 [14.2K]

Answer:

1.9\times 10^{-4}

1.2\times 10^{-4}\ m

Explanation:

r = Radius = 2.7 cm

F = Force = 3.2\times 10^4\ N

A = Area = \pi r^2

\sigma = Stress = \frac{F}{A}

E = Young's modulus = 7\times 10^{10}\ Pa

\epsilon = Strain

L_0 = Original length = 67 cm

\Delta L = Change in length

Young's modulus is given by

E=\frac{\sigma}{\epsilon}\\\Rightarrow \epsilon=\frac{\sigma}{E}\\\Rightarrow \epsilon=\frac{\frac{3.2\times 10^4}{\pi 0.027^2}}{7\times 10^{10}}\\\Rightarrow \epsilon=0.0001996=1.9\times 10^{-4}

Strain is 1.9\times 10^{-4}

Strain is given by

\epsilon=\frac{\Delta L}{L_0}\\\Rightarrow \Delta L=\epsilon\times L_0\\\Rightarrow \Delta L=1.9\times 10^{-4}\times 0.67\\\Rightarrow \Delta L=0.0001273\\\Rightarrow \Delta L=1.2\times 10^{-4}\ m

The cylinder height decreases by 1.2\times 10^{-4}\ m

3 0
3 years ago
A satellite is launched to orbit the Earth at an altitude of 2.90 x10^7 m for use in the Global Positioning System (GPS). Take t
marin [14]

Answer:

T=66262.4s

Explanation:

From the question we are told that:

Altitude A=2.90 *10^7

Mass m=5.97 * 10^{24} kg

Radius r=6.38 *10^6 m.

Generally the equation for Satellite Speed is mathematically given by

V=(\frac{GM}{d} )^{0.5}

V=(\frac{6.67*10^{-11}*5.97 * 10^{24}}{6.38 *10^6+2.90 *10^7} )^{0.5}

V=3354.83m/s

Therefore

Period T is Given as

T=\frac{2 \pi *a}{V}

T=\frac{2 \pi *(6.38 *10^6+2.90 *10^7}{3354.83}

T=66262.4s

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Answer:

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