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Artist 52 [7]
3 years ago
11

a 2kg marble is moving at 3 m/s when it strikes another 4kg marble moving in the opposite direction at -3 m/s. What will be the

velocity of the 4kg marble after the collision of the 2kg marble's new velocity is -4 m/s. Also state the direction of the 2kg marble after they hit.
Physics
1 answer:
Elan Coil [88]3 years ago
4 0
Momentum is conserved after the collision 

Momentum of 2 Kg before collision = 2 * 3 = 6
Momentum of 4 kg  before collision = 4 * -3 = -12

so 6 + -12 = 2 * -4 + 4 *x   where x is velocity of 4kg marble.

4x - 8 = -6
4x = 2
x = 0.5 

Velocity of 4 kg marble is 0.5 m/s after collision

The 2 kg marble will move in the opposite direction to which it was moving before the collision.
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The magnitude of the electric field = 4KQ / L^2

direction = 45° east to south

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Lightning produces a maximum air temperature on the order of 104K, whereas a nuclear explosion produces a temperature on the ord
gtnhenbr [62]

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tex]2.898\times 10^{-7}\ \text{m}[/tex] ultraviolet region

2.898\times 10^{-10}\ \text{m} x-ray region

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T = Temperature

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\lambda = Wavelength

T=10^4\ \text{K}

From Wein's law we have

\lambda=\dfrac{b}{T}\\\Rightarrow \lambda=\dfrac{2.898\times 10^{-3}}{10^4}\\\Rightarrow \lambda=2.898\times 10^{-7}\ \text{m}

The wavelength of the radiation will be 2.898\times 10^{-7}\ \text{m} and it is in the ultraviolet region.

T=10^7\ \text{K}

\lambda=\dfrac{2.898\times 10^{-3}}{10^7}\\\Rightarrow \lambda=2.898\times 10^{-10}\ \text{m}

The wavelength of the radiation will be 2.898\times 10^{-10}\ \text{m} and it is in the x-ray region.

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A car travels 150 meters East of its original starting position in 15s. What is the cars velocity?​
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Read 2 more answers
61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A
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Answer:

Part a)

percentage = 21.3%

Part b)

percentage = 2.13 \times 10^{-5}%

Explanation:

As we know that total power used in the room is given as

P = P_1 + P_2 + P_3 + P_4

here we have

P_1 = (110)(3) = 330 W

P_2 = 100 W

P_3 = 60 W

P_4 = 3 W

P = 330 + 100 + 60 + 3

P = 493 W

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

110\times i = 493

i = 4.48 A

now resistance of transmission line

R = \frac{\rho L}{A}

R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}

R = 5.23 \ohm

now power loss in line is given as

P = i^2 R

P = (4.48)^2(5.23)

P = 105 W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{105}{493} \times 100

percentage = 21.3%

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

110 \times 10^3 (i ) = 493

i = 4.48\times 10^{-3} A

now power loss in line is given as

P = i^2 R

P = (4.48 \times 10^{-3})^2(5.23)

P = 1.05 \times 10^{-4} W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{1.05 \times 10^{-4}}{493} \times 100

percentage = 2.13 \times 10^{-5}%

6 0
3 years ago
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