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alexgriva [62]
3 years ago
13

A car with a mass of 1.1 × 103 kilograms hits a stationary truck with a mass of 2.3 × 103 kilograms from the rear end. The initi

al velocity of the car is +22.0 meters/second. After the collision the velocity of the car is -11.0 meters/second. What is the velocity of the truck after this elastic collision?
A. +13 meters/second
B. +20 meters/second
C. +15.7 meters/second
D. +7.5 meters/second
E. +9.8 meters/second
Physics
1 answer:
snow_lady [41]3 years ago
4 0

Answer:

The velocity of the truck after this elastic collision is 15.7 m/s            

Explanation:

It is given that,

Mass of the car, m_1=1.1\times 10^3\ kg

Mass of the truck, m_2=2.3\times 10^3\ kg

Initial velocity of the car, u_1=22\ m/s

Initial velocity of the truck, u₂ = 0

After the collision the velocity of the car is, v₁ = -11 m/s

Let v₂ is the velocity of the truck after this elastic collision. Using the conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

1.1\times 10^3\times 22+0=1.1\times 10^3\times (11)+2.3\times 10^3\times v_2    

v_2=15.7\ m/s

So, the velocity of the truck after this elastic collision is 15.7 m/s. Hence, the correct option is (c).

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Answer:

How far will the electron travel beforehitting a plate is 248.125mm

Explanation:

Applying Gauss' law:

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E= 1.13×10^5N/C

Force on electron F=qE

Where q=charge of electron=1.6×10^-19C

Therefore F=1.6×10^-19×1.13×10^5

F=1.808×10^-14N

Acceleration on electron a = Force/Mass

Where Mass of electron = 9.10938356 × 10^-31

Therefore a= 1.808×10^-14 /9.11 × 10-31

a= 1.985×10^16m/s^2

Time spent between plate = Distance/Speed

From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2

Therefore Time = 0.01/2×10^6

Time =5×10^-9s

How far the electron would travel S =ut+ at^2/2 where u=0

S= 1.985×10^16×(5×10^-9)^2/2

S=24.8125×10^-2m

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4 0
3 years ago
The cube with 2.00m wide and 2.00m long and 2.00m high has a weight of 960.00 N what pressure does it exert
andre [41]
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3 years ago
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
Mademuasel [1]

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

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here we have

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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

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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

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now power loss in line is given as

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

6 0
3 years ago
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Thus, in order to sense motion of objects, humans beings are required to see an object change its position with respect to other objects.

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4 years ago
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