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satela [25.4K]
3 years ago
9

A 60kg student traveling in a 1000kg car with a constant velocity has a kinetic energy of 1.2 x 10^4 J. What is the speedometer

reading of the car in km/hr?
Physics
1 answer:
777dan777 [17]3 years ago
4 0

Answer:

17.64 km/h

Explanation:

mass of car, m = 1000 kg

Kinetic energy of car, K = 1.2 x 10^4 J

Let the speed of car is v.

Use the formula for kinetic energy.

K = \frac{1}{2}mv^{2}

By substituting the values

1.2\times 10^{4} = \frac{1}{2}\times 1000\times v^{2}

v = 4.9 m/s

Now convert metre per second into km / h

We know that

1 km = 1000 m

1 h = 3600 second

So, v = \left (\frac{4.9}{1000}   \right )\times \left ( \frac{3600}{1} \right )

v = 17.64 km/h

Thus, the reading of speedometer is 17.64 km/h.

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Explanation: Hope this helped. Have a great day!!!!

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3 years ago
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What is the correct path of electrons?
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3 0
2 years ago
A porter can carry 40 bricks of 10 N load of each. He can carry up to 75m in 40 sec, calculate his power.​
alexira [117]

Answer:

750W

Explanation:

40×10= 400N

work done= force × distance

=400 × 75

=30000 J

Power= work done/ time

= 30000 ÷ 40

= 750 W

4 0
1 year ago
How much thermal energy is created when a 3000 kg suv brakes to a stop from 20 m/s on a level road?
JulsSmile [24]

Answer:

b. 600,000 J

Explanation:

Applying the law of conservation of energy,

The thermal energy created = Kinetic energy of the suv.

Q' = 1/2(mv²)............... Equation 1

Where Q' = Thermal energy, m = mass of the suv, v = velocity of the suv.

From the question,

Given: m = 3000 kg, v = 20 m/s

Substitute these values into equation 1

Q' = 1/2(3000×20²)

Q' = 600000 J

Hence the right option is b. 600,000 J

5 0
3 years ago
The coefficient of linear expansion of copper is 17 x 10^-6 K-1. A block of copper 30 cm wide, 45 cm long, and 10 cm thick is he
schepotkina [342]

Answer:

The change in volume is 6.885\times 10^{- 5}\

Solution:

As per the question:

Coefficient of linear expansion of Copper, \alpha = 17\times 10^{- 6}\ K^{- 1}

Initial Temperature, T = 0^{\circ} = 273 K

Final Temperature, T' = 100^{\circ} = 273 + 100 = 373 K

Now,

Initial Volume of the block, V = 30\times 45\times 10\times 10^{- 6}\ m^{3} = 0.0135\ m^{3}

V' = V(1 + \gamma \Delta T)

\gamma = 3\alpha

V' = V(1 + 3\alpha \Delta T)

where

V' = Final volume

V' - V= 0.0135\times 17\times 10^{- 6} \times (T' - T))

\Delta V= 0.0135\times 3\times 17\times 10^{- 6} \times (373 - 273)) = 6.885\times 10^{- 5}\

7 0
3 years ago
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