You visited us 1 times! Enjoying our articles? Unlock Full Access!

V-t to a-t Graph

A truck of ma...

Question

A truck of mass $10000 k g$ moves up an inclined plane rising $s i n^{- 1} 50$ with a speed of $36 k m / h$ , then the power of engine: ( $g = 10 m / s^{2}$ )

10 k W

No worries! We‘ve got your back. Try BYJU‘S free classes today!

20 k W

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

30 k W

No worries! We‘ve got your back. Try BYJU‘S free classes today!

40 k W

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $20 k W$
Power $P = m g V sin θ$
Putting all the values in the equation will get,
$P = 10000 \times 10 \times \frac{1}{50} \times 36 \times \frac{5}{18} = 20 k W$

Suggest Corrections

Similar questions

30, 000 k g

1

100

30 k m / h

(G i v e n g = 10 m s^{- 1})

30, 000

1

100

30

θ = {tan}^{- 1} (\frac{1}{2})

36 k m / h

(\frac{1}{\sqrt{3}})

500

1

100

10 m s^{- 1}

2000

1

= 10^{3}

g = 9.8 m

s^{- 2}

1000 k g

{sin}^{- 1} (2 / 25)

18 k m p h

g = 10 m s^{- 2}

Join BYJU'S Learning Program