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2 years ago

How much work must be done to stop an 1 800kg vehicle travelling at 30 m/s?

Physics

1 answer:

mars1129 [50]2 years ago

5 0

Answer:

810 kJ

Explanation:

Work done is equal to change in kinetic energy

W=∆KE

Change in kinetic energy is given by

∆KE=½m(v²-u²)

Where ∆KE is the change in kinetic energy, m is mass of the vehicle, v is final velocity while u is initial velocity.

Since the vehicle must come to rest, the final velocity is zero.

Substituting 1800 kg for m, 0 m/s for v and 30 m/s for u then

∆KE=½*1800(0²-30²)=-810, 000J

Converted into kJ

∆KE=-810 kJ

Therefore, work done is 810 kJ

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2 years ago

A volcano erupts and launches a chunk of hot magma horizontally with a speed of 252 m/s. The magma travels a horizontal distance

ArbitrLikvidat [17]

Answer:

The value is $v_y = -48.61 \ m/s$

Explanation:

From the question we are told that

The horizontal speed is $u_x = 252 \ m/s$

The horizontal distance is $d = 1250 \ m$

Generally the time taken by the hot magma in air before landing is mathematically represented as

$t = \frac{d}{u_x}$

=> $t = \frac{ 1250 }{252}$

=> $t = 4.96 \ s$

Generally the initial vertical velocity of the magma when it was lunched is

$u_y = 0 \ m/ s$

Then the final velocity of the magma when it hits the ground is mathematically represented s

$- v_y = u_y + gt$

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$- v_y = 48.61 \ m/s$

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2 years ago