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3 years ago

if a firebird travels at a velocity of 0 to 60 mph in 4 seconds traveling east what was the acceleration of the firebird

1 answer:

4 0

22 ft/s^2.

explanation: In this case, the velocity has increased by 60 mph in 4 seconds, so the average increase is 60/4 = 15 mph every second, or 15 mph/s.

Usually, you will want to use the same units of time throughout. In this case, you could multiply by the number of seconds in an hour to find out the change in velocity if it carried on travelling for that length of time. This would give the acceleration in units of mph/h, or mph^2.

If you want the answer in feet/s^2, then take your 15 mph/s answer, multiply by the number of feet in a mile (to get it to feet per hour per second) then divide by the number of seconds in an hour to get it to feet per second per second, or feet/s^2.

You should get 22 ft/s^2.

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Explanation:

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$\lambda=30.2m$

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3 years ago

The magnetic field in the region between the poles of an electromagnet is uniform at any time, (1 point) but is increasing at th

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Answer:

The magnitude of induced emf in the conducting loop is 0.99 mV.

Explanation:

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4 years ago

if a firebird travels at a velocity of 0 to 60 mph in 4 seconds traveling east what was the acceleration of the firebird​

if a firebird travels at a velocity of 0 to 60 mph in 4 seconds traveling east what was the acceleration of the firebird