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erastovalidia [21]
3 years ago
12

A man weighs himself twice in an elevator. When the elevator is at rest, he weighs 824 N; when the elevator starts moving upward

, he weighs 932 N. Most nearly how fast is the elevator accelerating, assuming constant acceleration?
a. 0.64 m/s
b. 1.1 m/s^2
c. 1.3 m/s
d. 9.8 m/s^2
Physics
1 answer:
kifflom [539]3 years ago
3 0

Answer: c. 1.3 m/s^2

Explanation:

When he is at rest, is weight can be calculated as:

W = g*m

where:

m = mass of the man

g = gravitational acceleration = 9.8m/s^2

We know that at rest his weight is W = 824N, then we have:

824N = m*9.8m/s^2

824N/(9.8m/s^2) = m = 84.1 kg

Now, when the elevators moves up with an acceleration a, the acceleration that the man inside fells down is g + a.

Then the new weight is calculated as:

W = m*(g + a)

and we know that in this case:

W = 932N

g = 9.8m/s^2

m = 84.1 kg

Then we can find the value of a if we solve:

932N = 84.1kg*(9.8m/s^2 + a)

932N/84.1kg = 11.1 m/s^2 = 9.8m/s^2 + a

11.1 m/s^2 - 9.8m/s^2 = a = 1.3 m/s^2

The correct option is C

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Answer:

A) v_average = - 10 km / h,  B)    v = 1.6 m / s, v = 17.6 m / s

Explanation:

A) the average speed is the average speed of a body, if we assume that the direction of going up the hill is positive

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    v₂ = - 60 km / h

the average speed is

         v_average = \frac{v_1 + v_2}{2}

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B) in this case they indicate the acceleration a = 3.2 m / s² and the velocity vo = 9.6 m / s

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         v = v₀ + a t

as the time is earlier

          t = - 2.5 s

we substitute

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ii) the velocity for a subsequent time of 2.5 s

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