All categories

3 years ago

What is the force on a 250 kg elevator that is falling freely at 9.8 m/sec2?

Physics

2 answers:

Anettt [7]3 years ago

7 0

Answer:

$\boxed {\boxed {\sf 2450 \ Newtons }}$

Explanation:

The elevator is falling freely and the only force acting on it is weight. So, calculate the weight.

We can do this by multiplying the mass by the acceleration of gravity.

$W=m*g$

The mass of the elevator is 250 kilograms and the acceleration of gravity is 9.8 meters per second squared.

$m= 250 \ kg \\g= 9.8 \ m/s^2$

Substitute the values into the formula.

$W= 250 \ kg * 9.8 \ m/s^2$

Multiply.

$W=2450 \ kg*m/s^2$

1 kilogram meter per second squared is equal to 1 Newton
So, our answer of 2450 kg*m/s² is equal to 2450 N

$W= 2450 \ N$

The force of weight on the elevator is <u>2450 Newtons</u>

Viefleur [7K]3 years ago

6 0

Answer:

Since it is falling freely, the only force on it is its weight, w.

w = m × g = 250 kg × 9.8 m/s^2 = 2450 Newton/N

You might be interested in

In what subject could we see cross cutting concepts

meriva

A Framework for K–12 Science Education: Practices, Crosscutting Concepts, and Core Ideas (Framework) recommends science education in grades K–12 be built around three major dimensions: science and engineering practices, crosscutting concepts that unify the study of science and engineering through their common application across fields, and core ideas in the major disciplines of natural science.

4 0

3 years ago

A baseball has mass 0.147 kg. If the velocity of a pitched ball has a magnitude of 44.5 m/s and the batted ball's velocity is 55

Anit [1.1K]

Explanation:

We have,

Mass of a baseball is 0.147 kg

Initial velocity of the baseball is 44.5 m/s

The ball is moved in the opposite direction with a velocity of 55.5 m/s

It is required to find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Change in momentum,

$\Delta p=mv-mu\\\\\Delta p=m(v-u)\\\\\Delta p=0.147\times ((-55.5)-44.5)\\\\\Delta p=-14.7\ kg-m/s\\\\|\Delta p|=14.7\ kg-m/s$

Impulse = 14.7 kg-m/s

Therefore, the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat is 14.7 kg-m/s

4 0

3 years ago

An infant's toy has a 120 g wooden animal hanging from a spring. If pulled down gently, the animal oscillates up and down with a

Morgarella [4.7K]

Answer:

0.37 m

Explanation:

The angular frequency, ω, of a loaded spring is related to the period, T, by

$\omega = \dfrac{2\pi}{T}$

The maximum velocity of the oscillation occurs at the equilibrium point and is given by

$v = \omega A$

A is the amplitude or maximum displacement from the equilibrium.

$v = \dfrac{2\pi A}{T}$

From the the question, T = 0.58 and A = 25 cm = 0.25 m. Taking π as 3.142,

$v = \dfrac{2\times3.142\times0.25\text{ m}}{0.58\text{ s}} = 2.71 \text{ m/s}$

To determine the height we reached, we consider the beginning of the vertical motion as the equilibrium point with velocity, v. Since it is against gravity, acceleration of gravity is negative. At maximum height, the final velocity is 0 m/s. We use the equation

$v_f^2 = v_i^2+2ah$