All categories

3 years ago

A roller coaster goes from 2.00 m/s [forward] to 10.0 m/s [forward) in 4.50 s. What is its acceleration?

1 answer:

8 0

The answer would be 1.77778:

We must use the kinematic equation v = v0 + at, then fill in the elements given and solve which equals 1.77778

You might be interested in

Is Bytes a scalar or vector?

melisa1 [442]

It is a scaler because it’s only fully describes by a magnitude and a numerical alone

8 0

3 years ago

Read 2 more answers

Which statement describes the law of conservation of energy?

dmitriy555 [2]

Answer:

Explanation:

cuz it transforms from one to another can't be created not destroyed.PERIOD!

6 0

3 years ago

Read 2 more answers

Human centrifuges are used to train military pilots and astronauts in preparation for high-g maneuvers. A trained, fit person we

jeka57 [31]

(a) $4.14 rad/s^2$

The relationship beween centripetal acceleration and angular speed is

$a=\omega^2 r$

where

$\omega$ is the angular speed

r is the radius of the circular path

Here we gave

$a = 9g = 88.2 m/s^2$ is the centripetal acceleration

r = 5.15 m is the radius

Solving for $\omega$ , we find:

$\omega = \sqrt{\frac{a}{r}}=\sqrt{\frac{88.2 m/s^2}{5.15 m}}=4.14 rad/s^2$

(b) 21.3 m/s

The relationship between the linear speed and the angular speed is

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular speed

r is the radius of the circular path

In this problem we have

$\omega=4.14 rad/s$

r = 5.15 m

Solving the equation for v, we find

$v=(4.14 rad/s)(5.15 m)=21.3 m/s$

7 0

3 years ago

Read 2 more answers

A geosynchronous satellite orbits Earth at a distance of 42,250 km from the center of Earth and has a period of 1 day. What is t

yan [13]

Answer:

The centripetal acceleration of the satellite is $a=0.22\ m/s^2$ .

Explanation:

Given that,

The distance covered by a geosynchronous satellite, d = 42250 km

The time taken by the satellite to covered distance, t = 1 day = 24 hours

Since, 24 hours = 86400 seconds

Let v is the speed of the satellite. It is given by the total distance divided by total time taken such that :

$v=\dfrac{d}{t}$

$v=\dfrac{2\pi d}{t}$

$v=\dfrac{2\pi\times 42250 \times 10^3}{86400 }$

v = 3072.5 m/s

The centripetal acceleration of the satellite is given by :

$a=\dfrac{v^2}{d}$

$a=\dfrac{(3072.5)^2}{42250 \times 10^3}$

$a=0.22\ m/s^2$

So, the centripetal acceleration of the satellite is $a=0.22\ m/s^2$ . Hence, this is the required solution.

7 0

3 years ago

A certain ocean wave has a frequency of 0.07 hertz and a wavelength of 10 meters. what is the wave's speed? a 0.07 m/s

Elis [28]

V=f*wavelength
v=0.07hz*10m
v=0.7 m/s

8 0

3 years ago