A(n) _________________ is a push or pull that one object exerts on another

Who’s going faster in the attached image?

ioda

Answer:

Art

Explanation:

Polly's line is linear, while arts line is going up with constant velocity. There for art is going faster.

8 0

3 years ago

Read 2 more answers

What formed as a result of a hot spot?

masha68 [24]

A the San Andreas fault

7 0

4 years ago

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A(n) 1100 kg car is parked on a 4◦ incline. The acceleration of gravity is 9.8 m/s 2 . Find the force of friction keeping the ca

igomit [66]

Answer:

Force of friction, f = 751.97 N

Explanation:

it is given that,

Mass of the car, m = 1100 kg

It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.

From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.

$f=mg\ sin\theta$

$f=1100\ kg\times 9.8\ m/s^2\ sin(4)$

f = 751.97 N

So, the force of friction on the car is 751.97 N. Hence, this is the required solution.

3 0

3 years ago

An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the

garik1379 [7]

(a) 6.04 rev/s

The speed of the ball is given by:

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the ball from the centre of the circle

In situation 1), we have

$\omega=8.13 rev/s \cdot 2\pi = 51.0 rad/s$

r = 0.600 m

So the speed of the ball is

$v=(51.0 rad/s)(0.600 m)=30.6 m/s$

In situation 2), we have

$\omega=6.04 rev/s \cdot 2\pi = 37.9 rad/s$

r = 0.900 m

So the speed of the ball is

$v=(37.9 rad/s)(0.900 m)=34.1 m/s$

So, the ball has greater speed when rotating at 6.04 rev/s.

(b) $1561 m/s^2$

The centripetal acceleration of the ball is given by

$a=\frac{v^2}{r}$

where

v is the speed

r is the distance of the ball from the centre of the trajectory

For situation 1),

v = 30.6 m/s

r = 0.600 m

So the centripetal acceleration is

$a=\frac{(30.6 m/s)^2}{0.600 m}=1561 m/s^2$

(c) $1292 m/s^2$

For situation 2 we have

v = 34.1 m/s

r = 0.900 m

So the centripetal acceleration is

$a=\frac{v^2}{r}=\frac{(34.1 m/s)^2}{0.900 m}=1292 m/s^2$

5 0

3 years ago

An electrical motor spins at a constant 2035 rpm. if the armature radius is 5.862 cm, what is the acceleration of the edge of th

Zepler [3.9K]

An object that moves in a uniform circular motion would have a constant speed. However, we would not say that acceleration would be zero. The object would be accelerating due to the changes in the direction. Acceleration or this case would be the ratio of the square of its velocity an the radius of the circular motion. We would calculate as follows:

Acceleration = velocity^2 / radius
velocity = 2035 rpm (1 min / 60 s) (5.862 cm)
velocity = 198.8195 cm/s
Acceleration = (198.8195 cm/s)^2 / 5.862 cm = 33.92 cm/s^2

Therefore, the acceleration of the edge of the rotor would be 33.92 cm per s^2.

3 0

4 years ago