Ask question

Ask question

All categories

3 years ago

6

What happens when velocity and acceleration are at right angles to each other

1 answer:

photoshop1234 [79]3 years ago

7 0

You get circular motion, where the acceleration is pointing towards the center of the circle, as long as they are constant, and not fluctuating.

You might be interested in

8. Humans need oxygen to break down food in order to provide energy for cells. The oxygen needed for this process is delivered b

Marina86 [1]

Answer:

The answer is B. Lungs

7 0

3 years ago

Read 2 more answers

Which two conditions would result in the weakest electric force between objects?

Novay_Z [31]

The objects are far apart the weakest the result the two conditions

3 0

3 years ago

Which of the following demonstrates a translation?

vampirchik [111]

A spinning wheel and the blade of a kitchen blender both illustrate rotation.

A child swinging on a swing illustrates oscillation, or 'harmonic' motion.

A balloon being blown up is an example of dilation or inflation.

A sliding hockey puck demonstrates the concept of translation.

5 0

2 years ago

Read 2 more answers

A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th

photoshop1234 [79]

Answer:

The height is $h_c = 42.857$

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

From the question we are told that

The height is $h_s = 30 \ cm$

The angle of the slope is $\theta = 15^o$

According to the law of conservation of energy

The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

$mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2$

Where I is the moment of inertia which is mathematically represented as this for a sphere

$I = \frac{2}{5} mr^2$

The angular velocity $w$ is mathematically represented as

$w = \frac{v}{r}$

So the equation for conservation of energy becomes

$mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2$

$mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]$

$mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]$

$gh_s =[\frac{7}{10} ] v^2$

$v^2 = \frac{10gh_s}{7}$

Considering a circular hoop

The moment of inertial is different for circle and it is mathematically represented as

$I = mr^2$

Substituting this into the conservation equation above

$mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2$

Where $h_c$ is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

$mgh_c = mv^2$

$gh_c = v^2$

$h_c = \frac{v^2}{g}$

Recall that $v^2 = \frac{10gh_s}{7}$

$h_c= \frac{\frac{10gh_s}{7} }{g}$

$= \frac{10h_s}{7}$

Substituting values

$h_c = \frac{10(30)}{7}$

$h_c = 42.86 \ cm$

8 0

2 years ago

I can’t figure it out! <br> Please help I have a test tomorrow

nignag [31]

it's 20m/s i.e. no. (2)

7 0

3 years ago