Why doesn't the motor work?

Describe why using the simulation is a good method for studying projectiles. Clearly identify the error sources the simulation e

DiKsa [7]

Am not sure
Have a nice day

6 0

3 years ago

A uniform marble rolls down a symmetrical bowl, start- ing from rest at the top of the left side. The top of each side is a dist

Paha777 [63]

Answer:

Part a)

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

Explanation:

As we know by energy conservation the total energy at the bottom of the bowl is given as

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

here we know that on the left side the ball is rolling due to which it is having rotational and transnational both kinetic energy

now on the right side of the bowl there is no friction

so its rotational kinetic energy will not change and remains the same

so it will have

$\frac{1}{2}mv^2 = mgh'$

now we know that

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

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

so we have

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh$

$\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh$

$\frac{7}{10}mv^2 = mgh$

$\frac{1}{2}mv^2 = \frac{10}{14}mgh$

so the height on the smooth side is given as

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

6 0

3 years ago

Read 2 more answers

24 How long does it take for the Earth to orbit the Sun?

Zarrin [17]

Answer:

365 days

So The Final Answer is a Year

D is the answer

Explanation:

7 0

2 years ago

Read 2 more answers

A 45-kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. Neglect air resistan

Lubov Fominskaja [6]

The altitude or height of the pole vaulter as she crosses the bar is 4.04 m.

<h3>What is the height of the pole vaulter?</h3>

The height of the pole vaulter is determined from the change in kinetic energy which is equal to the potential energy at that height.

Potential energy = Change in kinetic energy

mgh = m(v - u)²/2

h = (v - u)²/2g

h = (10 - 1.1)²/2 * 9.8

h = 4.04 m.

In conclusion, the height is determined from the potential energy at that height.

Learn more about potential energy at: brainly.com/question/14427111

#SPJ1

7 0

1 year ago

What is the acceleration of a block on a ramp inclined 35o to the horizontal if µk = 0.4?

lina2011 [118]

We can solve the problem by applying Newton's second law, which states that the resultant of the forces acting on an object is equal to the product between its mass and its acceleration:
$\sum F = ma$

We should consider two different directions: the direction perpendicular to the inclined plane and the direction parallel to it. Let's write the equations of the forces along the two directions, decomposing the weight of the object (mg):

$mg \sin \theta - \mu_K N = ma$ (parallel direction) (1)
$mg \cos \theta - N =0$ (perpendicular direction) (2)
where
$\theta=35^{\circ}$ is the angle of the inclined plane, N is the normal reaction of the plane, $\mu_K N$ is the frictional force, with $\mu_K=0.4$ being the coefficient of friction.

From eq.(2), we find
$N=mg \cos \theta$
and if we substitute into eq.(1), we can find the acceleration of the block:
$mg \sin \theta - \mu_k mg \cos \theta = ma$
from which
$a=g(\sin \theta - \mu_K \cos \theta)=(9.81 m/s^2)(\sin 35^{\circ} - 0.4 \cos 35^{\circ})=2.41 m/s^2$

7 0

3 years ago