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3 years ago

Why is your velocity continuously changing as you ride on a carousel?

2 answers:

5 0

Since carousels go in circles, the direction of travel is constantly changing and velocity is defined as constant change. This is the reason why velocity is continuously changing when riding a carousel for velocity is involve in change of direction.

Moreover, a change in velocity is an acceleration and velocity has two properties that comprises direction and magnitude. Whereas the direction of velocity is the direction of travel and speed is the magnitude of velocity.

nydimaria [60]3 years ago

5 0

<span>To make this more obvious (and give it some relevance beyond being just a HW question), imagine that the ride is going super, super, duper fast. You'd better hold on tight or else you'll fly off. There's obviously a force acting on you, then, to keep you on that ride. Therefore, you are accelerating, which means your velocity is changing.</span>

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A planet exerts a gravitational force of magnitude 9e22 N on a star. If the planet were 2 times closer to the star (that is, if

Dmitrij [34]

To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.

Planet gravitational force

$F_p = 6*10^{22}N$

$F_p = \frac{GMm}{R^2}$

$F_p = 9*10^{22}N$

Distance between planet and star

$r = \frac{R}{2}$

Gravitational force is

$F = \frac{GMm}{r^2}$

Applying the new distance,

$F = \frac{GMm}{(\frac{R}{2})^2}$

$F = 4\frac{GMm}{R^2}$

Replacing with the previous force,

$F = 4F_p$

Replacing our values

$F= 4(9*10^{22}N)$

$F = 36*10^{22}N$

Therefore the magnitude of the force on the star due to the planet is $36*10^{22}N$

5 0

3 years ago

Answer these two questions please

Grace [21]

1.The answer is True
2.The answer is False

8 0

3 years ago

Select to show the energy of pendulum 1. Be sure that friction is set to none. Drag the pendulum to an angle (with respect to th

iragen [17]

Answer:

it have Potential energy

Explanation:

given data

Drag the pendulum to an angle 30∘

to find out

what form of energy does it have

solution

we know that pendulum start no kinetic energy when it release from any rest position then in starting it have potential energy only so that when pendulum is angle 30∘ at some height from ground so when it start it have potential energy same as in starting.

we know that the total energy is always conserve

so it have potential energy

3 0

3 years ago

A metal cylinder with a mass of 4.20 kg is attached to a spring and is able to oscillate horizontally with negligible friction.

kherson [118]

Answer:

a) k = 120 N / m

, b) f = 0.851 Hz

, c) v = 1,069 m / s

, d) x = 0

, e) a = 5.71 m / s²

, f) x = 0.200 m

, g) Em = 2.4 J

, h) v = -1.01 m / s

Explanation:

a) Hooke's law is

F = k x

k = F / x

k = 24.0 / 0.200

k = 120 N / m

b) the angular velocity of the simple harmonic movement is

w = √ k / m

w = √ (120 / 4.2)

w = 5,345 rad / s

Angular velocity and frequency are related.

w = 2π f

f = w / 2π

f = 5.345 / 2π

f = 0.851 Hz

c) the equation that describes the movement is

x = A cos (wt + Ф)

As the body is released without initial velocity, Ф = 0

x = 0.2 cos wt

Speed is

v = dx / dt

v = -A w sin wt

The speed is maximum for sin wt = ±1

v = A w

v = 0.200 5.345

v = 1,069 m / s

d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is

x = A cos wt = 0

x = 0

e) the acceleration is

a = d²x / dt² = dv / dt

a = - Aw² cos wt

The acceleration is maximum when cos wt = ± 1

a = A w²

a = 0.2 5.345

a = 5.71 m / s²

f) the position for this acceleration is

x = A cos wt

x = A

x = 0.200 m

g) Mechanical energy is

Em = ½ k A²

Em = ½ 120 0.2²

Em = 2.4 J

h) the position is

x = 1/3 A

Let's calculate the time to reach this point

x = A cos wt

1/3 A = A cos 5.345t

t = 1 / w cos⁻¹(1/3)

The angles are in radians

t = 1.23 / 5,345

t = 0.2301 s

Speed is

v = -A w sin wt

v = -0.2 5.345 sin (5.345 0.2301)

v = -1.01 m / s

i) acceleration

a = -A w² sin wt

a = - 0.2 5.345² cos (5.345 0.2301)

a = -1.91 m / s²

5 0

3 years ago

The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the magnitude of the

PIT_PIT [208]

The drag force is directly proportional to the square of the velocity of motion of the object. So as the speed is doubled, the magnitude of drag force will get quadrupled.

<u>Explanation: </u>

Drag force is the opposing or resisting force acting on any object by the medium in which it is moving. So in this case, you are riding a bicycle, thus the medium can be considered as air.

The formula for calculating drag force is as below:

$\text {Drag force }=\frac{1}{2} \rho v^{2} C_{D} A$

Here, ρ is the density of the air molecules, v is the velocity or speed of the bicycle, CD is the drag coefficient and A is the area of the bicycle.

In the above equation, only the term velocity will be a varying quantity with respect to time and other quantities will remain constant throughout the single situation of riding of bicycle.

So, the equation can be,

$\text { Drag force }=k v^{2}$