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3 years ago

Which type of force is changing the object's speed and direction when a toy clown pops out of a jack-in-the-box?

Physics

2 answers:

pochemuha3 years ago

4 0

The correct answer is

B. push

To be more precise, the force that changes the speed and direction of the toy clown is the elastic force due to the spring: when the box is closed, the spring is compressed, so all the energy is stored in the elastic potential energy store of the spring. As the box opens, the spring is free to stretch, therefore it pushes the toy out of the box, transferring all the elastic potential energy into kinetic energy of motion of the toy clown.

Murrr4er [49]3 years ago

3 0

I think it would be d because the spring or whatever was pushing until it reached the farthest it could then it would pull down but idrk

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A 26.2-kg dog is running northward at 3.02 m/s, while a 5.30-kg cat is running eastward at 2.74 m/s. Their 65.1-kg owner has the

REY [17]

Answer:

Angle with the +x axis is θ = 79.599degree

Then the velocity of owner = 1.235m/s

Explanation:

Given that the mass of dog is m1 =26.2 kg

velocity of dog is u1 = 3.02 m/s (north)

mass of cat is m2 = 5.3 kg

velocity is u2 = 2.74 m/s (east )

Mass of owner is M = 65.1 kg

Consider the east direction along +x axis andnorth along +y

momentum of dog is Py = m1 x u1

= 79.124 kg.m/s (j)

momentum of cat is Px = m2 x u2

= 14.522 kg.m/s (i)

Then the net magnitude of momentum is P = (Px2 + Py2)1/2

= 80.445

Angle with the +x axis is θ =tan-1(Py / Px ) = 79.599 degree

Then the velocity of owner is v = P / M = 1.235 m/s

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3 years ago

What is kelvin scale?

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4 years ago

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Question 1(Multiple Choice Worth 4 points) The star named Canopus has a declination of approximately –52°. Which of these statem

Y_Kistochka [10]

Answer:

It is 52° below the celestial equator.

Explanation:

The declination is the angle in degrees measured north (+) or south (-) of the an imaginary line called the celestial equator.

The celestial equator is a projection of the earth's equator on the celestial sphere. imaginary

The star named Canopus has a declination of approximately –52°.

Since the angle is negative, this shows that it is south or below the celestial equator and at 52° south of the celestial equator.

Thus, the star named Caponus is 52° below the celestial equator.

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3 years ago

How will creat thunderstrom

emmasim [6.3K]

Answer:

the air has to be unstable as well as it needs to be moved upwards.

Explanation:

it needs to be moved upwards and also needs to have unstable air.

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2 years ago

A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u

uranmaximum [27]

Answer:

a) $t = \sqrt{\frac{h}{2g}}$

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

$y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{1}}$ : is the initial height = h

$v_{0_{1}}$ : is the initial speed of ball 1 = 0 (it is dropped from rest)

$y = h - \frac{1}{2}gt^{2}$ (1)

Now, for ball 2 we have:

$y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{2}}$ : is the initial height of ball 2 = 0

$y = v_{0_{2}}t - \frac{1}{2}gt^{2}$ (2)

By equating equation (1) and (2) we have:

$h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}$

$t=\frac{h}{v_{0_{2}}}$

Where the initial velocity of the ball 2 is:

$v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh$

Since $v_{f_{2}}^{2}$ = 0 at the maximum height (h):

$v_{0_{2}} = \sqrt{2gh}$

Hence, the time when they pass each other is:

$t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$

b) When they are passing the speed of each one is:

For ball 1:

$v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}$

The minus sign is because ball 1 is going down.

For ball 2:

$v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}$

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

$y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

For ball 2:

$y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!

7 0

3 years ago