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3 years ago

Bill Nye help.. I didn’t get link to vid, hoping y’all have seen it. 20 POINTS!

Physics

1 answer:

Lyrx [107]3 years ago

6 0

Answer:

Potential, Kinetic and Chemical energy.

Explanation:

btw, congratulations for turning into an expert.

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As rotational speed increases, thrust____?

never [62]

Increases exponentially is your correct answer

6 0

3 years ago

Which line represents a stationary object?

sineoko [7]

Answer:

I think its B but I may be wrong

3 0

3 years ago

Read 2 more answers

A boy pushes a cart with a constant velocity of 0.5m/s by applying a force of 60 N. What is the total frictional force acting on

Nutka1998 [239]

Answer:

60N

Explanation:

in this case the minimum amount of force required must be equal to the friction Force. i.e Newton's first law of motion.

therefore the maximum amount of frictional force is equal to the applied force which is 60N.

because of the net force acting on the object is zero the object is in constant motion . i.e equal and opposite force must be applied so that the object is in constant velocity therefore the total frictional force must be 60N

8 0

3 years ago

The question is in the picture

Sedbober [7]

Answer:

e) 120m/s

Explanation:

When the ball reaches its highest point, its velocity becomes zero, meaning

$v_0-gt = 0$ .

where $v_0$ is the initial velocity.

Solving for $t$ we get

$t = \dfrac{v_0}{g}$

which is the time it takes the ball to reach the highest point.

Now, after the ball has reached its highest point, it turns around and falls downwards. After time $t_0$ since it had reached the highest point, the ball has traveled downwards and the velocity $v_f$ it has gained is

$v_f = gt_0$ ,

and we are told that this is twice the initial velocity $v_0$ ; therefore,

$v_f = 2v_0 = gt_0$

which gives

$t_0 = \dfrac{2v_0}{g}.$

Thus, the total time taken to reach velocity $2v_0$ is

$t_{tot} = t+t_0 = \dfrac{v_0}{g}+\dfrac{2v_0}{g}$

$t_{tot} = \dfrac{3v_0}{g}.$

This $t_{tot}$ , we are told, is 36 seconds; therefore,

$36= \dfrac{3v_0}{g},$

and solving for $v_0$ we get:

$v_0 = \dfrac{36g}{3}$

$v_0 = \dfrac{36s(10m/s^2)}{3}$

$\boxed{v_0 = 120m/s}$

which from the options given is choice e.

7 0

3 years ago

Why the greater the depth, the greater the pressure?

Tatiana [17]

Well, the reason behind this is because water depth (I assume you're talking about water/liquid) increases the pressure because of the weight and volume of the water above. I hope this helps! ~Mia

7 0

3 years ago