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4 years ago

What happens to the demand for butter, if the price of margarine increases? For Economics.

Physics

2 answers:

damaskus [11]4 years ago

8 0

People will buy more butter, and then eventually the margarine would lower and this cycle would keep repeating

Anastaziya [24]4 years ago

7 0

Supply and demand goes up .

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if a freely falling object were somehow equipped with a speedometer on a planet where the acceleration due to gravity is 20 m/s/

Pie

Answer: The speed reading would increase each second by 20m/s

Explanation:

Acceleration is the change in velocity per unit change in time.

Acceleration = velocity/time

∆v = a∆t

Given a = g = 20m/s/s (acceleration due to gravity)

∆t = 1 sec

∆v = 20m/s/s × 1s

∆v = 20m/s

Therefore, the speed reading would increase each second by 20m/s

3 0

4 years ago

What an object at rest stays at rest and an object in motion with the same speed and in the same direction unless acted upon by

GalinKa [24]

Answer:

True.

Explanation:

Newton's First Law of Motion states that every object continues in it's state of rest or of uniform motion in a straight line unless acted upon by an external force.

7 0

3 years ago

Read 2 more answers

A rocket moves away from the surface of earth. As the distance r from the center of the earth the planet increases, what happens

Blizzard [7]

The force of gravity will decrease.
More the distance between the objects, less the force of gravity between them.
They are inversely related.
Hope this help :)

6 0

4 years ago

A 57 kg skier starts from rest at a height of H = 27 m above the end of the ski-jump ramp. As the skier leaves the ramp, his vel

Hatshy [7]

Answer:

(a) $h=5.95m$

(b) h is the same

Explanation:

According to the law of conservation of energy:

$E_i=E_f\\U_i+K_i=U_f+K_f$

The skier starts from rest, so $K_i=0$ and we choose the zero point of potential energy in the end of the ramp, so $U_f=0$ . We calculate the final speed, that is, the speed when the skier leaves the ramp:

$mgH=\frac{mv^2}{2}\\v=\sqrt{2gH}\\v=\sqrt{2(9.8\frac{m}{s^2})(27m)}\\v=23\frac{m}{s}$

Finally, we calculate the maximum height h above the end of the ramp:

$v_f^2=v_i^2-2gh\\$

The initial vertical speed is given by:

$v_i=vsin\theta$

and the final speed is zero, solving for h:

$h=\frac{v_i^2}{2g}\\h=\frac{((23\frac{m}{s})sin(28^\circ))^2}{2(9.8\frac{m}{s^2})}\\h=5.95m$

(b) We can observe that the height reached does not depend on the mass of the skier

3 0

4 years ago

What do a mirror and a magnifying glass have in common? What makes them different?

jeka57 [31]

Answer:

They both have a reflection, but a mirror just reflects anything in front of it, while a magnifying glass helps you see closer through the reflection of something.

Explanation:

6 0

4 years ago