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slega [8]
3 years ago
9

A man stands on a scale and holds a heavy object in his hands. What happens to the scale reading if the man quickly lifts the ob

ject upward and then stops lifting it?
Physics
1 answer:
Novosadov [1.4K]3 years ago
6 0

Answer:

Explanation:

When he accelerates the heavy object up , the reading increases because an extra downward normal force acts on it, then scale reading returns to the same reading as when standing stationary, and then decreases as although he is lifting the heavy object , the acceleration is decreasing ,so the extra upward normal force acts.

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You should select Choice-4 .

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A small spinning asteroid is in a circular orbit around a star, much like the earth's motion around our sun. The asteroid has a
Fudgin [204]

Answer:

Temperature will be 305 K  

Explanation:

We have given The asteroid has a surface area A=7.70m^2

Power absorbed P = 3800 watt

Boltzmann constant \sigma =5.67\times 10^{-8}Wm/K^4

According to Boltzmann rule power radiated is given by

P=\sigma AT^4

3800=5.67\times 10^{-8}\times 7.70\times T^4

T^4=87.0381\times 10^8

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So temperature will be 305 K  

8 0
3 years ago
(8c8p49) A 115g Frisbee is thrown from a point 1.00 m above the ground with a speed of 12.00 m/s. When it has reached a height o
IgorLugansk [536]

Answer:

The work done on the Frisbee is 1.36 J.

Explanation:

Given that,

Mass of Frisbee, m = 115 g = 0.115 kg

Initial speed of Frisbee, u = 12 m/s at a point 1 m above the ground

Final speed of Frisbee , v = 10.9674 m/s when it has reached a height of 2.00 m. Let W is the work done on the Frisbee by its weight. According to work energy theorem, the work done is equal to the change in its kinetic energy. So,

W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times0.115\times\left((10.9674)^{2}-(12)^{2})\right)\\\\W=-1.36\ J

So, the work done on the Frisbee is 1.36 J. Hence, this is the required solution.

3 0
3 years ago
Read 2 more answers
Inside a NASA test vehicle, a 3.50-kg ball is pulled along by a horizontal ideal spring fixed to a friction-free table. The forc
erastova [34]
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3 0
3 years ago
a shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. th
Savatey [412]

Answer:

15.7 m

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The range (horizontal distance) of the projectile is determined only by its horizontal motion.

The horizontal motion is a motion with constant speed, which is equal to the initial horizontal velocity of the object:

v_x = v cos \theta

where

v = 12.0 m/s is the initial velocity

\theta=51.0^{\circ} is the angle between the direction of v and the horizontal

Substituting,

v_x = (12.0 m/s)(cos 51.0^{\circ} )=7.55 m/s

We know that the projectile hits the ground in a time of

t = 2.08 s

so the horizontal distance covered is

d = v_x t = (7.55 m/s)(2.08 s)=15.7 m

8 0
3 years ago
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