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ANEK [815]
4 years ago
11

Please help ASAP. There are 3 question it would be a big help if you can answer any.

Physics
2 answers:
sesenic [268]4 years ago
7 0

Answer:

number 1 is falling freely number 2 I think the 2nd option number 3 3rd option

erastova [34]4 years ago
7 0

1).  If you're standing on a scale in an elevator, and the scale reads much less than your real weight, then the elevator must be<em> accelerating downward.</em>

(Think of the extreme limit: If the elevator is in free-fall, accelerating down at 9.8 m/s², then you'll apear to be weightless ... the scale will read zero.)

2).  A 100 kg man is standing in an elevator, accelerating up at 2.0 m/s².

The net force acting on him is F = m·a = (100 kg) x (2.0 m/s²) = 200 N up.

The forces acting on  him are:

-- gravity, m·g = (100 kg) x (9.8 m/s²) = 980 N down

-- elevator floor force . . . EF up

Net force = (980N down) + (EF up)

(200N up) = (980N down) + (EF up)

Add (980N up) to each side:

(200N up + 980N up) = EF up

<em>Elevator force = 1180N up</em>.

Pick the second choice:  <em>1.2 x 10³ N up</em>

<em></em>

3).  My computer is resting on my desk.

<em>There are many forces acting on my computer, and they are all balanced.</em>

That means the NET force acting on my computer is zero.

If the net force were NOT zero, then my computer would be accelerating in the direction of the net force. But it's at rest. So the net force must be zero.

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A solid sphere rolls along a horizontal, smooth surface at a constant linear speed without slipping. What is the ratio between t
dsp73

Answer:

option E

Explanation:

given,

Rotational Kinetic Energy, KE_r = \dfrac{1}{2}I\omega^2

Moment of inertia of the solid,

I = \dfrac{2}{5}MR^2

\omega = \dfrac{V}{R}

now,

KE_r = \dfrac{1}{2}I\omega^2

KE_r = \dfrac{1}{2}\times \dfrac{2}{5}MR^2 (\dfrac{V}{R})^2

KE_r =\dfrac{1}{5}MV^2......(1)

transnational kinetic energy

KE_t =\dfrac{1}{2}MV^2

Total kinetic energy

KE = \dfrac{1}{2}MV^2 + \dfrac{1}{5} MV^2

KE = \dfrac{7}{10}MV^2

ratio of rotational kinetic energy to the total kinetic energy

\dfrac{KE_r}{KE_t}=\dfrac{\dfrac{1}{5}MV^2}{\dfrac{7}{10}MV^2}

\dfrac{KE_r}{KE_t}=\dfrac{2}{7}

hence, the correct answer is option E

4 0
3 years ago
A 0.320 kg ball approaches a bat horizontally with a speed of 14.0 m/s and after getting hit by the bat, the ball moves in the o
katen-ka-za [31]

Answer:

<h2>42.67N</h2>

Explanation:

Step one:

<u>Given </u>

mass m= 0.32kg

intital velocity, u= 14m/s

final velocity v= 22m/s

time= 0.06s

Step two:

<u>Required</u>

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the expression for the force is

F=mΔv/t

F=0.32*(22-14)/0.06

F=(0.32*8)/0.06

F=2.56/0.06

F=42.67N

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4 0
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The speed of a 4-kg ball with a momentum of 12 kg m/s is
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Answer:

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Explanation:

The momentum of an obejct is given by

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v is the velocity of the object

For the ball in this problem, we know:

p = 12 kg m/s is the momentum

m=4 kg is the mass

Therefore, we can re-arrange the equation to find v, the velocity of the ball:

v=\frac{p}{m}=\frac{12 kg m/s}{4 kg}=3 m/s

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