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lukranit [14]
2 years ago
15

Angelina jumps off a stool. As she is falling, the Earth’s gravitational force on her is larger in magnitude than the gravitatio

nal force she is exerting on the earth. 1. False 2. This may happen in many cases, but not in every case. 3. True 4. None of these 5. This may happen in a few cases, but it is very rare.
Physics
2 answers:
valentinak56 [21]2 years ago
8 0
The answer is True. The amount force exerted by any object is directly proportional to its mass. This means that our planet is exerting more gravitational force to Angelina, and Angelina is also exerting a gravitational force on our planet directly proportional to her mass. Angelina is actually falling towards the center of the earth,and also our planet is also moving towards Angelina, but it seems negligible with respect to Angelina.Our Sun is so massive that it held our planet in its orbit because of its gravitational force.
STatiana [176]2 years ago
5 0

Answer:

1. False

Explanation:

As per law of gravitational attraction force we know that

F = \frac{Gm_1m_2}{r^2}

where we know that

m_1,m_2 = masses of two point mass

r = distance between two masses

so here we know that as the force does not depend on individual mass but it will depend on the product of two masses

so here the gravitational force is same in magnitude on both the bodies

so when Angelina jumps off the stool the gravitational force on Earth and on Angelina must be of same magnitude

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A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t
Semenov [28]

Answer:

a. I=2.77x10^{-8} kg*m^2

b. K=4.37 x10^{-6} N*m

Explanation:

The inertia can be find using

a.

I = m*r^2

m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg

r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m

I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2

I=2.77x10^{-8} kg*m^2

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

T=2\pi *\sqrt{\frac{I}{K}}

Solve to K'

K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}

K=4.37 x10^{-6} N*m

3 0
2 years ago
A sailboat starts from rest and accelerates at a rate of 0.21 m/s^2 over a distance of 280 m. find the magnitude of the boat's f
sasho [114]

We use the kinematic equations,

v=u+at                                          (A)

S= ut + \frac{1}{2} at^2                  (B)

Here, u is initial velocity, v is final velocity, a is acceleration and t is time.

Given,  u=0, a=0.21 \ m/s^2 and s= 280 m.

Substituting these values in equation (B), we get

280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s.

Therefore from equation (A),

v = 0 + (0.21) \times (51.63 s)= 10.84 \ m/s

Thus, the magnitude of the boat's final velocity is 10.84 m/s and the time taken by boat to travel the distance 280 m is 51.63 s



8 0
2 years ago
Three balls are in water. Ball 1 floats, with half of it exposed above the water level. Ball 2, with a density less than the den
Ymorist [56]

Answer:

The magnitude of buoyancy force is equal to that of ball's weight.

Explanation:

Ball 1 is floating on water. Weight of ball 1 is Fg=m1g  is acting vertically downward

Force of buoyancy FB = ρVdisg is acting vertically upward.

Net force acting on the ball is zero, FB=Fg

Answer

The magnitude of buoyancy force is equal to that of ball's weight.

4 0
2 years ago
Read 2 more answers
A 4kg block is sliding on a horizontal frictionless floor at a speed of 2.5ms and runs into a horizontal spring. The spring has
alexgriva [62]

Answer:

Explanation:

Given

mas of block m=4\ kg

speed of block v=2.5\ m/s

spring constant k=30\ N-m

As the mass collides with the spring its kinetic energy is converted to the Elastic Potential energy of the spring

\frac{1}{2}mv^2=\frac{1}{2}kx^2

x=v\sqrt{\frac{m}{k}}

x=2.5\times \sqrt{\frac{4}{30}}

x=0.912\ m

8 0
2 years ago
Read 2 more answers
A physics student shoves a 0.50-kg block from the bottom of a frictionless 30.0° inclined plane. The student performs 4.0 j of w
Musya8 [376]

Answer:1.63 m

Explanation:

Given

mass of block m=0.5 kg

inclination \theta =30^{\circ}

Amount of work done W=4 J

block slides a distance s along the Plane

Work done =change in Potential Energy

Increase in height of block is s\sin \theta

Change in Potential Energy =mg(\sin \theta -0)

\Delta P.E.=0.5\times 9.8\times s\sin 30

4=0.5\times 9.8\times s\sin 30

s=\frac{4}{2.45}      

s=1.63 m

5 0
2 years ago
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