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Mazyrski [523]
3 years ago
12

The weight of an object is the product of its mass m, and acceleration of gravity , g . If an objects mass is m=10 . Kg what is

its weight?
Physics
1 answer:
zzz [600]3 years ago
8 0

Its weight is 10g .  The 'g' is the acceleration of gravity
wherever the object happens to be at the time.

On Mars, g=3.71 m/s², the object weighs 37.1 newtons.

On the Moon, g=1.62 m/s², the object weights16.2 newtons.

On Earth, g=9.81 m/s², the object weighs 98.1 newtons.
.
.
etc.

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What can you conclude about the pendulum at the instant shown by the graph below (just past 2 s)?
Semenov [28]
Below are the chocies 
A. The pendulum is at its highest point. 
<span>B. The pendulum is 0.9 m above its lowest point. </span>
<span>C. The pendulum is between its highest and lowest points. </span>
<span>D. The pendulum is at its lowest point. 
</span>
I think the answer is C.. 

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
7 0
4 years ago
Choose the correct statement regarding the behavior of water. a. The water phase with the highest heat capacity is gas vapor. b.
e-lub [12.9K]

Answer:

D. The ice-to-liquid phase change of water requires less energy than the liquid-to-vapor phase.

Explanation:

In the phase change from liquid to gas, the bonds between atoms are completely broken. The phase change from liquid to gas requires more energy because the bonds must be completely broken for it to take place, rather than just loosened as in the phase change of solid to liquid.

Phase changes can have a strong stabilizing effect on temperatures that are not near the melting and boiling points, since evaporation and condensation occur even at temperatures below the boiling point.

More energy is required to evaporate water below the boiling point than at the boiling point, because the kinetic energy of water molecules at temperatures below 100°C is less than that at 100°C, so less energy is available from random thermal motions.

5 0
3 years ago
A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm
bogdanovich [222]

Answer:

A) 2.75 m/s  B) 0.1911 m    C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = \sqrt\frac{k}{M}

                                                       = \sqrt\frac{21}{0.1}

                                                       = 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic  Motion is given as:

                                x = A sin (\omega t + \phi)\\

Differentiating this to find speed of the block immediately before the collision:

                    v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\

As bullet strikes at equilibrium position so,

                                  φ = 0

                                   t= 2nπ

                             ⇒ cos (ω₀t + φ) = 1

                             ⇒ v= A\omega_{o}

                                       v=(.19)(14.49)\\v= 2.75 ms^{-1}

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

                              x= Bsin(\omega t + \phi)

To find B, consider law of conservation of energy

K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2}  \\P.E = \frac{1}{2} kB^{2}

\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec

5 0
4 years ago
How much force is required to accelerate 3 kg at 3 m/sec^2
bogdanovich [222]

Answer:

F=?

a=3m/s2

m=3kg

F=ma=[3][3]=9 N

Explanation:

7 0
3 years ago
Formula for finding displacement
FrozenT [24]
Displacement = (final position) - (initial position) = change in position.

HOPE IT HELPS YOU
5 0
3 years ago
Read 2 more answers
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