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vovangra [49]
4 years ago
7

How much work is done when a 48-kg stack of books are lifted 2m with a net force of 25 N ?

Physics
1 answer:
alexdok [17]4 years ago
7 0
Work=Force x Distance
2x25=50
The correct answer is C
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Two objects, one having three times the mass of the other, are dropped from the same height in a vacuum. At the end of their fal
gogolik [260]

Answer:

Two objects will have the equal velocities but the forces on both of them will not be equal. The equal velocities of these objects are due to their equal acceleration.

Explanation:

From the newton's equation

v^{2} -u^{2} = 2as

so here we can say that velocity does not depends on the mass.

The acceleration of both objects will be same but not the forces because

F = Ma

As the force is depending on the mass so it will not be the same for both objects.

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Chromosomes are a form of blood cells.
Inessa05 [86]

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false im just trying to get it if you'd like to give it to me

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What is the best way to increase the solubility of something?
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On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
irinina [24]

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

3 0
3 years ago
Someone who is good at physics please help me out! i’ll give out brainliest answers
alexgriva [62]

Answer:

450m

Explanation:

You would use the equation x-x0=0.5(v0+v)t as you have the time and velocities. x-x0(Change in position/displacement)=0.5(0+30)30. The distance will be 450m.

6 0
3 years ago
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