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Olenka [21]
2 years ago
12

Calculate the accleration of a 25-Kg object that moved with force of 300 N

Physics
1 answer:
inysia [295]2 years ago
4 0
F=ma
a=F/m=300/25=12 m/s^2
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The linear expansivity of metal P is twice that of another metal Q. When these materials are heated through the same temperature
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It’s p as it’s to the 19th century
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3 years ago
A hailstone traveling with a velocity of 43 meters/second comes to a virtual stop 0.28 seconds after hitting water. What is the
lisabon 2012 [21]

Magnitude of acceleration = (change in speed) / (time for the change) .

Change in speed = (ending speed) - (starting speed)

                            =       zero            - (43 m/s)

                            =          -43 m/s .

Magnitude of acceleration = (-43 m/sec) / (0.28 sec)

                                          =  (-43 / 0.28)  (m/sec) / sec

                                          =        153.57...  m/s²

                                          =        1.5...  x 10²  m/s²  .

5 0
3 years ago
Read 2 more answers
Help it’s due tomorrow
Sati [7]

Answer:

A., the variables have a direct relationship.

Explanation:

As K rises, L rises.

It's not B. because one isn't rising as the other is lowering.

It's not C. because undefined would be a vertical line.

4 0
3 years ago
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

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2 years ago
Before being engulfed, matter that is pulled into a black hole should become very hot and emit _____.
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4 0
3 years ago
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