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bearhunter [10]
3 years ago
5

What two factors affect the gravitational force between two objects

Physics
1 answer:
Afina-wow [57]3 years ago
4 0
Mass and distance is the correct answer
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Convert gravitational potential energy to kinetic energy.
Digiron [165]

Answer:

1.7 J

Explanation:

The energy carried by a single photon is given by

E=hf

where h is the Planck's constant and f is the frequency of the photon.

The photon of our exercise has a frequency of f=1.7 \cdot 10^{17} Hz, therefore its energy is

E=hf=(6.63 \cdot 10^{-34}Js)(1.7 \cdot 10^{17} Hz)=1.1 \cdot 10^{-16} J

4 0
3 years ago
What can you conclude about this product?
aksik [14]
What’s the question
6 0
4 years ago
Temperature is a measure of the average ____________ energy of an object's particles. light mechanical potential kinetic
Zielflug [23.3K]
Temperature is a measure of the average kinetic energy of the particles of a substance.
Hope this helps! :)
6 0
4 years ago
Read 2 more answers
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
A child is sledding. He starts at the top of the hill with a velocity of zero, and 3 seconds later he is speeding down the hill
Flauer [41]

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

                                             (12 m/s  -  0)   /   (3 sec)  =

                                               12/3  =  <em>4 m/s²</em>

What's a challenge question ?  Have we all passed the event horizon
and been spaghettified without knowing it ?


8 0
3 years ago
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