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Dvinal [7]
3 years ago
6

All forces on the bullets cancel so that the net force on a bullet is zero, which means the bullet has zero acceleration and is

in a state known as _____.
Physics
1 answer:
andre [41]3 years ago
7 0
The acceleration is defined as the rate of change of velocity.
So, if the acceleration is zero, this means that the rate of change of velocity is zero, which also means that the body is moving with constant velocity.
Since we are given that the net forces acting on the body is zero, this means that the body is at equilibrium

Based on this:
<span>All forces on the bullets cancel so that the net force on a bullet is zero, which means the bullet has zero acceleration and is in a state known as equilibrium.

Note that if this constant velocity is equal to zero, then the body would be at rest (not moving)</span>
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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
A 1900kg airplane is flying at an altitude of 510 above the ground. What is the gravitational potential energy in Joules?
Natalija [7]

Answer: 9496200 joules

Explanation:

Gravitational potential energy, GPE is the energy possessed by the moving plane since it moves against gravity.

Thus, GPE = Mass m x Acceleration due to gravity g x Height h

Since Mass = 1900kg

g = 9.8m/s^2

h = 510 metres (units of height is metres)

Thus, GPE = 1900kg x 9.8m/s^2 x 510m

GPE = 9496200 joules

Thus, the gravitational potential energy of the airplane is 9496200 joules

8 0
3 years ago
In this relationship, the barnacles derive benefit. There is not benefit or harm to the whale. This relationship is an example o
klio [65]
Commensalism: only one species benefits while the other is neither helped nor
harmed
so the relationship is Commensalism
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4 years ago
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Raphael refers to a wave by noting its wavelength. lucinda refers to a wave by noting its frequency. which student is correct an
blsea [12.9K]
They are both right because you can note both things, I mean Raphael and Lucinda, both has a right statement or explanation about the wave. Wave by nothing is both for its wavelength and for its frequency. So Raphael and Lucinda are both correct because you can note both wavelength and frequency.
7 0
4 years ago
Which of the following best describes chemical weathering?
dexar [7]

Answer:

<h3>ir id n becuswe ig gyst makres sensre</h3>

Explanation:

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3 years ago
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