2 years ago

Waves that move through space are called

Physics

2 answers:

egoroff_w [7]2 years ago

8 0

Answer:

Electromagnetic waves are waves that can travel through matter or through empty space

lawyer [7]2 years ago

6 0

Electromagnetic waves

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Since when was the light we see now emanating from the quasar? Note that the distance between the Earth and the quasar is 598 Mp

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2 years ago

The type of energy that depends on position is called

attashe74 [19]

The type of energy that depends on position is called
kinetic energy

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3 years ago

The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge

lesya [120]

Answer:

Loss, $\Delta E=-10.63\ J$

Explanation:

Given that,

Mass of particle 1, $m_1=m =0.66\ kg$

Mass of particle 2, $m_2=7.4m =4.884\ kg$

Speed of particle 1, $v_1=2v_o=2\times 6=12\ m/s$

Speed of particle 2, $v_2=v_o=6\ m/s$

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

$m_1v_1+m_2v_2=(m_1+m_2)V$

$V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}$

$V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}$

V = 6.71 m/s

Initial kinetic energy before the collision,

$K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)$

$K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)$

$K_i=135.43\ J$

Final kinetic energy after the collision,

$K_f=\dfrac{1}{2}(m_1+m_2)V^2$

$K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2$

$K_f=124.80\ J$

Lost in kinetic energy,

$\Delta K=K_f-K_i$

$\Delta K=124.80-135.43$

$\Delta E=-10.63\ J$

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

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3 years ago

As players speed up, their kinetic energy will?

igomit [66]

Kinetic energy is proportional to the square of the speed. So when anything or anybody speeds up, its kinetic energy increases.

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3 years ago

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A jumbo jet of mass 100,000 kg during takeoff experience a thrust for each of its four engines of 50,000 N. producing an acceler

PtichkaEL [24]

Answer:

F=MA

5OOOO=100000×A

50000/100000=A

A=5/10 M/ S SQ

A=0.5 M/S SQ.

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3 years ago