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Eva8 [605]
3 years ago
6

If electromagnetic radiation a has a lower frequency than electromagnetic radiation b the wavelength of a is

Physics
1 answer:
Jobisdone [24]3 years ago
5 0
Inversely proportional to its frequency. If electromagnetic radiation A has a lower frequency than electromagnetic B, then compared to B, the wavelength of A is...? - equal - shorter - longer - exactly half the length of
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Waves with higher amplitude transfer ____energy.
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Answer:

- Waves with higher amplitude transfer HIGHER energy.

- Waves with higher frequency transfer HIGHER energy.

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A: How far did she travel? <br><br><br><br> B: How long did she take?
allochka39001 [22]
A. 60 miles
B. 5 hours

Unless you are looking for slope, in which case the answer is different
7 0
3 years ago
A wedge with a mechanical advantage of 0.78 is used to raise a house corner from its foundation. If the output force is 7500 N,
Zigmanuir [339]

Answer:

5850

Explanation:

MA = Force out/ Force in

.78 = out/ 7500

.78 times 7500 = input

6 0
3 years ago
What is refraction?
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B- light bends as it passes through an object ( a would be reflect)
4 0
3 years ago
A 91.0-kg hockey player is skating on ice at 5.50 m/s. another hockey player of equal mass, moving at 8.1 m/s in the
never [62]

The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2=  91.0-kg

(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.

(u₂) is the velocity before the collision of hockey player 2=?

a)

Momentum before the collision;

\rm  m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

1237.6 kg m/s² =(2 ×91.0-kg )V

V=6.8 m/sec.

Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

8 0
2 years ago
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