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lozanna [386]
3 years ago
7

What is an example of velocity?

Physics
1 answer:
Andrei [34K]3 years ago
3 0
Number 2 i am pretty sure
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How can submarines use echolocation to tell how close they are to the bottom of the ocean?
Lapatulllka [165]
To locate a specific target or to determine how close submarines are to the seafloor, they use active and passive sound navigation and ranging (or a SONAR, in simple terms.) It emits pulses of sound waves that travel through the water, reflect off the target and relayed back to the ship. By determining how fast the sound wave travels back, the computers on the sub calculate how far they are from the target.

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3 0
3 years ago
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When a golf ball is hit, it travels at 41 meters per second. The mass of a golf ball is 0.045kg. Calculate the kinetic energy of
Fittoniya [83]

Answer:

  75.645 J

Explanation:

The kinetic energy is related to the mass and velocity by the formula ...

  KE = 1/2mv²

For the given mass of 0.045 kg, and velocity of 41 m/s, the kinetic energy is ...

  KE = 1/2(0.045 kg)(41 m/s)² = 75.645 J

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The unit of energy, joule, is a derived unit equal to 1 kg·m²/s².

4 0
2 years ago
Explain how a projectile might be modified to decrease the air resistance impacting its trajectory.
Vitek1552 [10]

Answer:

Explanation:

Explain how a projectile might be modified to decrease the air resistance impacting its trajectory.

5 0
2 years ago
Two charged objects separated by some distance attract each other. If the charges on both objects are doubled with no change in
Serggg [28]

Answer:

(a) The force between them quadruples

Explanation:

According to coulomb's law, initial force between the two charged objects is given as;

F_1=\frac{Kq_1q_2}{r^2}

where;

k is coulomb's constant

q₁ is the charge on the first object

q₂ is the charge on the second object

r is the distance between the two objects

When the charges on both objects are doubled, then;

q₁ = 2q₁

q₂ = 2q₂

Force between the two charged objects will become

F_2 = \frac{K2q_12q_2}{r^2} =  \frac{4Kq_1q_2}{r^2} = 4(\frac{Kq_1q_2}{r^2}) = 4F_1

Therefore, the force between them quadruples

4 0
3 years ago
How much pressure is created when you apply a
Alekssandra [29.7K]
Pressure
= Force/Area

Area = π(d^2)/4
= π(0.4^2)/4
=0.126 m2
Pressure
= 50/0.126
= 396.825 Pa
5 0
2 years ago
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