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Sauron [17]
3 years ago
10

A student looks at ocean waves coming into the beach. An ocean wave with more energy will A) have a greater height. B) have a gr

eater period. C) travel toward the beach faster. D) strike the beach with greater frequency.
Physics
2 answers:
attashe74 [19]3 years ago
3 0

It would be A: Have a greater height.


The higher the wave the higher the energy!

Nitella [24]3 years ago
3 0

Answer:

An ocean wave with more energy will have a greater height

Explanation:

A student looks at ocean waves coming into the beach. The energy of wave depends on the amplitude of wave. Amplitude of a wave is the distance from equilibrium position to the crest or trough of the wave.

So, an ocean wave with more energy will have a greater height. The height of wave is equal to twice the amplitude of wave. Hence, the correct option is (A).

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3. A roller-coaster car is moving over the top of a hill. At the instant it is on the hill top, it has a kinetic energy of E and
aleksandrvk [35]

Answer:

The correct option is (b).

Explanation:

Given that,

At the top of the hill, the kinetic energy is E and the gravitational potential energy is 3E.

We need to find the kinetic energy of the car on the ground.

We know that,

Mechanical energy = kinetic energy + potential energy

According to the law of conservation of energy, the total mechanical energy is conserved.

It means, when it coasts down to ground level, the kinetic energy is same as that on the top of the hill. Hence, the required kinetic energy on the ground is equal to 3E.

8 0
3 years ago
If the effects of heat and friction are ignored, the amount of work output is always _______ the amount of work input, even when
vodomira [7]

Answer:

A

Explanation:

3 0
3 years ago
An ideal monatomic gas initially has a temperature of T and a pressure of p. It is to expand from volume V1 to volume V2. If the
yawa3891 [41]

Answer:

Isothermal :   P2 = ( P1V1 / V2 ) ,  work-done pdv = nRT * In( \frac{V2}{v1} )

Adiabatic : : P2 = \frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }  , work-done =

W = (3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)

Explanation:

initial temperature : T

Pressure : P

initial volume : V1

Final volume : V2

A) If expansion was isothermal calculate final pressure and work-done

we use the gas laws

= PIVI = P2V2

Hence : P2 = ( P1V1 / V2 )

work-done :

pdv = nRT * In( \frac{V2}{v1} )

B) If the expansion was Adiabatic show the Final pressure and work-done

final pressure

P1V1^y = P2V2^y

where y = 5/3

hence : P2 = \frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }

Work-done

W = (3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)

Where    T2 = T1V1^(2/3)/V2^(2/3)

3 0
3 years ago
A small charged bead has a mass of 1.0 g. It is held in a uniform electric field of magnitude E = 200,000 N/C, directed upward.
Art [367]

Answer:

10^-7 C

Explanation:

m = 1 g = 10^-3 kg, E = 200,000 N/C, a = 20 m/s^2, u = 0

Let q be the charge on bead

Force = m a = q E

a = q E / m

q = m a / E = (10^-3 x 20) / 200000 = 10^-7 C

4 0
3 years ago
Which of the following is a Nobel gas?<br> Ba2+<br> OA<br> OB. Ci<br> Kr<br> Ос.<br> Ca<br> D.
stepan [7]

Answer:

Kr

Explanation:

the element that is in group 18 is noble gasses. the elements are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og).

5 0
3 years ago
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