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Semmy [17]
3 years ago
10

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Physics
1 answer:
Ivanshal [37]3 years ago
7 0

If the wavelength increases (gets longer), then the frequency <em>decreases</em>.

(A wave occurs less often.)

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The diagram shows a ball resting at the top of a hill.
WARRIOR [948]

Answer:

a) The potential energy in the system is greatest at X.

Explanation:

Let be X the point where a ball rests at the top of a hill. By applying the Principle of Energy Conservation, the total energy in the physical system remains constant and gravitational potential energy at the top of the hill is equal to the sum of kinetic energy, a lower gravitational energy and dissipated work due to nonconservative forces (friction, dragging).

U_{grav, X} = U_{grav, Y} + K_{Y} + \Delta W_{X \rightarrow Y} = U_{grav,Z}+\Delta W_{X \rightarrow Z}

Conclusions are showed as follows:

a) The potential energy in the system is greatest at X.

b) The kinetic energy is the lowest at X and Z.

c) Total energy remains constant as the ball moves from X to Y.

Hence, the correct answer is A.

7 0
3 years ago
Read 2 more answers
A power plant has two types of transformers. The first type of transformer triples voltage and the second type of transformer re
Stella [2.4K]

Answer:

d . 6 times the secondary turns in the second type of transformer

Explanation:

In transformer , voltage is increased or decreased according to ratio of no of turns in secondary to no of turns in primary coil . The relation is as follows

V₂ / V₁ = n₂/n₁

n₁ and n₂ are no of turns in primary and secondary coil and V₁ and V₂ are voltage in primary and secondary coil.

for first type , Let no of turn in primary = n and no of turn in secondary = n₁

V₂ / V₁ = 3

so

n₁/n  = 3

n = n₁ /3

n₁ = 3n

For second type , let no of turn in secondary = n₂

V₂ / V₁ = .5

n₂/n = .5

n₂ = .5n

n₂ / n₁ = .5n / 3n

n₂ / n₁ = 1 / 6

n₁ = 6n₂

option d is correct .

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3 years ago
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A plane travels at a speed of 205mph in still air. Flying with a tailwind, the plane is clocked over a distance of 1000 miles. F
vaieri [72.5K]

While plane is moving under tailwind condition it took time "t"

so here we will have

t = \frac{d}{v_{net}}

here net speed of the plane will be given as

v_{net} = v + v_w

t = \frac{1000}{205 + v_w}

similarly when it moves under the condition of headwind its net speed is given as

v_{net} = v - v_w

now time taken to cover the distance is 2 hours more

t + 2 = \frac{1000}{205 - v_w}

now solving two equations

\frac{1000}{205 + v_w} + 2 = \frac{1000}{205 - v_w}

solving above for v_w we got

v_w = 40.4 mph

6 0
3 years ago
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