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Rasek [7]
3 years ago
8

The battery charger for an mp3 player contains a step-down transformer with a turns ratio of 1:24, so that the voltage of 120 v

available at a wall socket can be used to charge the battery pack or operate the player. what voltage does the secondary coil of the transformer provide? (no response) seenkey 5 v
Physics
1 answer:
miskamm [114]3 years ago
8 0
Considering that we are talking about a stepdown transformer, and a turn ration of 1:24

Then

Vsecondary coil = 120 V / 24 = 5V

(But lets remember that the power must be conserved in the transformer, so the voltage is 24 times less, but the current is 24 times higher)

It provides 5 volts to operate the player or charge the batteries
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Anvisha [2.4K]

Answer:

14.2 m

Explanation:

Using conservation of energy:

PE at top = KE at bottom

mgh = ½ mv²

h = v² / (2g)

h = (16.7 m/s)² / (2 × 9.8 m/s²)

h = 14.2 m

Using kinematics:

Given:

v₀ = 16.7 m/s

v = 0 m/s

a = -9.8 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (16.7 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 14.2 m

7 0
3 years ago
A 403.0-kg copper bar is melted in a smelter. The initial temperature of the copper is 320.0 K. How much heat must the smelter p
aleksklad [387]
The answer is 2.49 x 10^5 KJ. This was obtained (1) use the formula for specific heat to achieve Q or heat then (2) get the energy to melt the copper lastly (3) Subtract both work and the total energy required to completely melt the copper bar is achieved.
7 0
3 years ago
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Work done in taking charge from one point of a conductor to is another point is called ​
Yuliya22 [10]

Answer:

⁸

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6 0
2 years ago
What is the Kinetic Energy of a tennis ball after 40.0cm of freefall?
OlgaM077 [116]
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3 0
3 years ago
The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of
adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = ma_a (0.35)

1.8v^2(1.1) - 1200(9.8)(1.25) = 1200a(0.35)

1.8v^2(1.1) - 14700 = 420 a   ------- equation (1)

F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a             --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

{1.1 * 1200 \ a} - 14700 = 420 a

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900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

5 0
2 years ago
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