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2 years ago

A car engine applies a force of 65,000 N, how much work is done by the engine as it pushed a car a distance of 75 m?

Physics

1 answer:

kupik [55]2 years ago

3 0

Answer:

$workdone = force \times distance \\ = 65000 \times 75 \\ = 4,875,000 \: J$

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A uniformly charged ring of radius 10.0 cm has a total charge of 71.0 μC. Find the electric field on the axis of the ring at the

oee [108]

Answer:

General Expression: E = kql/(l² + r²)^(3/2)

(a) 6.3 MN/C

(b) 22.8 MN/C

(d) 0.63 MN/C

Explanation:

The general expression for electric field along axis of a uniformly charged ring is:

E = kqL/(L² + r²)^(3/2)

where,

E = Electric Field Strength = ?

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q = Total Charge = 71 μC = 71 x 10⁻⁶ C

L = Distance from center on axis

r = radius of ring = 10 cm = 0.1 m

(a)

L = 1 cm = 0.01 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.01 m)/[(0.01 m)² + (0.1 m)²]^(3/2)

E = (6390 N.m³/C)/(0.00101 m³)

E = 6.3 x 10⁶ N/C = 6.3 MN/C

(b)

L = 5 cm = 0.05 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.05 m)/[(0.05 m)² + (0.1 m)²]^(3/2)

E = (31950 N.m³/C)/(0.00139 m³)

E = 22.8 x 10⁶ N/C = 27.4 MN/C

(c)

L = 30 cm = 0.3 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.3 m)/[(0.3 m)² + (0.1 m)²]^(3/2)

E = (191700 N.m³/C)/(0.03162 m³)

E = 6.1 x 10⁶ N/C = 6.1 MN/C

(d)

L = 100 cm = 1 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(1 m)/[(1 m)² + (0.1 m)²]^(3/2)

E = (639000 N.m³/C)/(1.015 m³)

E = 0.63 x 10⁶ N/C = 0.63 MN/C

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2 years ago

What is the output voltage from a step up transformer with 25 turns in the primary coil and 75 turns in the secondary coil if th

Greeley [361]

Answer:

360 V

Explanation:

We can use the transformer equation:

$\frac{V_p}{N_p}=\frac{V_s}{N_s}$

where

$V_p$ is the primary (input) voltage

$N_p$ is the number of turns in the primary coil

$V_s$ is the secondary (output) voltage

$N_s$ is the number of turns in the secondary coil

In this problem,

$V_p = 120 V\\N_p = 25\\N_s = 75$

Substituting into the equation, we can find the output voltage:

$V_s = N_s \frac{V_p}{N_p}=(75)\frac{120}{25}=360 V$

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Answer:

Every object in space exerts a gravitational pull on every other

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What are the four forms that water takes when it returns to earth from the clouds?

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Air is composed of many gases, while water is composed of a pure substance. Which best compares air and water?

natta225 [31]

Answer: c

Explanation:

C Air is a compound of two or more components that keep their own identifying properties, while water is composed of mixtures that combine to form a compound.

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1 year ago

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