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3 years ago

A car of mass 1167 kg accelerates on a flat highway from 10 m/s to 28.0 m/s. How much work does the car's engine do on the car?

Physics

1 answer:

denis23 [38]3 years ago

7 0

Answer:

Workdone = 465766038 Joules.

Explanation:

<u>Given the following data;</u>

Mass = 1167

Initial velocity = 10m/s

Final velocity =28m/s

To find the workdone;

We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.

Mathematically, it is given by the equation;

W = Kf - Ki

W = ½MVf² - ½MVi²

Substituting into the equation

W = ½(1167)*28² - ½(1167)*10²

W = ½ * 1361889* 784 - ½ * 1361889 * 100

W = 533860488 - 68094450

Workdone = 465766038 Joules.

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Answer:

n = 1.42

Explanation:

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When operated on a household 110.0 V line, typical hair dryers draw about 1650 W of power. The current can be modeled as a long,

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Explanation:

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(c) The magnetic field produced by the dryer at the user's hand is given by :

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alexandr1967 [171]

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3 years ago

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A supertanker filled with oil has a total mass of 6.1 x108 kg. If the dimensions of the ship are those of a rectangular box 300

IrinaVladis [17]

Answer:

The bottom of the sea is 25 m below sea level.

Explanation:

Given data

Mass = 6.1 × $10^{8} \ kg$

$\rho_{sea} = 1020\ \frac{kg}{m^{3} }$

We know that Buoyant force on the tank is equal to gravity force of the tank.

$F_B = F_g$

$(\rho_{Fluid}) (g) (V_{disp}) = m g$

$(\rho_{Fluid}) (V_{disp}) = m$

1020 × $V_{disp}$ = 6.1 × $10^{8}$

$V_{disp}$ = 598039.21 $m^{3}$

We know that

$V_{disp}$ = W × L × H

598039.21 = 300 × 80 × H

H = 25 m

Therefore the bottom of the sea is 25 m below sea level.

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3 years ago

One end of a rope is tied to the handle of a horizontally-oriented and uniform door. a force f is applied to the other end of th

Vlada [557]

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now we will have

$F = \frac{mg}{2}$

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3 years ago