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2 years ago

I've had several mental breakdowns please help me

Physics

1 answer:

sesenic [268]2 years ago

4 0

Answer:

The efficiency of the system is 63.7 %

Explanation:

Given;

input power of the motor, = 1.5 kW = 1,500 W

mass of the car lifted, m = 1300 kg

height through which the car was lifted, h = 1.8 m

time, t = 24 s

The output power of the motor is calculated as;

Output Power = F x v

= (mg) x (d/t)

= (1300 x 9.8) x (1.8 / 24)

= 12,740 x 0.075

= 955.5 W

The efficiency of the system is calculated as;

$E = \frac{0utput \ power}{1nput \ power} \times \ 100\%\\\\E = \frac{955.5}{1500} \times \ 100\% \\\\E = 63.7 \ \%$

The correct answer is 63.7%

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If the mass of a planet is 0.231 mE and its radius is 0.528 rE, estimate the gravitational field g at the surface of the planet.

crimeas [40]

Answer:

$8.1 m/s^2$

Explanation:

The strength of the gravitational field at the surface of a planet is given by

$g=\frac{GM}{R^2}$ (1)

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

For the Earth:

$g_E = \frac{GM_E}{R_E^2}=9.8 m/s^2$

For the unknown planet,

$M_X = 0.231 M_E\\R_X = 0.528 R_E$

Substituting into the eq.(1), we find the gravitational acceleration of planet X relative to that of the Earth:

$g_X = \frac{GM_X}{R_X^2}=\frac{G(0.231M_E)}{(0.528R_E)^2}=\frac{0.231}{0.528^2}(\frac{GM_E}{R_E^2})=0.829 g_E$

And substituting g = 9.8 m/s^2,

$g_X = 0.829(9.8)=8.1 m/s^2$

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3 years ago

A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on

g100num [7]

Answer:

(a) 81.54 N

(b) 570.75 J

(d) 0 J, 0 J

(e) 0 J

Explanation:

mass of crate, m = 32 kg

distance, s = 7 m

coefficient of friction = 0.26

(a) As it is moving with constant velocity so the force applied is equal to the friction force.

F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N

(b) The work done on the crate

W = F x s = 81.54 x 7 = 570.75 J

W' = - W = - 570.75 J

(d) Work done by the normal force

W'' = m g cos 90 = 0 J

Work done by the gravity

Wg = m g cos 90 = 0 J

(e) The total work done is

Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J

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3 years ago

An astronaut on a small planet wishes to measure the local value of g by timing pulses traveling down a wire which has a large o

Alekssandra [29.7K]

Answer:

$0.53m/s^2$

Explanation:

We are given that

Mass of wire=m=3.6 g= $3.6\times 10^{-3} kg$

1 kg=1000 g

Length of wire=l=1.6 m

Mass of object=m'=3 kg

Time,t=60.1 ms= $60.1\times 10^{-3} s$

$1 ms=10^{-3} s$

Speed,v= $\frac{distance}{time}=\frac{1.6}{60.1\times 10^{-3}}=26.62 m/s$

$g=\frac{v^2m}{m'l}$

Using the formula

$g=\frac{(26.62)^2\times 3.6\times 10^{-3}}{3\times 1.6}=0.53m/s^2$

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3 years ago

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Hallar el valor numerico de 24m2n3p

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What is the angular speed of (a) the second hand, (b) the minute hand, and (c) the hour hand of a smoothly running analog watch?

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3 years ago