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Zina [86]
4 years ago
10

A box weighing 100 n is pushed up an incline plane that is 5 m long. It takes a force of 75 n to push it to the top, which has a

height of 3m. Find the efficiency
Physics
1 answer:
SVETLANKA909090 [29]4 years ago
3 0

Answer:

0.8

Explanation:

Work = force × distance

W = Fd

W = 75 N × 5 m

W = 375 J

Potential energy = weight × height

PE = mgh

PE = 100 N × 3 m

PE = 300 J

Efficiency = energy out / energy in

e = 300 J / 375 J

e = 0.8

The efficiency is 0.8, or 80%.

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Radiant heat makes it impossible to stand close to a hot lava flow. Calculate the rate of heat loss by radiation from 1.00 m^2 o
myrzilka [38]

Answer:

259.274 kW

Explanation:

Given:

Area of the lava, A = 1.00 m²

Temperature of the surrounding, T₁ = 30.0° C = 303 k

Temperature of the lava, T₂ = 1190° C = 1463 K

emissivity, e = 1

Now,

from the Stefan-Boltzmann law of radiation the rate of heat loss is given as,

u = σeA(T₂⁴ - T₁⁴)

where,

u = rate of heat loss

σ = Stefan-Boltzmann constant =  5.67 × 10⁻⁸ W/m²∙K⁴

on substituting the respective values, we get

u = 5.67 × 10⁻⁸ × 1 × 1 × (1463⁴ - 303⁴)

or

u = 259274.957 W

or

u = 259.274 kW

7 0
3 years ago
Suppose your car accelerates from rest to 9 m/s in 2 s. Assume that the acceleration is constant in this time interval. A. What
Stolb23 [73]

Answer :

(a) The acceleration of the car is, 4.5m/s^2

(b) The distance covered by the car is, 9 m

Explanation :  

By the 1st equation of motion,

v=u+at ...........(1)

where,

v = final velocity = 9 m/s

u = initial velocity  = 0 m/s

t = time = 2 s

a = acceleration of the car = ?

Now put all the given values in the above equation 1, we get:

9m/s=0m/s+a\times (2s)

a=4.5m/s^2

The acceleration of the car is, 4.5m/s^2

By the 2nd equation of motion,

s=ut+\frac{1}{2}at^2 ...........(2)

where,

s = distance covered by the car = ?

u = initial velocity  = 0 m/s

t = time = 2 s

a = acceleration  of the car = 4.5m/s^2

Now put all the given values in the above equation 2, we get:

s=(0m/s)\times (2s)+\frac{1}{2}\times (4.5m/s^2)\times (2s)^2

By solving the term, we get:

s=9m

The distance covered by the car is, 9 m

3 0
3 years ago
A 75.0 kg man pushes on a 500,000 kg wall for 250 s but it does not move.
TiliK225 [7]

Answer:

he does no work on the wall coz wall didn't move

the wall doesn't move so the energy that he use will be wasted

the force that he put on the wall is also zero since the wall didn't move

8 0
3 years ago
Planet-X has a mass of 3.42×1024 kg and a radius of 8450 km. What is the First Cosmic Speed i.e. the speed of a satellite on a l
tino4ka555 [31]

Answer:

First cosmic speed = 5195.74m/s

Second cosmic speed = 7346.05m/s

The raduis of the synchronous 0rbit of satellite is 2.80×10^7m

Explanation:

The first cosmic speed Is determined using the Orbital speed equation which is given by:

V = Sqrt(GM/r)

Where G = gravitational constant = 6.67 ×10^-11

M = Mass of planet

r = radius of the planet

V = Sqrt (6.67×10^-11)(3.42×10^24)/(8450×10^3)

V = Sqrt (2.28×10^14)/(8450×10^3)

V = Sqrt ( 26995739.64)

V = 5195.75m/s

The second cosmic speed is given by :

V = Sqrt(2 × GM)/r

V = Sqrt (2 × (6.67×10^-11)(3.42×10^24)/(8450×10^3)

V = Sqrt( 4.5×10^14)/ (8450×10^3)

V = Sqrt(53964497.04)

V = 7346.05m/s

The raduis of the synchronous orbit if the satellite around the planet is given by:

r = Cuberoot(T^2GM/4 pi r^2where T is the period of rotation of the planet in second

Given :

T = 17.1 hours converting to seconds

T = 17.1 ×60 ×60 = 61560 seconds

Substituting into the equation

r = Cuberoot ([(61560)^2×(6.67×10^-11)(3.42×10^24)/ (4 ×3.142×r^2)]

r = 2.80×10^7m

5 0
3 years ago
Read 2 more answers
Braking distance is the distance a car travels from the time a person applies the brakes to when the car comes to a complete sto
Dvinal [7]
The kinetic energy of an object increases as its velocity increases. Moreover, we see that the braking distance of the car increases as its velocity increases. Therefore, the breaking distance increases when the kinetic energy of the car is higher. This is evident in the graph where when the velocity is lower, the kinetic energy is lower and the braking distance is less.
6 0
4 years ago
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