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aleksley [76]
2 years ago
5

What is the horse power of an electric motor which can do by 1250 joule of work in 5 seconds​

Physics
1 answer:
Ksju [112]2 years ago
5 0
1250 J in 5 sec= 250 Joule(s) per second (1250/5 0

250 Joules per second = 250 Watts ( 1J/s = 1 Watt per definition)

250 Watts output = 250/0.65 efficiency = 384 Watts input

1 Horsepower = 732 Watts

Motors 1 Horsepower and under are made in certain step sizes like

3/4 , 1/2 , 1/3, 1/4, 1/16 1/20 of a Horsepower.

3/4 Horsepower is 549 Watts

1/2 Horsepower is 366 Watts

so you need to 3/4 horsepower motor to achieve 1250 J of work in 5 seconds.
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denpristay [2]

Answer:

2.32 s

Explanation:

Using the equation of motion,

s = ut+g't²/2............................ Equation 1

Where s = distance, u = initial velocity, g' = acceleration due to gravity of  the moon, t = time.

Note: Since Onur drops the basket ball from a height, u = 0 m/s

Then,

s = g't²/2

make t the subject of the equation,

t = √(2s/g')...................... Equation 2

Given: s = 10 m, g' = 3.7 m/s²

Substitute this value into equation 2

t = √(2×10/3.7)

t = √(20/3.7)

t = √(5.405)

t = 2.32 s.

4 0
3 years ago
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Work out the kinetic energy of a 2.5 kg remote-controlled car that is moving at 2 m/s.
lbvjy [14]

Answer: 5 joules

Explanation:

mass=m=2.5kg

Velocity=v=2m/s

Kinetic energy=ke

ke=(m x v x v)/2

ke=(2.5 x 2 x 2)/2

Ke=10/2

Ke=5

Kinetic energy=5 joules

8 0
3 years ago
Suzy drives 10 miles north then drives 24 miles west. What is her total distance traveled?
anzhelika [568]

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34

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3 years ago
The electric field in a certain region of Earth’s atmo- sphere is directed vertically down. At an altitude of 300 m the field ha
I am Lyosha [343]

Answer:

Net charge contained in the cubeq= 3.536×10^-6C

Explanation:

Formular for total flux in a cube is given as:

Total flux= E300Acos(180) + E200Acos(0)

Where A is crossectional area

Total flux= A(E200-E300)

Total flux= q/Eo

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3 years ago
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At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 V and 16.2 V/m
son4ous [18]

Answer:

A. 30.7cm

B. 1.7*10^{-10}C

C. The electric field is directed away from the point of charge

Explanation:

A.

 because, E=\frac{v}{d} \\\\d= \frac{4.98}{16.2}\\\\ d = 0.307m\\\\d = 30.7 cm

B.

Considering Gauss's law

EA = \frac{Q}{e}\\\\ where, e = pertittivity. space= 8.85* 10^{-12} Fm^{-1} \\\\A = surface. area. with.radius 0.307m\\Q= eEA = (8.85*10^{-12})(16.2)(4\pi)(0.307)^{2}\\\\= 1.7*10^{-10}C

C. The electric field directed away from the point of charge when the charge is positive.

3 0
3 years ago
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