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Oksi-84 [34.3K]
3 years ago
11

A pump is to lift 18 kg of water per minute tthrough a height of 3.6 m. what output rating (watts) should the pump have?

Physics
1 answer:
MrRa [10]3 years ago
7 0
Given: Mass m = 18 Kg;   Height h = 3.6 m;   Time t = 60 s

Required:  Power output  in unit of Watts

Formula: P = mgh/t

               P = (18 Kg)(9.8 m/s²)(3.6 m)/60 s

               P = 10.58 J/s or Watts  


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A car is traveling at a constant speed on the highway. Its tires have a diameter of 65.0 cm and are rolling without sliding or s
lesya692 [45]

Answer:

Explanation:

The car is rolling without slipping so Vcm= R×ω = 0.325×49 = 16

4 0
3 years ago
An archer shoots a 150. gram arrow straight up in the air. (Do not try this at home.) The bow was drawn back 75.0 cm by a 175 N
Dmitry [639]

Answer:

The arrow rise 44.64 m high assuming air friction is negligible

Explanation:

According to the law of conservation of energy

Kinetic energy =  Potential energy at highest

\frac{1}{2}kx^2=mgh----1

We know that F=kx

So,k = \frac{F}{x}\\k=\frac{175}{0.75}

Substitute the value in 1

\frac{1}{2}(\frac{175}{0.75})(0.75)^2=0.15 \times 9.8 \times h\\\frac{\frac{1}{2}(\frac{175}{0.75})(0.75)^2}{0.15 \times 9.8}=h\\44.64 = h

Hence  the arrow rise 44.64 m high assuming air friction is negligible

4 0
3 years ago
If a car is taveling with a speed 6 and comes to a curve in a flat road with radius ???? 13.5 m, what is the minumum value the c
Lemur [1.5K]

Answer:

\mu_s \geq 0.27

Explanation:

The centripetal force acting on the car must be equal to mv²/R, where m is the mass of the car, v its speed and R the radius of the curve. Since the only force acting on the car that is in the direction of the center of the circle is the frictional force, we have by the Newton's Second Law:

f_s=\frac{mv^{2}}{R}

But we know that:

f_s\leq \mu_s N

And the normal force is given by the sum of the forces in the vertical direction:

N-mg=0 \implies N=mg

Finally, we have:

f_s=\frac{mv^{2}}{R}  \leq \mu_s mg\\\\\implies \mu_s\geq \frac{v^{2}}{gR}  \\\\\mu_s\geq \frac{(6\frac{m}{s}) ^{2}}{(9.8\frac{m}{s^{2}})(13.5m) }\\\\\mu_s\geq0.27

So, the minimum value for the coefficient of friction is 0.27.

4 0
3 years ago
A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is
loris [4]

Answer:

The acceleration of the crate is 1.82\ m/s^2.

Explanation:

Given that,

Force, F = 750 N

Mass of the crate, m = 250 kg

The coefficient of friction is 0.12.

We need to find the acceleration of the crate. The net force acting on the crate is given by :

F=ma\\\\F-f=ma

f is frictional force, f=\mu N=\mu mg

F-\mu mg=ma\\\\a=\dfrac{F-\mu mg}{m}\\\\a=\dfrac{750-0.12\times 250\times 9.8}{250}\\\\a=1.82\ m/s^2

So, the acceleration of the crate is 1.82\ m/s^2. Hence, this is the required solution.

4 0
3 years ago
A laser beam enters a 16.5 cm thick glass window at an angle of 58.0° from the normal. The index of refraction of the glass is 1
ivann1987 [24]

Answer:

using Snells law

Oi = angle of incidence = 58.0°

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    nr = index of refraction of glass = 1.47

    c = speed of light in vacuum = 3 x 10^8 m/s

    Or = angle of refraction = ?

ni(sinOi) = nR (sinOr)

ni( sinOi)/ nR = sinOr

arcsin(ni(sin0i))/nR = Or

arcsin( 1.0003(sin58.0)) / 1.47

Or = 35.25°

Explanation:

5 0
3 years ago
Read 2 more answers
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