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2 years ago

A pump lifts 400 kg of water per hour a height of 4.5 m .

Physics

1 answer:

nasty-shy [4]2 years ago

6 0

Answer:

Power = Work / Time

P = 400 kg * 9.8 m/s * 4.5 m / 3600 sec = 4.9 J/s = 4.9 Watts

Also, 4.9 Watts / (746 Watts / Horsepower) = .0066 Hp

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A 5.5-kW resistance heater in a water heater runs for 3 hours to raise the water temperature to the desired level. Determine the

Stella [2.4K]

Answer:

16.5 kwh and 59400 kJ.

Explanation:

kWh is a measure of energy that is equivalent to the power in kw times the number of hours the device worked.

In this case, it would be equal to:

$E_{kwh}=5.5kw*3h=16.5kwh$

1 kw also means 1kj of energy spent per second. With this, we calculate the amount of energy in kJ spent by the resistance:

$E_{kJ}=5.5\frac{kJ}{s}*3h*\frac{3600s}{1h} = 59400 kJ$

5 0

3 years ago

Two sound waves have equal displacement amplitudes, but wave 1 has two-thirds the frequency of wave 2. What is the ratio of the

zlopas [31]

Answer:

$\dfrac{I_1}{I_2}=\dfrac{4}{9}$

Explanation:

c = Speed of wave

$\rho$ = Density of medium

A = Area

$\nu$ = Frequency

$\nu_1=\dfrac{2}{3}\nu_2$

Intensity of sound is given by

$I=\dfrac{1}{2}\rho c(A\omega)^2\\\Rightarrow I=\dfrac{1}{2}\rho c(A2\pi \nu)^2$

So,

$I\propto \nu^2$

We get

$\dfrac{I_1}{I_2}=\dfrac{\nu_1^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{\dfrac{2}{3}^2\nu_2^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{4}{9}$

The ratio is $\dfrac{I_1}{I_2}=\dfrac{4}{9}$

8 0

2 years ago

Because the soles of your shoes have cleats, you can exert a forward force of 100 N even on slippery ice. A 10-kg picnic cooler

Brilliant_brown [7]

Answer:

you must throw 3 snowballs

Explanation:

We can solve this exercise using the concepts of conservation of the moment, let's define the system as formed by the refrigerator and all the snowballs. Let's write the moment

Initial. Before bumping that refrigerator

p₀ = n m v₀

Where n is the snowball number

Final. When the refrigerator moves

pf = (n m + M) v

The moment is preserved because the forces during the crash are internal

n m v₀ = (n m + M) v

n m (v₀ - v) = M v

n = M/m v/(vo-v)

Let's look for the initial velocity of the balls, suppose the person throws them with the maximum force if it slides in the snow (F = 100N), let's use the second law and Newton

F = m a

a = F / m

The distance the ball travels from zero speed to maximum speed is the extension of the arm (x = 1 m), let's look kinematically for the speed of the balls when leaving the arm

v₁² = v₀² + 2 a x

v₁² = 0+ 2 (100/1) 1

v₁ = 14.14 m / s

This is the initial speed for the crash

v₀ = v = 14.14 m / s

Let's calculate

n = M/m v/ (v₀-v)

n = 10/1 3 / (14.14 -3)

n = 2.7 balls

you must throw 3 snowballs

7 0

3 years ago

How do you rationalize the tension being used in Tennis Racket strings using the concept of impulse and momentum?

zheka24 [161]

Answer:

The momentum, ΔP, and therefore, kinetic energy given to the ball in a serve is the result of the product of the tension force, 'F', in the string and the time of contact, Δt, between the ball and the string

ΔP = F × Δt

Explanation:

The impulse, ΔP, is the produce of the force, 'F', applied to a body for a given period of time, Δt', that gives motion to the body, and it is equal to the change of momentum of the body

ΔP = F × Δt

The momentum, 'P', of a body is the product of the mass, 'm', of the body and its velocity, 'v'

P = m × v

Tension is the axial pulling force of a string

T = Axial Force, F $_{axial}$

The tension used in Tennis Racket strings is between 40 to 65 lbs.

When high tension is used in the string, the string is taut, and the contact duration between the Racket string and the ball is minimal, and the player needs to use more force to obtain a high momentum, and therefore, energy in the ball, which reduces control, and increase stress, as force is more emphasized

When low tension is used in the string, the Tennis Racket strings are more elastic. During a serve, the ball pushes the strings further back into the racket, such that the ball spends more time in contact with the string, (Δt is larger), and therefore, the impulse, F·Δt = ΔP, given to the ball is larger, therefore, the ball has a larger change in momentum, and therefore more energy in the intended direction.

However, a very slackened string will increase the increase area and time (large Δt) of contact of the ball and the racket such that the force given to the ball, F = ΔP/(large Δt) is reduced and therefore reduce the likelihood of gaining points from a serve against an opponent with a much forceful return of a serve.

3 0

2 years ago

A bullet is fired from a gun at 45° angle to the horizontal with a velocity of 500 m/s. Find the

rjkz [21]

answer :

D. 6370.92 m

Explanation:

pls refer to the attachment...

7 0

2 years ago