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4 years ago

How long will it take a 400 W engine to lift a 100 N object up to a height of 80 m?

Physics

2 answers:

Norma-Jean [14]4 years ago

4 0

<h3>Answer:</h3>

20 seconds

<h3>Explanation:</h3>

<u>We are given</u>;

Power of the engine as 400 watts
Force as 100 N
Distance the object is lifted up as 80 m

We are required to determine the time taken.

We need to know that power is the rate of work done

Therefore;

Power = Work done ÷ time

But, work done = Force × distance

Therefore; In this case;

Work done = 100 N × 80 m

= 8000 Joules

Since , Power = Work done ÷ time

Then, time = Work done ÷ Power

Thus;

Time = 8000 J ÷ 400 W

= 20 s

Therefore, the time taken by the engine to lift the object is 20 seconds

Kitty [74]4 years ago

3 0

Answer:

The time will be 20 seconds

Explanation:

The work is:

W=F*d

where:

F=force[N]

d=distance[m]

W=100*80=8000[J]

The time will be given by:

t= Work/power

where:

work=work done by the force[J]

Power=work done in an interval of time [Watts]

t=8000/400

Analysis of units (J / J/s) = [s]

t=20[s]

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A 2.0 kg particle moves in a circle of radius 3.1 m. As you look down on the plane of its orbit, the particle is initially movin

Ghella [55]

Answer

given,

L(t) = 10 - 3.5 t

mass of particle = 2 Kg

radius of the circle = 3.1 m

a) torque

τ = $\dfrac{dL}{dt}$

τ = $\dfrac{d}{dt}(10 - 3.5 t)$

τ = -3.5 N.m

Particle rotates clockwise as i look down the plane. Hence, its angular velocity is downward.

L decreases the angular acceleration upward. so, net torque is upward.

b) Moment of inertia of the particle

I = m R^2

I = 2 x 3.1²

I = 19.22 kg.m²

L = I ω

ω = $\dfrac{L}{I}$

ω = $\dfrac{10 - 3.5 t}{19.22}$

ω = $0.520 - 0.182 t$

A = 0.52 rad/s B = -0.182 rad/s²

5 0

3 years ago

Read 2 more answers

A tuning fork with a frequency of 335 Hz and a tuning fork of unknown frequency produce beats with a frequency of 5.3 when struc

Ganezh [65]

Answer:

Explanation:

Unknown fork frequency is either

335 + 5.3 = 340.3 Hz

335 - 5.3 = 329.7 Hz

After we modify the known fork, the unknown fork frequency equation becomes either

(335 - x) + 8 = 340.3

(335 - x) = 332.3

x = 2.7 Hz

(335 - x) + 8 = 329.7

(335 - x) = 321.7

x = 13.3 Hz

IF the unknown fork frequency was 340.3 Hz,

THEN the 335 Hz fork was detuned to 335 - 2.7 = 332.3 Hz

IF the unknown fork frequency was 329.7 Hz,

THEN the 335 Hz fork was detuned to 335 - 13.3 = 321.7 Hz

3 0

3 years ago

The early workers in spectroscopy (Fraunhofer with the solar spectrum, Bunsen and Kirchhoff with laboratory spectra) discovered

Anestetic [448]

The hot gases produce their own characteristic pattern of spectral lines, which remain fixed as the temperature increases moderately.

<h3><u>Explanation: </u></h3>

A continuous light spectrum emitted by excited atoms of a hot gas with dark spaces in between due to scattered light of specific wavelengths is termed as an atomic spectrum. A hot gas has excited electrons and produces an emission spectrum; the scattered light forming dark bands are called spectral lines.

Fraunhofer closely observed sunlight by expanding the spectrum and a huge number of dark spectral lines were seen. "Robert Bunsen and Gustav Kirchhoff" discovered that when certain chemicals were burnt using a Bunsen burner, atomic spectra with spectral lines were seen. Atomic spectral pattern is thus a unique characteristic of any gas and can be used to independently identify presence of elements.

The spectrum change does not depend greatly on increasing temperatures and hence no significant change is observed in the emitted spectrum with moderate increase in temperature.

8 0

3 years ago

Can somebody help me with this quiz please

mezya [45]

Sure what do u need help with

4 0

3 years ago

Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4

Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

3 0

3 years ago