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Tems11 [23]
3 years ago
13

If it takes 600 N to move a box 5 meters, how much work is done on the box?

Physics
2 answers:
erastovalidia [21]3 years ago
8 0
C it is 600 x 5 = 3000 J hope this helped!
Anika [276]3 years ago
3 0

Answer:

<h2>3000 J</h2>

Option C is the correct option.

Explanation:

Given,

Force = 600 N

Distance = 5 meters

Work = ?

Now,

Work = Force \times distance

= 600  \times 5

Calculate the product

= 3000 \: Joule

Hope this helps...

Good luck on your assignment..

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Monochromatic light of a given wavelength is incident on a metal surface. However, no photoelectrons are emitted. If electrons a
mrs_skeptik [129]

Answer:

Light of a shorter wavelength should be used.

Explanation:

This is studied in the phenomenon called photoelectric effect, in which light is able to release electrons from a metal, said electrons are called photoelectrons .

The experiments that have been carried out show that <u>increasing  or decreasing the intensity of the light will not cause the photoelectrons to be emitted</u>, what will cause the photoelectrons to be emitted is to increase the frequency of the incident light.

And a higher frequency corresponds to a shorter wavelength according to the equation:

f=\frac{c}{\lambda}

(where f is frequency, c the speed of light, and \lambda the wavelength)

So the answer is that the wavelength of the light must be shortened to cause the emission of electrones.

4 0
3 years ago
How far does light travel in the time it takes sound to travel 1 cm in air at 20°c?
kykrilka [37]
The speed of sound at 20^{\circ}C is approximately v=343 m/s. The distance covered by the sound wave is
s=1 cm=0.01 m
And the time it takes is
t= \frac{S}{v}= \frac{0.01 m}{343 m/s}=2.9 \cdot 10^{-5} s

Now we want to find how far the light travels during this time. Light travels at speed c=3 \cdot 10^8 m/s, therefore the distance it covers during this time is
S=ct = (3 \cdot 10^8 m/s)(2.9 \cdot 10^{-5} s)=8700 m
8 0
3 years ago
A small lead ball, attached to a 1.10-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled
hjlf

Answer:

1.84 m

Explanation:

For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.

So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,

a = g + rω²

= 9.8 m/s² + 1.10 m × (18.85 rad/s)²

= 9.8 m/s² + 390.85 m/s²

= 400.65 m/s²

Now, using v² = u² + 2a(h₂ - h₁)  where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.

v² = u² + 2a(h₂ - h₁)

So, v² - u² = 2a(h₂ - h₁)

h₂ - h₁ =  (v² - u²)/2a

h₂ =  h₁ + (v² - u²)/2a

substituting the values of the variables into the equation, we have

h₂ =  1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)

h₂ =  1.3 m + [-430.15 (m/s)²]/-801.3 m/s²

h₂ =  1.3 m + 0.54 m

h₂ =  1.84 m

7 0
2 years ago
Supposing you were in space in a weightless environment, would it require a force to set an object in motion
Jobisdone [24]

Yes.  If your smartphone was floating in front of your face, motionless
relative to you, it would require a force to start it moving toward you or
away from you. 

But there's no minimum force required.  ANY force, no matter how small,
even smaller than the smallest force that you can imagine, would set it in
motion. 

The thing is, though, that the smaller the force acting on it, the smaller
acceleration it would get, and the slower it would move away from where
it is. 

So if, say, you wanted to send it across the crew compartment and over
to your sleeping bag on the wall, and you had all day to watch it mope
along over there, you might breathe on it, and the force of your breath
would set it in slow motion in that direction.  But if you wanted to throw it
at your crewmate, you'd need to give it more force.
  
7 0
2 years ago
A flux density of 1.2Wb/m^2 is required in the 1 mm air gap of an electromagnet having an iron path of length 1.5 m. Calculate t
Mandarinka [93]

Answer:

The mmf required is 1.125×10^{-3} A

Explanation:

The Magnetomotive force (mmf) is given by the formula below

F_{M} = Hl\\

where F_{M} is the Magnetomotive force (mmf)

H is the Magnetic field strength

l is the magnetic length

The magnetic permeability μ is given by

μ = B / H

Where B is the Magnetic flux density

and H is the Magnetic field strength

From the question,

B = 1.2Wb/m^2

μ = 1600m

From μ = B / H

∴H = B/μ

H = 1.2 / 1600\\

H = 7.5 × 10^{-4}A/m

Now, for the Magnetomotive force (mmf)

F_{M} = Hl\\

From the question

l = 1.5 m

∴ F_{M} = 7.5×10^{-4} × 1.5

F_{M} = 1.125×10^{-3} A

Hence, The mmf required is 1.125×10^{-3} A

3 0
3 years ago
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