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gulaghasi [49]
4 years ago
10

What is the net work done on the 20kg block while it moves the 4 meters?

Physics
1 answer:
Vinil7 [7]4 years ago
5 0

Answer:

The answer to your question is 784.8 J. None of your answer, did you forget some information?

Explanation:

Data

mass = 20 kg

distance = 4 m

work = ?

Formula

Work = force x distance

Force = mass x gravity

Process

1.- Calculate the weight of the block

     Weight = 20 x 9.81

     Weight = 196.2 N

2.- Calculate the work done

     Work = 196.2 x 4

     Work = 784.8 J

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What quantity is represented by a unit called newton (N) ?
bixtya [17]

Newton is the SI unit of force . Newton is the name of a British scientist and the name of unit is to honour him. The unit is actually Kg.m/s 2 The unit can be derived by the formula. Take the example of weight. It's formula is W = mg . We know that the unit of mass is kg and gravity is m/s 2 so the unit of weight becomes kg.m/s 2 This unit is known as a Newton. It is always given a capital letter because it is someone's name. Other units that are always capitalised (upper case) are Ampere (Amp), Watt, Volt, Coulomb, Kelvin, Celsius, Fahrenheit, Curie, Roentgen because they are also people's names.

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3 years ago
The horizon moon appears to shrink in size if it is viewed through a narrow tube that eliminates the perception of distance cues
Sonja [21]

Answer: (c) Context Effect

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4 years ago
A circular rod has a radius of curvature R = 9.09 cm and a uniformly distributed positive charge Q = 6.49 pC and subtends an ang
Digiron [165]

Answer:

E = 1.19 N/C

Explanation:

Let's first determine the length of the arc which can be given as:

L= Rθ

where:

L = length of the arc

R = radius of curvature

θ = angle in radius

L = (9.09×10⁻²m)(2.59)

L = (0.0909)(2.59)

L = 0.235431 m

Then, the magnitude of electric field that Q produces at the center of curvature can be calculated by using the formula:

E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}-sin(-\frac{\theta}{2})]

E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}+sin(\frac{\theta}{2})]

E= \frac{2\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}]

Since \lambda = \frac{Q}{L}

where;

L = length

Q = charge

λ =  density of the charge;

then substituting \frac{Q}{L} for λ, we have :

E= \frac{2(\frac{Q}{L})}{4 \pi E_oR}[sin\frac{\theta}{2}]

E= \frac{2Q[sin\frac{\theta}{2}]}{4 \pi E_oLR}

substituting our given parameter; we have:

E= \frac{2(6.26*10^{-12}C)[sin\frac{2.59rad}{2}]}{4 \pi (8.85*10^{-12}C^2/N.m^2)(0.235431)(0.0909)}

E = 1.1889 N/C

E = 1.19 N/C

∴ the magnitude of the electric field that Q produces at the center of curvature = 1.19 N/C

4 0
3 years ago
a circuit element having terminals a and b has vab = 10v and iba = 2a over a period of 20s, how much charge moves throught the e
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Answer: charge Q = 40Coulombs

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Here the 2amp current is same as the amount of charge in coulomb that flows through the terminal in ONE second.

So, if 2 coulomb of charge flows in 1sec then in 20 secs we will have

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In what scenario might a red ball appear black?
zhuklara [117]
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