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PilotLPTM [1.2K]
3 years ago
14

. How much work is done in lifting an object that has a mass of 5kg a vertical distance of 2m?

Physics
1 answer:
malfutka [58]3 years ago
7 0
<h3>Answer:</h3>

100 Joules

<h3>Explanation:</h3>

The question is testing on how to calculate work done;

We are given;

  • Mass of an object is 5 kg
  • Distance moved is 2 m

We are required to calculate the amount of work done.

How do we calculate work done?

  • Work done is the product of force and distance.
  • Therefore without distance covered no work is done.
  • Work done = Force × distance
  • Additionally, it is measured in joules or Nm

In this case;

Force = mass × 10 N/kg

         = 5 kg × 10 N/kg

         = 50 N

Therefore;

work done = 50 N × 2 m

                  = 100 Joules

Thus, the work done in lifting the object is 100 Joules

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The current that charges a capacitor transfers energy that is stored in the capacitor’s electric field. Consider a 2.0 μF capaci
lapo4ka [179]

Answer:

the capacitor voltage is V = 20 V

Explanation:

Given,

Capacitance of the capacitor =  2.0 μF

energy stored = 200 W

time (t) =2.0 μs

Capacitor voltage = ?

\dfrac{dE}{dt}=200 W

E =200\times 2 \times 10^{-6}

E =400 \times 10^{-6}\ J

we know,

E = \dfrac{1}{2}CV^2

V =\sqrt{\dfrac{2E}{C}}

V =\sqrt{\dfrac{2\times 400 \times 10^{-6}}{2\times 10^{-6}}}

V =\sqrt{400}

V = 20 V

so , the capacitor voltage is V = 20 V

7 0
3 years ago
Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol
klasskru [66]

The current flowing in silicon bar is 2.02 \times 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

     J = (1.6 \times 10^-19) / 0.001 (104 \times 1200 + 1016 \times 500)

     J = 1012480 \times 10^-16 A / m^2.

or

J = 1.01 \times 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 \times 10^-9 \times 0.002 = 2.02 \times 10^-12

So, the current flowing in silicon bar is 2.02 \times 10^-12 A.  

6 0
3 years ago
A helicopter ambulance flew from one hospital to another in a straight line. The pilot had to change speed several times due to
Olenka [21]

Answer:

Explanation:

All the rest of the information is extraneous. The only 2 things you have to know are

d = 20 km

t = 8 minutes = 8/60 hours = 0.13333333

So the speed is s = d/t

s = 20/0.1333333 = 150 km/hour

Note: you have not specified what units the speed is. I suppose you could answer 20/8 = 2.5 km/min

4 0
3 years ago
A kangaroo can jump straight up to a height of 2.0 m. What is its takeoff speed
12345 [234]
7.17m/s glad I could help
5 0
3 years ago
A machine that operates a ride at the fair requires 2500 J to lift a 294 N child 5.0 m. What is the efficiency of this machine?
NeX [460]

Answer:

η = 58.8%

Explanation:

Work is defined as the force applied by the distance traveled by the body.

W =F*d

where:

W = work [J] (units of joules)

F = force = 294 [N]

d = distance = 5 [m]

W = 294*5\\W = 1470 [J]\\

Efficiency is defined as the energy required to perform an activity in relation to the energy actually added to perform some activity. This can be better understood by means of the following equation.

efficiency = W_{done}/W_{required}\\efficiency = 1470/2500\\efficiency = 0.588 = 58.8%

5 0
3 years ago
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