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Svetllana [295]
3 years ago
10

Using a force of 28.0 Newton, a student pulls a 70.0 Newton weight along the tabletop for a distance of 15.0 meters in 3.0 secon

ds. Compute the power developed by the student.
Physics
1 answer:
Musya8 [376]3 years ago
8 0

Answer:

140 watt

Explanation:

We are given that

Force applied by student ,F=28 N

Weight pulled  by students=70 N

Displacement,s=15 m

Time=3 s

We have to find the power developed  by the student.

Work done=w=F\times s

Work done by the student=28\times 15=420 J

Power=\frac{work\;done}{time}

Using the formula

Power=\frac{420}{3}=140watt

Hence, the power developed by the students=140 watt

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What is the net work done on the 20kg block while it moves the 4 meters?
Vinil7 [7]

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

5 0
4 years ago
A 4.60 gg coin is placed 19.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with
Komok [63]

Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\  g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s

\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM

Therefore the speed in RPM will be 62.64 RPM.

5 0
3 years ago
what will be the focal lenght of a combined lens made by contact of two lenses of power +3D and -2D.​
4vir4ik [10]
  • P_1=+3D
  • P_2=-2D

\\ \bull\tt\dashrightarrow P=P_1+P_2

\\ \bull\tt\dashrightarrow P=+3D-2D=+1D

Now

\\ \bull\tt\dashrightarrow f=\dfrac{1}{P}

\\ \bull\tt\dashrightarrow f=\dfrac{1}{1}

\\ \bull\tt\dashrightarrow f=1m

3 0
2 years ago
A swimming pool, 10.0 m by 4.0 m, is filled with water to a depth of 3.0 m at a temperature of 20.2°c. if the energy needed to r
Tomtit [17]

You need to find the mass of water in the pool.
Find the volume (10 x 4 x 3) = 120 m3

Water has a density of 1000g/m3,so 120 m3 = 120 x 1000 = 120 000 kg

[delta]H = 4.187 x 120 000 x 3.4 (and the units will be kJ)

You then use the heat of combustion knowing that each mole of methane releases 891 kJ of heat so if you divide 891 into the previous answer, you will get the number of moles of CH4


8 0
3 years ago
Read 2 more answers
A wire 6.60 m long with diameter of 2.05 mm has a resistance of 0.0310 Ω.
Alex73 [517]

Answer:

1.551×10^-8 Ωm

Explanation:

Resistivity of a material is expressed as shown;.

Resistivity = RA/l

R is the resistance of the material

A is the cross sectional area

l is the length of the wire.

Given;

R = 0.0310 Ω

A = πd²/4

A = π(2.05×10^-3)²/4

A = 0.000013204255/4

A = 0.00000330106375

A = 3.30×10^-6m

l = 6.60m

Substituting this values into the formula for calculating resistivity.

rho = 0.0310× 3.30×10^-6/6.60

rho = 1.023×10^-7/6.60

rho = 1.551×10^-8 Ωm

Hence the resistivity of the material is 1.551×10^-8 Ωm

6 0
3 years ago
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