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stich3 [128]
3 years ago
7

A student with a mass of 62.0 kg walks up a flight of stairs with a height of 3.40 meters. If it takes him 7.00 seconds to get t

o the top, how much power did he generate?
Physics
1 answer:
Inessa05 [86]3 years ago
3 0

Answer:

Power generate by generator = 265 W (Approx.)

Explanation:

Given:

Mass of student = 62 kg

Height of stairs = 3.4 meter

Time taken = 7 second

Find:

Power generate by generator

Computation:

Power = Force x [Distance / Time]

Power = [Mass x gravitational acceleration] x [Distance / Time]

Power = [62 x 9.8][3.4/7]

Power = [607.6][3.4/7]

Power = 265.12

Power generate by generator = 265.12

Power generate by generator = 265 W (Approx.)

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Two particles each of mass m and charge q are suspended by strings of length / from a common point. Find the angle e that each s
ozzi

Answer:

\theta =\left (\frac{kq^{2}}{4L^{2}\times mg}  \right )^{\frac{1}{3}}

Explanation:

Let the length of the string is L.

Let T be the tension in the string.

Resolve the components of T.

As the charge q is in equilibrium.

T Sinθ = Fe       ..... (1)

T Cosθ = mg     .......(2)

Divide equation (1) by equation (2), we get

tan θ = Fe / mg

tan\theta =\frac{\frac{kq^{2}}{AB^{2}}}{mg}

tan\theta =\frac{\frac{kq^{2}}{4L^{2}Sin^{\theta }}}}{mg}

tan\theta =\frac{kq^{2}}{4L^{2}Sin^{2}\theta \times mg}

tan\theta\times Sin^{2}\theta =\frac{kq^{2}}{4L^{2}\times mg}

As θ is very small, so tanθ and Sinθ is equal to θ.

\theta ^{3} =\frac{kq^{2}}{4L^{2}\times mg}

\theta =\left (\frac{kq^{2}}{4L^{2}\times mg}  \right )^{\frac{1}{3}}

7 0
3 years ago
Ask Your Teacher In the air over a particular region at an altitude of 500 m above the ground, the electric field is 120 N/C dir
strojnjashka [21]

Answer:

1.475\times 10^{-13}\ C/m^3

Explanation:

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

A = Area

h = Altitude = 600 m

Electric flux through the top would be

-110A (negative as the electric field is going into the volume)

At the bottom

120A

Total flux through the volume

\phi=120-110\\\Rightarrow \phi=10A

Electric flux is given by

\phi=\dfrac{q}{\epsilon_0}\\\Rightarrow q=\phi\epsilon_0\\\Rightarrow q=10A\epsilon_0

Charge per volume is given by

\rho=\dfrac{q}{v}\\\Rightarrow \rho=\dfrac{10A\epsilon_0}{Ah}\\\Rightarrow \rho=dfrac{10\epsilon_0}{h}\\\Rightarrow \rho=\dfrac{10\times 8.85\times 10^{-12}}{600}\\\Rightarrow \rho=1.475\times 10^{-13}\ C/m^3

The volume charge density is 1.475\times 10^{-13}\ C/m^3

7 0
3 years ago
What is the average power output pf a weight lifter who can lift 250 kg 2.0 m in 2.0 s
liq [111]

Answer:

Power = 2.45Kw or 2450 Watts.

Explanation:

<u>Given the following data;</u>

Mass, m = 250kg

Height, h = 2m

Time, t = 2secs

We know that acceleration due to gravity, g is equal to 9.8m/s²

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

Power = \frac {Energy}{time}

But Energy = mgh

Substituting into the equation, we have

Power = \frac {mgh}{time}

Power = \frac {250*9.8*2}{2}

Power = \frac {4900}{2}

Power = 2450 Watts

To convert to kilowatt (Kw), we would divide by 1000

Power = 2450/1000

Power = 2.45Kw.

Therefore, the average power output of the weightlifter is 2.45 Kilowatts.

8 0
3 years ago
If an object is projected horizontally from a height of 5 m with an initial velocity of 7 m/s, what is the value of x0?
tatiyna
ANSWER IS D X0=9.8m/s^2
5 0
2 years ago
A manometer is used to measure the air pressure in a tank. the fluid used has a specific gravity of 1.25, and the differential h
BartSMP [9]
Specific Gravity of the fluid = 1.25 
Height h = 28 in
 Atmospheric Pressure = 12.7 psia
 Density of water = 62.4 lbm/ft^3 at 32F
 Density of the Fluid = Specific Gravity of the fluid x Density of water = 1.25 x 62.4
 Density of the Fluid p = 78 lbm/ft^3
 Difference in pressure as we got the differential height, dP = p x g x h  dP = (78 lbm/ft^3) x (32.174 ft/s^2) x (28/12 ft) [ 1 lbf / 32.174 ft/s^2] [1 ft^2 /
144in^2]
 Difference in pressure = 1.26 psia
 (a) Pressure in the arm that is at Higher 
 P = Atmospheric Pressure - Pressure difference = 12.7 - 1.26 = 11.44 psia
 (b) Pressure in the tank that is at Lower
 P = Atmospheric Pressure + Pressure difference = 12.7 + 1.26 = 13.96psia
4 0
3 years ago
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