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Scilla [17]
3 years ago
6

What is the potential energy of the bowling ball as it sits on top of a building ?​

Physics
1 answer:
tangare [24]3 years ago
8 0

Answer:

The potential energy of the bowling ball will be mgh

Explanation:

Let the mass of bowling ball =m

The height of building on which bowling ball sits=h

So,

The potential energy of the bowling ball =P.E.= mgh

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A student hooks up a voltmeter and an ammeter in a circuit to find the resistance of a light bulb. The ammeter read 3 amps and t
yuradex [85]

Answer:

2\Omega

Explanation:

(Assuming the cell in the circuit has 0 internal resistance)

Ohm's Law is given as:

V=IR

Voltage is Current multiplied by Resistance.

We can rearrange this formula to give us:

R=\frac{V}{I}

Now we can plug in our values

R=\frac{6}{3}=2\Omega

7 0
3 years ago
Read 2 more answers
It has been proposed that a spaceship might be propelled in the solar system by radiation pressure, using a large sail made of f
Anestetic [448]

Answer:

A = 8.34 x 10^(5) m²

Explanation:

The intensity of the solar radiation is the average solar power per unit area. Thus,

I = P/4A = P/(4(πr²))

Where;

P is average solar power.

r is it's distance from centre of sun

A is area

Now, The rate at which the Sun emits energy has a standard value of 3.90 × 10^(26) W

Thus;

I = [3.90 × 10^(26)]/(4πr²)

I = [3 x 10^(25)]/r²

Now, the formula for radiation pressure is;

P_rad = 2I/c

Where c is speed of light and has a value of 3 x 10^(8) m/s

Thus,

P_rad = 2([3 x 10^(25)]/r²)/(3 x 10^(8))

P_rad = [2.07 x 10^(17)]/r² N/m²

Also, Radiation pressure on ship; P_rad = F/A

Where Force on ship and A is area.

Thus;

F = P_rad x A

So, F = [2.07 x 10^(17)•A]/r²

Now,

F_grav = GMm/r²

Where;

G is gravitational constant with a value = 6.67 x 10^(-11) Nm²/kg²

M is mass of sun with a value of 1.99 x 10^(30)

m is mass of ship and sail = 1300 kg

Thus, plugging in the relevant values to obtain;

F_grav = (6.67×10^(-11) × 1.99 x 10^(30) × 1300)/r²

F_grav = [17.255 x 10^(22)]/r²

Now, equating F to F_grav, we get;

[2.07 x 10^(17)•A]/r² = [17.255 x 10^(22)]/r²

r² will cancel out to give;

2.07 x 10^(17)•A= [17.255 x 10^(22)]

A = [17.255 x 10^(22)]/2.07 x 10^(17)

A = 8.34 x 10^(5) m²

7 0
4 years ago
A Gaussian surface in the form of a hemisphere of radius R = 5.51 cm lies in a uniform electric field of magnitude E = 1.08 N/C.
Ksju [112]

Answer:

Flux_{base}=-0.0103Nm^2/C

Flux_{curvePortion}=-Flux_{base}=0.0103Nm^2/C

Explanation:

a)At the base of the surface the Electric flux is easy to find, because the Electric Field is constant in magnitude and perpendicular to the flat surface:

Flux_{base}=-E*S=-E*\pi*R^2=-1.08*\pi*0.0551^2=-0.0103Nm^2/C

The Flux is negative because the Electric field goes into the surface

b) The Gaussian surface in form of a hemisphere encloses no net charge. The Gauss law says that the Flux of the electric field is proportional to the net charge enclosed. At this case, the charge is zero, then the total Flux is zero too.

Flux_{total}=Flux_{base}+Flux_{curvePortion}=0

Then:

Flux_{curvePortion}=-Flux_{base}=0.0103Nm^2/C

The Flux is positif because the Electric field goes out of the surface.

4 0
3 years ago
A 3-ft3 rigid tank initially contains saturated water vapor at 300°F. The tank is connected by a valve to a supply line that car
zhannawk [14.2K]

Explanation:

Given\\V=3ft^3\\T_{1} =300^oF\\T_{L} =400^oF\\P_{L} =200psia\\V_{f} =0.5V

We assume kinetic and potential energy changes are negligible and there is no work interactions.

a) Taking tank as a system, The energy balance can be define as

E_{in}- E_{out}= E_{sys}

m_{i} h_{L} -Q_{out}=m_{2}  u_{2}-m_{1}  u_{1}

The mass balance could be written as

m_{in}- m_{out}= m_{sys} \\m_{i}= m_{2}- m_{1}

The final pressure in the tank could be defined as following

P_{2} =P_{sat300^oF}

from standard steam table we know at

T_{2}= T_{1}=300^oF\\ P_{2}=67.028psia

b)

From steam table at

T_{1} =300^oF\\v_{1}= v_{g}=6.4663ft^3/lbm\\v_{1}=  v_{g}=1099.8Btu/lbm\\ v_{f} =0.01745ft^3/lbm\\v_{f} =269.51Btu/lbm\\

at\\P_{L} =200psia\\T_{L}=400^oF\\h_{L}=1210.9Btu/lbm

initial mass in the tank could be define as

m_{1}=\frac{V}{v_{1} }  \\m_{1} =\frac{3}{6.4663} =0.464lbm\\

Final mass in the tank could be define as

m_{2}=\frac{V_{f} }{v_{f} }+\frac{V_{g} }{v_{g} }  \\ m_{2} =\frac{1.5}{0.01745} +\frac{1.5}{6.4663} =86.2lbm

The amount of steam that has entered the tank

m_{i}=m_{2}-  m_{1}\\ m_{i}=86.2-0.464=85.74lbm

c)

The internal energy in final state could be defined as following

U_{2}=m_{f}  u_{f}+ m_{g} u_{g}\\ U_{2} =85.96*269.51+0.232*1099.8=23422Btu\\

The heat transfer could be defined as following

Q_{out}=m_{i}  h_{L}+m_{1}  u_{1}-m_{2}  u_{2}\\ Q_{out}=85.74*1210.9+0.464*1099.8-23422=80910Btu

6 0
3 years ago
Read 2 more answers
A taxi is travelling at 15 m/s. Its driver accelerates with
Hitman42 [59]
If we assume that the acceleration is constant, we can use on the kinematic equations:

Vf = Vi + a*t = 15 + 3*4 = 27 m/s
6 0
3 years ago
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