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2 years ago

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A 4kg ball that is moving straight upward has 20J of kinetic energy and its total mechanical energy is 32J. Find the gravitation

al potential energy of the ball

1 answer:

Dovator [93]2 years ago

3 0

Answer:

h = 0.25 [m]v = 3.26 [m/s]

Explanation: In order to solve this problem, we must remember that mechanical energy is defined as the sum of kinetic energy plus potential energy.

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The volume of the lung 0.0024m^3 following exhalation and the pressure is 101.70KPa. Calculate the volume of the lungs during in

Mekhanik [1.2K]

According to Boyle-Mariotte:

p₁V₁=p₂V₂=>V₂=p₁V₁/p₂= 0.0024*101.70/ 84.16=0.0028 m³

5 0

3 years ago

Which type of energy is water held back by a dam?

stepladder [879]

Answer: hydroelectric dams

Explanation: Enormous energy is stored in water held back by hydroelectric dams. The energy is transformed into (a form of kinetic energy) as it falls across the dam. The falling water strikes the blades of a turbine and makes them spin, and the then turns the shaft of a generator.

8 0

3 years ago

Read 2 more answers

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated

Lelu [443]

Incomplete question as we have not told to find what quantity.The complete question is here

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 5.00 cm.calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

Answer:

(a) $C=16.7pF$

(b) $r_{a} =3.749cm$

(c) $E=2.24*10^{4} N/C$

Explanation:

Given data

$Q=3.50nC\\V=210V\\r_{b}=5.0cm$

For part (a)

The Capacitance given by:

$C=\frac{Q}{V}\\ C=\frac{3.50*10^{-9} C}{210V}\\C=1.6666*10^{-11}F\\or\\C=16.7pF$

For part (b)

The Capacitance of coordinates is given as

$C=\frac{4\pi e}{\frac{1}{r_{a} }-\frac{1}{r_{b} } }\\ So\\{\frac{1}{r_{a} }-\frac{1}{r_{b} } }=\frac{4\pi *8.85*10^{-12} }{1.666*10^{-11}}=6.672m^{-1} \\ \frac{1}{r_{a} }=6.672+(1 /0.05)\\\frac{1}{r_{a} }=26.672\\r_{a} =1/26.672\\r_{a} =0.0375m\\r_{a} =3.749cm$

For part (c)

The electric field according to Gauss Law is given by:

$EA=\frac{Q}{e}\\ E=\frac{Q}{4\pi er_{a}^{2} }=\frac{kQ}{r_{a}^{2}}\\ E=\frac{9*10^{9}*3.50*10^{-9} }{(0.0375m)^{2} }\\ E=2.24*10^{4} N/C$

7 0

3 years ago

Twist-on connectors without the spring-steel coils (plastic threads only) are suitable for making branch-circuit connections.

Gnoma [55]

Answer:

if it is a plastic connector it wont work but if there is metal or steel it will work

Explanation:

6 0

3 years ago

A power plant produces 1000 MW to suply a city 40Km away.Current flows from the power plant on a single wire of resistance0.050

Westkost [7]

Answer:

The current in wire resistance 2Ω

a). 8696 A

b). fraction power 15.1% a 115kV

Explanation:

Resistance

$R=0.05$ Ω/Km*40km

R=2Ω

P=1000 MW

a).

$P=V*I\\I=\frac{P}{V}=\frac{1000x10^{6}W}{115x10^{3}k } =8696.65A$

Using law ohm

b).

$V=I*R\\I=\frac{V}{R}$

$P=I*I*R\\P=I^{2} *R\\P=8696.65^{2}*2\\P=151.228 x10^{6} W$

$e=\frac{151.228x10^{6} }{1000x10^{6} }*100= 15.12$ %

8 0

3 years ago