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mrs_skeptik [129]
3 years ago
12

16. At a particular point, a satellite in an elliptical orbit has a gravitational potential energy of 7000 MJ with respect to Ea

rth’s surface and a kinetic energy of 4000 MJ. At another point in its orbit, the satellite’s potential energy is 2000 MJ. What is its kinetic energy at that point? Show all calculations leading to your answer.
Physics
1 answer:
WARRIOR [948]3 years ago
8 0
Assuming conservation of energy:

Energy before = 7000 + 4000 = 11 000 MJ

Energy after = 2000 + Kinetic = 11 000

Therefore Kinetic = 11 000 - 2 000

Kinetic = 9 000 MJ
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When an electron moves 2.5 m in the direction of an electric field, the change in electrical potential energy of the electron is
anygoal [31]
In the field of electromagnetism, when two charged plates that are situated opposite to each other by a certain distance, it forms an energy called the electric field. This energy is due to the difference in potential energy with respect to distance. Thus,

E = V/d

However, the voltage in volts is energy per coulomb. Thus,
V = (8x10-17 J/electron)*(1electron/1.60218x10^-19 C)
V = 499.32 volts

Therefore,

E = 499.32 volts /2.5 m
E = 199.73 N/C

The electric field that caused the change in potential energy is equal to 199.73 Newtons per Coulomb.
7 0
3 years ago
All stars in a stellar cluster have roughly the same:___.
3241004551 [841]

All stars in a stellar cluster have roughly the same distance.

<h3>What coloration are celebrity clusters?</h3>

Open clusters have a tendency to be blue in color. They frequently include glowing gas and dust. The stars in an open cluster are young stars that all formed from the equal nebula. These warm blue stars are in an open cluster known as the Jewel Bo

<h3>Are stars in the identical cluster?</h3>

Though stellar associations are grouped in with megastar clusters, they're pretty a bit different. "Stellar associations are companies of tens to hundreds of stars that have similar a while and metallicities, and are moving in roughly the equal direction within the galaxy, but are no longer gravitationally bound," Geller said.

Learn more about star cluster here:

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4 0
1 year ago
A square coil ℓ = 2cm on a side with 30 turns rotates in a uniform magnetic field, B~ = B0zˆ = 0.1Tˆz, such that the normal of t
kow [346]

Answer:

a) 1.2*10^{-3}cos(1.25t)

b) 0.49mV

Explanation:

a) The coil rotates periodically with period T. Hence, we can write the variation of the magnetic flux with a sinusoidal function, and with max flux NAB. Thus, we have that:

\Phi_B(t)=NABcos(\omega t)\\\\\omega=\frac{2\pi}{T}=1.25\frac{rad}{s}\\\\A=l^2=(0.02m)^2=4*10^{-4}m^2\\\\B=0.1T\\\\\Phi_B(t)=1.2*10^{-3}cos(1.25 t) W

where we have used the values given by the information of the problem for N B and A.

b)

the emf is given by:

emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV

hope this helps!!

5 0
3 years ago
A piston/cylinder contains 2 kg of water at 20◦C with a volume of 0.1 m3. By mistake someone locks the piston, preventing it fro
morpeh [17]

Answer:

Hi

Final temperature = 250.11 °C

Final volume = 0,1 m3.

Process work = 0

Explanation:

The specific volume in the initial state is: v = 0.1m3/2 kg = 0.05 m3/kg.

This volume is located between the volumes as saturated liquid and saturated steam at 20 °C. For this reason the water is initially in a liquid vapor mixture. As the piston was blocked the volume remains constant and the process is isometric, also known as isocoric process, so the final temperature will be the water temperature at a saturated steam of v=0.05m3/kg, which is obtained by using steam tables for water, by linear interpolation. As follows, using table A-4 of the Cengel book 7th Edition:

v=0.05 m3/kg

v1=0.057061 m3/kg

T1=242.56°C

v2=0.049779 m3/kg

T2=250.35°C

T=\frac{T2-T1}{v2-v1} x(v-v1)+T1=\frac{250.35°C-242.56°C}{0.049779m3/kg-0.057061m3/kg}x(0.05m3/kg-0.057061m3/kg)+242.56°C=250.11°C

The process work is zero because there is no change in volume during heating:

W=PxΔv=Px0=0

where

W=process work

P=pressure

Δv=change of volume, is zero because the piston was blocked so the volume remains constant.

7 0
3 years ago
a chamber with a fixed volume of 1.0 meters cubed contains a monatomic gas at 3.00 *10^K. The chamber is heated to a temperature
igomit [66]

Answer:

Explanation:

Given

Volume of fixed chamber V=1 m^3

Initial Temperature T_1=300 K

Final Temperature T_2=400 K

Heat Supplied Q=10 J

From First law of thermodynamics

Change in internal energy of the system is equal to heat added minus work done by the system

\Delta U=Q-W

as the volume is fixed therefore work

W=\int PdV=0

thus \Delta U=mc_v\Delta T=Q

c_v for mono-atomic gas is 12.471 J/K-mol

n\times 12.471\times (400-300)=10

n=0.008018 mol

and 1 mole contains 6.022\times 10^{23} molecules

thus  No of molecules=0.008018\times 6.022\times 10^{23}

No of molecules=4.82\times 10^{21} molecules

3 0
3 years ago
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