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ella [17]
3 years ago
10

How far away is the furthest artificial satellite orbiting earth?

Physics
1 answer:
rusak2 [61]3 years ago
8 0
Real-time distance and velocity data is provided by NASA and JPL. At a distance of 153.88 AU (23.020 billion km; 14.304 billion mi) from Earth as of September 5, 2021, it is the most distant artificial object from Earth. The probe made successful flybys of Jupiter, Saturn, and Saturn's largest moon, Titan.
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A college student is working on her physics homework in her dorm room. Her room contains a total of 6.0 x 10^26 gas molecules. A
IceJOKER [234]

Answer:

Temperature, T = 3.62 kelvin

Explanation:

It is given that,

Total number of gas molecules, N=6\times 10^{26}

Her body is converting chemical energy into thermal energy at a rate of 125 W, P = 125 W

Time taken, t = 6 min = 360 s

Energy of a gas molecules is given by :

\Delta E =\dfrac{3}{2}NkT

T=\dfrac{2E}{3Nk}, k is Boltzmann constant

T=\dfrac{2\times P\times t}{3Nk}

T=\dfrac{2\times 125\times 360}{3\times 6\times 10^{26}\times 1.38\times 10^{-23}}

T = 3.62 K

So, the temperature increases by 3.62 kelvin. Hence, this is the required solution.

4 0
3 years ago
PLSSS HELP MEEEE
vekshin1

Answer:

Cu2+

Explanation:

7 0
3 years ago
Read 2 more answers
A commuter backs her car out of her garage with an acceleration of 1.40 m/s^2. (a) How long does it take her to reach a speed of
jonny [76]

Answer:

a) It takes her 1.43 s to reach a speed of 2.00 m/s.

b) Her deceleration is - 2.50 m/s²

Explanation:

The equation of velocity for an object that moves in a straight line with constant acceleration is as follows:

v = v0 + a · t

Where:

v = velocty.

v0 = initial velocity.

a = acceleration.

t = time.

a) Using the equation of velocity, let´s consider that the car moves in the positive direction. Then:

v = v0 + a · t

2.00 m/s = 0 m/s + 1.40 m/s² · t

t = 2.00 m/s / 1.40 m/s²

t = 1.43 s

It takes her 1.43 s to reach a speed of 2.00 m/s

b) Let´s use again the equation of velocity, knowing that at t = 0.800 s the velocity is 0 m/s:

v = v0 + a · t

0 = 2.00 m/s + a · 0.800 s

-2.00 m/s / 0.800 s = a

a = -2.50 m/s²

Her deceleration is - 2.50 m/s²

7 0
3 years ago
A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 kg particle has the xy coordinates (0.600 m, -0.750 m).
just olya [345]

Answer:

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

Explanation:

The center of mass of a system of particles (\vec r_{cm}), measured in meters, is defined by this weighted average:

\vec r_{cm} = \frac{\Sigma_{i=1}^{n}\,m_{i}\cdot \vec r_{i}}{\Sigma_{i=1}^{n}\,m_{i}} (1)

Where:

m_{i} - Mass of the i-th particle, measured in kilograms.

\vec r_{i} - Location of the i-th particle with respect to origin, measured in meters.

If we know that \vec r_{cm} = (-0.500\,m,-0.700\,m), m_{1} = 1\,kg, \vec r_{1} = (-1.20\,m, 0.500\,m), m_{2} = 4.50\,kg, \vec r_{2} = (0.600\,m, -0.750\,m) and m_{3} = 4\,kg, then the coordinates of the third particle are:

(-0.500\,m, -0.700\,m) = \frac{(1\,kg)\cdot (-1.20\,m,0.500\,m)+(4.50\,kg)\cdot (0.600\,m,-0.750\,m)+(4\,kg)\cdot \vec r_{3}}{1\,kg+4.50\,kg+4\,kg}

(-4.75\,kg\cdot m, -6.65\,kg\cdot m) = (-1.20\,kg\cdot m, 0.500\,kg\cdot m) + (2.7\,kg\cdot m, -3.375\,kg\cdot m) +(4\cdot x_{3},4\cdot y_{3})

(4\cdot x_{3}, 4\cdot y_{3}) = (-6.25\,kg\cdot m,-3.775\,kg\cdot m)

(x_{3},y_{3}) = (-1.562\,m,-0.944\,m)

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

5 0
3 years ago
A typical student listening attentively to a physics lecture has a heat output of 100 w . how much heat energy does a class of 1
nexus9112 [7]

Since each student emits 100 W, so 170 students will emit:

total heat = 100 W * 170 = 17,000 W

 

Convert minutes to seconds:

time = 50 min * (60 s / min) = 3000 s

 

The energy is therefore:

E = 17,000 W * 3000 s

<span>E = 51 x 10^6 J = 51 MJ</span>

7 0
3 years ago
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