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dalvyx [7]
3 years ago
14

6. A satellite is orbiting Earth just above its surface. The centripetal force making the satellite follow a circular trajectory

is just its weight, so its centripetal acceleration is about 9.8 m/s2 (the acceleration due to gravity near Earth’s surface). If Earth’s radius is about 6375 km, how fast must the satellite be moving? How long will it take for the satellite to complete one trip around Earth?
Physics
1 answer:
Svetllana [295]3 years ago
5 0

Answer:

Engular velocity: w=1,24*10^{-3}/s

Linear velocity: V=7905 m/s

The time it takes:

t=5060s=84min

Explanation:

The magnitude of the centripetal acceleration can be related to the angular velocity and radius as:

(1)a=r*w^{2}

Solving for w:

(2)w=\sqrt{\frac{a}{r} }

Replacing a=9,8m/s2 and r=6,375,000m:

(3)w=\sqrt{\frac{9,8m/s^{2}}{6375000m} }=1,24*10^{-3}/s

And the angular velocity relates to the linear velocity:

V=w*r=1,24*10^{-3}/s*6375000m=7905 m/s

The perimeter of the orbit is:

P=\pi *2*r=\pi *2*6375000m=40.05*10^{6}m

The time it takes:

t=\frac{P}{V} =\frac{40.05*10^{6}m}{7905 m/s}=5060s=84min

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Which of the following statements about sidewall markings is correct? a. The load index is given as a letter. b. The traction an
vagabundo [1.1K]

Answer: all the above options are correct.

Explanation:

In sidewall markings,the load index is given as a letter,traction and temperature ratings are based on the speed rating of the tire,the tire's recommended inflation pressure and load are indicated and the DOT code indicates when and where the tire was made.

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3 years ago
What is the acceleration of an object that goes from 45m/s to 10 m/s in 5 seconds?
kipiarov [429]

Answer:

\boxed {\boxed {\sf a= -7 \ m/s^2}}

Explanation:

Acceleration is the change in velocity over time.

a= \frac {v_f-v_i}{t}

The object accelerates <em>from</em> 45 meters per second <em>to </em>10 meters per second in 5 seconds. Therefore,

v_f=10 \ m/s \\v_i= 45 \ m/s \\t= 5 \ s

Substitute the values into the formula.

a= \frac{ 10 \ m/s - 45 \ m/s}{5 \ s}

Solve the numerator.

a= \frac { -35 \ m/s}{5 \ s}

Divide.

a= -7 \ m/s/s

a= -7 \ m/s^2

The acceleration of the object is -7 meters per square second. The acceleration is negative because the object's velocity decreases and the object slows down.

5 0
3 years ago
The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at
denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

A)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m  _____ eqn (1)

<u>FOR HYDROGEN:</u>

v = √(3KT)/m = 2000 m/s  _____ eqn (2)

<u>FOR OXYGEN:</u>

velocity of oxygen = √(3KT)/(mass of oxygen)  

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

<u>velocity of oxygen = (1/4)(2000 m/s) = 500 m/s</u>

B)

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

<u>K.E of Oxygen = 4000 J</u>

C)

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

<u>v = 2149.24 m/s</u>

<u></u>

8 0
3 years ago
It is determined that a certain light wave has a wavelength of 3.012 ×10−12 m. The light travels at 2.99 ×108 m/s. What is the f
horsena [70]

Answer:

Explanation:

velocity=frequency*wavelength

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2.99*10^8m/s = (f)(3.012*10^-12)

f=9.58*10^19 Hertz    

5 0
3 years ago
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