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hoa [83]
3 years ago
15

An armadillo and Earth attract each other gravitationally. Which experiences the greater gravitational force, or do they experie

nce the same force magnitude?
a. Same force magnitude
b. Earth
c. Armadillo
Physics
1 answer:
Taya2010 [7]3 years ago
6 0

Answer:

a. Same force magnitude

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It takes 40 J of energy to push a large box 4m across a floor. Assuming the push is the same direction as the movement, what is
Jobisdone [24]

Answer:

10 N

Explanation:

Work done is the dot product of the force magnitude with distance and cosine of the angle betweenthem.

Equation to use: W=|F|*|d|*cosθ

Your unknown is the force F; You have one equation, oneunknown.

40j=(F)*(4m)*cos(0)

F=40j/4m

F=10N

3 0
3 years ago
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According to Einstein's prediction, which of the following happens to the mass of an object in a moving internal reference frame
avanturin [10]

Answer:

option (A)

Explanation:

The a body having rest mass is mo is moving with velocity v which is of the order of velocity of light in vacuum, so the mass of the body increases. According to the formula

m=\frac{m_{0}}{\sqrt{1-\left ( \frac{v^{2}}{c^{2}} \right )}}

As the velocity increases, the mass m of the body increases. As the velocity of the object is equal to the velocity of light in vacuum, then the mass of the body becomes infinite.

4 0
3 years ago
Which of the following account for the difference between radio waves , infrared waves, and ultraviolet rays?
gayaneshka [121]

Answer: B

The only difference between these different types of radiation is their wavelength or frequency

Explanation:

5 0
3 years ago
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A 165-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,015 A. If the conductor is copper
elena55 [62]

Answer:

22.1 years

Explanation:

Since the current in the wire is I = nevA where n = electron density = 8.50 × 10²⁸ electrons/cm³ × 10⁶ cm³/m³= 8.50 × 10³⁴ electrons/m³, e = electron charge = 1.602 × 10⁻¹⁹ C, v = drift velocity of electrons and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.00 cm = 2 × 10⁻² m

Making v subject of the formula, we have

v = I/neA

So, v = I/neπd²/4

v = 4I/neπd²

Since I = 1,015 A, substituting the values of the other variables into the equation, we have

v = 4I/neπd²

v = 4(1,015 A)/[8.50 × 10³⁴ electrons/m³ × 1.602 × 10⁻¹⁹ C × π ×(2 × 10⁻² m)²]

v = 4(1,015 A)/[8.50 × 10³⁴ electrons/m³ × 1.602 × 10⁻¹⁹ C × π × 4 × 10⁻⁴ m²]

v = (1,015 A)/[42.779 × 10¹¹ electronsC/m]

v = 23.73 × 10⁻¹¹ m/s

v = 2.373 × 10⁻¹⁰ m/s

Since distance d = speed, v × time, t

d = vt

So, the time it takes one electron to travel the full length of the cable is t = d/v

Since d = distance moved by free charge = length of transmission line = 165 km = 165 × 10³ m and v = drift velocity of charge = 2.373 × 10⁻¹⁰ m/s

t = 165 × 10³ m/2.373 × 10⁻¹⁰ m/s

t = 69.54 × 10⁷ s

t = 6.954 × 10⁸ s

Since we have 365 × 24 hr/day × 60 min/hr × 60 s/min = 31536000 s in a year = 3.1536 × 10⁷ s

So,  6.954 × 10⁸ s =  6.954 × 10⁸ s × 1yr/3.1536 × 10⁷ s = 2.21 × 10 yrs = 22.1 years

It will take one electron 22.1 years to travel the full length of the cable

3 0
3 years ago
What is blood ? Name its components<br>​
Bas_tet [7]
Plasma, red blood cells, white blood cells, and platelets.
3 0
3 years ago
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