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Vinil7 [7]
3 years ago
6

If you hold a 50 kilogram barbell above your head for 3 seconds and

Physics
1 answer:
aalyn [17]3 years ago
3 0

Answer:

c

if you calculate the net force you get 490 N

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Ask Your Teacher A basketball player shoots toward a basket 5.8 m away and 3.0 m above the floor. If the ball is released 1.7 m
const2013 [10]

Answer:

The answer to your question is    vo = 5.43 m/s

Explanation:

Data

distance = d= 5.8 m

height = 3 m

height 2 = 1.7 m

angle = 60°

vo = ?

g = 9.81 m/s²

Formula

              hmax = vo²sinФ/ 2g

Solve for vo²

              vo² = 2ghmax / sinФ

Substitution

              vo² = 2(9.81)(3 - 1.7) / 0.866

Simplification

              vo² = 19.62(1.3) / 0.866

              vo² = 25.51 / 0.866

              vo² = 29.45

Result

              vo = 5.43 m/s

               

5 0
3 years ago
What are the periodic variations in Earth's rotation and orbit around the sun that alter the way solar radiation is distributed
Dahasolnce [82]

Answer:

1. The precession of the equinoxes.

2. Changes in the tilt angle of Earth’s rotational axis relative to the plane of Earth’s orbit around the Sun.

3. Variations in the eccentricity

Explanation:

These variations listed above;  the precession of the equinoxes (refers, changes in the timing of the seasons of summer and winter), this occurs on  a roughly about 26,000-year interval; changes in the tilt angle of Earth’s rotational axis relative to the plane of Earth’s orbit around the Sun, this occurs roughly in a 41,000-year interval; and changes in the eccentricity (that is a departure from a perfect circle) of Earth’s orbit around the Sun, occurring on a roughly 100,000-year timescale. which influences the mean annual solar radiation at the top of Earth’s atmosphere.

5 0
3 years ago
Help me asap its due today
Margaret [11]

Answer:

3.54* 10^{22} N

Explanation:

Using the formula you gave:

F_g = \frac{6.67*10^{-11}*2.0*10^{30}*5.97^{24}  }{(1.5*10^{11})^2 }

3 0
3 years ago
Nodes and antinodes.are part of an _____ wave<br><br> A active <br> B standing
mel-nik [20]
It could be A :) not sure tho
5 0
3 years ago
What is the speed of sound at 33 °C (m/s)? For a frequency of 5 kHz, how large do you expect the wavelength to be (m)?
Vedmedyk [2.9K]

Answer:

-The speed of sound at 33°C is 362.8 m/s.

-The wavelength at a frequency at 5 kHz is 0.07256 m .

Explanation:

let v = 343 m/s be the speed of sound.

let T be the temperature.

then the speed of sound V, at 33°C is given by:

V = v + 0.6×T

   = 343 + 0.6×33

   = 362.8 m/s

Therefore, the speed of sound at 33°C is 362.8 m/s.

the wavelength at a frequency of f = 5kHz = 5000 Hz is given by:

λ = V/f

  = (362.8)/(5000)

  = 0.07256 m

Therefore, the wavelength at a frequency at 5 kHz is 0.07256 m .

7 0
3 years ago
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