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Julli [10]
3 years ago
14

Free fall is a situation in which the only force acting upon an object is gravity. Why do all objects in free fall have the same

acceleration?
Physics
2 answers:
Angelina_Jolie [31]3 years ago
5 0

free fall is a special type of motion in which the only force acting upon an object is gravity.  all objects will fall with the same rate of acceleration, regardless of their mass.

stich3 [128]3 years ago
5 0

Answer:

The force of gravity between an object of mass m and Earth M at distance r is as follows (G is gravitational constant):

F = (G) x (m) x (M) / (r^2)

Now, we also know that a force acting on the object is F = (mass) x (acceleration).

Rewrite the first formula as:

F = (m) x (G x M / r^2)

The (G x M / r^2) part is precomputed for the near-surface of Earth and is referred to a gravitational acceleration g:

F = (m) x (g)

As you can see g (or G x M / r^2) does not depend on the mass of the object, only on the mass of Earth but that is the same for all objects in free fall (towards Earth).


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Radar uses radio waves of a wavelength of 2.9 m . The time interval for one radiation pulse is 100 times larger than the time of
Mandarinka [93]

Answer:

145 m

Explanation:

Given:

Wavelength (λ) = 2.9 m  

we know,

c = f × λ  

where,

c = speed of light ; 3.0 x 10⁸ m/s

f = frequency  

thus,

f=\frac{c}{\lambda}

substituting the values in the equation we get,

f=\frac{3.0\times 10^8 m/s}{2.9m}

f = 1.03 x 10⁸Hz  

Now,

The time period (T) = \frac{1}{f}

or

T =  \frac{1}{1.03\times 10^8}  = 9.6 x 10⁻⁹ seconds  

thus,

the time interval of one pulse = 100T = 9.6 x 10⁻⁷ s  

Time between pulses = (100T×10) = 9.6 x 10⁻⁶ s  

Now,

For radar to detect the object the pulse must hit the object and come back to the detector.

Hence, the shortest distance will be half the distance travelled by the pulse back and forth.

Distance = speed × time = 3 x 10^8 m/s × 9.6 x 10⁻⁷ s) = 290 m {Back and forth}  

Thus, the minimum distance to target = \frac{290}{2} = 145 m

6 0
3 years ago
Is the bike rider in the picture above demonstrating kinetic energy or potential energy? you need to explain your answer.
soldi70 [24.7K]

Answer:

Kinetic energy

Explanation:

Kinetic energy is a function of velocity. Since the rider is moving at a certain speed, he's demonstrating kinetic energy. It can't be potential energy because potential energy encompass mgh

4 0
3 years ago
In the history of astronomy, which of the following is true? . A] Tycho Brahe built one of the first observatories. B] Ptolemy w
Gelneren [198K]
D. <span>Johannes Kepler argued that Earth was the center of the universe.

</span>
4 0
3 years ago
Read 2 more answers
I neeed help im so comfused Points Possible: 1, Points Correct: 0 Which group of numbers is listed from greatest to least? -3, -
Alona [7]

The groups aren't well formatted ;

The groups are ;

(-3, -1, 0, 2, 7) ; (9, 7, 6, -5, -4) ; (8, -6, 5, -4, 1) ; (-3, -4, -7, -8, -9)

Answer:

(-3, -4, -7, -8, -9)

Explanation:

Given the following group of numbers :

Evaluating each group of values for which is correctly arranged from greatest to least.

(-3, -1, 0, 2, 7) : - 1 is greater than - 3 (the group isn't arranged from greatest to least)

(9, 7, 6, -5, -4) : - 4 is greater than - 5 ; hence, the group isn't arranged from greatest to least

(8, -6, 5, -4, 1) : 5 is greater than - 6 ; hence, the group isn't arranged from greatest to least

(-3, -4, -7, -8, -9) ; the group of numbers here is arranged from greatest to least ;

(-3 > -4 > -7 > -8 > -9) ; hence, the correct group

5 0
3 years ago
A rotating wheel requires 5.00 s to rotate 28.0 revolutions. Its angular velocity at the end of the 5.00-s interval is 96.0 rad/
torisob [31]

Answer:

The constant angular acceleration of the wheel is 12.16 rad/s²

Explanation:

Given;

initial angular distance, θ = 28

time of the motion, t = 5 s

initial angular velocity is calculated as;

\omega _i = \frac{\theta}{t} = \frac{28}{5}.\frac{rev}{s}  \ \times \ \ \frac{2 \pi \ rad}{1 \ \ rev} = 35.19 \ rad/s

final angular velocity is given as, \omega _f = 96.0 \ rad/s

The constant angular acceleration is calculated as;

\alpha = \frac{\omega _f - \omega _i}{t} \\\\\alpha = \frac{96 - 35.19}{5} \\\\\alpha =  12.16 \ rad/s^2

Therefore, the constant angular acceleration of the wheel is 12.16 rad/s²

6 0
3 years ago
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