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Vsevolod [243]
3 years ago
11

Explain when a falling object is in free fall.

Physics
2 answers:
lapo4ka [179]3 years ago
5 0

A falling object is in free fall if the only force acting on it is gravity.


Ksenya-84 [330]3 years ago
3 0
There is no gravity
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Explain how Pascal's principle can be used to design a fluid power system and describe how a fluid power system works.
zhenek [66]
Pascal's law of fluid transfer states that when there is an increase in fluid pressure, the rest of the extrinsic variables also increases. For example, in a flow of liquid in an orifice, there is a contraction of diameter in the orifice part. The fluid that will go in there increases in pressure and thereby an increase in velocity as well.
4 0
3 years ago
A bullet fired vertically at a velocity of 36m/s .after 45 the bullet hit the top of a bulid how height is a bulid?​
zzz [600]

Answer:

The height of the building is 8,302.5 m

Explanation:

Given;

velocity of the projectile, u = 36 m/s

time of motion, t = 45 s

Let the upward direction of the bullet be negative,

The height of the building is calculated as;

h = ut - \frac{1}{2} gt^2\\\\h = (36\times 45) - (\frac{1}{2} \times 9.8 \times 45^2)\\\\h = 1620 - 9922.5\\\\h = -8,302.5 \ m\\\\The \ height \ of \ the \ building \ is \ 8,302.5  \ m

3 0
2 years ago
What is the period of a wave that has a frequency of 300 hz?
Butoxors [25]

Answer:

T = 0.003 s

(Period is written as T)

Explanation:

Period = time it takes for one wave to pass (measured in seconds)

frequency = number of cycles that occur in 1 second

(measured in Hz / hertz / 1 second)

Period : T

frequency : f

So, if we know that the frequency of a wave is 300 Hz, we can find the period of the wave from the relation between frequency and period

T =  \frac{1}{f}    f = \frac{1}{T}

to find the period (T) of this wave, we need to plug in the frequency (f) of 300

T = \frac{1}{300}

T = 0.00333333333

So, the period of a wave that has a frequency of 300 Hz is 0.003 s

[the period/T of this wave is 0.003 s]

3 0
1 year ago
If a cylindrical space station 275 m in diameter is to spin about its central axis, at how many revolutions per minute (rpm) mus
sergij07 [2.7K]

The centripetal acceleration is responsible for the artificial gravity because the acceleration of an object moving in constant circular motion causing from net external force is called centripetal acceleration. It defines to the center or seeking the center.

Given the following:

Cylindrical space station diameter           = 275 meters; 137.5 meters for the radius

Standard gravity                                       = 9.80665 m/s²

 

Using the formula:

w² x r =g

w² = g / r

w² = 9.80665 m/s² / 137.5 m

w² = 9.80665 m/s² / 137.5 m

w² = 0.0713 s²

Then take the roots

w = 0.267 this is radians per second / 2 x (3.1416 which is the pi)

w = 0.0424 rps convert to rpm

w = 0.0424 r/s (1minute / 60 seconds)

w = 7.08 x 10⁻⁴ revolutions per minute

4 0
3 years ago
Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m
Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}

Substituting,

V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

V_{cm} = 1.034m/s

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where here v_f is the velocity after the collision.

v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}

v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

v_f = 1.034m/s

8 0
3 years ago
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