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Mademuasel [1]
3 years ago
15

Factors that affect pressure in fluid​

Physics
1 answer:
satela [25.4K]3 years ago
5 0

Answer:

The factors that affect are depth of the fluid and its density

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My very eager mother just served us nine pizzas
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A gymnast of mass 63.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
Sergio [31]

Answer:

Explanation:

A ) When gymnast is motionless , he is in equilibrium

T = mg

= 63 x 9.81

= 618.03 N

B )

When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.

T = mg

= 618.03 N

C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2

Net force on it = T - mg   , acting in upward direction

T - mg = m a

T =  mg + m a

= m ( g + a )

= 63 ( 9.81 + .6)

= 655.83 N

D )  If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2

Net force acting in downward direction

mg - T = ma

T = m ( g - a )

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6 0
3 years ago
Ions can have a positive or negative charge. Please select the best answer from the choices provided T F
mixas84 [53]

From among the choices provided, the more appropriate
answer is ' T ', the initial letter often used to represent
words that include 'true', 'truth', 'trust', etc., (as well as
'tree', 'train', 'transmit', 'Transylvania', 'trachea', 'travesty',
and 'trick', which are irrelevant to the present discussion).

This response is the most fitting and appropriate, because
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5 0
3 years ago
A top-fuel dragster starts from rest and has a constant acceleration of 42.0 m/s2. What are (a) the final velocity of the dragst
disa [49]

Answer:

a)  Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) The displacement of the dragster at the end of 1.8 s = 68.04 m

d) The displacement of the dragster at the end of 3.6 s = 272.16 m

Explanation:

a) We have equation of motion v = u + at

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 1.8 s    

Substituting

  v = u + at

  v  = 0 + 42 x 1.8 = 75.6 m/s

Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) We have equation of motion v = u + at

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 3.6 s    

Substituting

  v = u + at

  v  = 0 + 42 x 3.6 = 75.6 m/s

Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) We have equation of motion s= ut + 0.5 at²

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 1.8 s    

Substituting

   s= ut + 0.5 at²

    s = 0 x 1.8 + 0.5 x 42 x 1.8²

    s = 68.04 m

The displacement of the dragster at the end of 1.8 s = 68.04 m

d) We have equation of motion s= ut + 0.5 at²

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 3.6 s    

Substituting

   s= ut + 0.5 at²

    s = 0 x 3.6 + 0.5 x 42 x 3.6²

    s = 272.16 m

The displacement of the dragster at the end of 3.6 s = 272.16 m

3 0
2 years ago
A wave with a speed of 9 m/s and a frequency of 0.5 Hz has a λ of what?
patriot [66]

Wave speed = (wavelength) x (frequency)

Wavelength = (wave speed) / (frequency)

Wavelength = (9 m/s) / (0.5 Hz)

<em>Wavelength = 18 m</em>

6 0
2 years ago
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