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elena-14-01-66 [18.8K]
2 years ago
11

How do fluids excert pressure​

Physics
1 answer:
Artyom0805 [142]2 years ago
5 0

All fluids exert pressure like the air inside a tire. The particles of fluids are constantly moving in all directions at random. As the particles move, they keep bumping into each other and into anything else in their path. These collisions cause pressure, and the pressure is exerted equally in all directions.

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To maintain the same amount of torque due to a mass on a balance as the mass is increased, how should the position of the mass c
olasank [31]

Answer:

the mass should be bring closer to the point about which we are finding torque

Explanation:

τ = Σr × F = rmg

where m is the mass, g is acceleration due to gravity, and r is the distance

Torque is directly proportional to -

1.mass, m , of object

2. distance, r, of the mass from the point about which we are finding the torque.

So if we increase or decrease them then the torque will also increase or decrease.

So if we increase the mass the torque will increase but since we have to maintain same torque therefore we have to decrease the distance of mass from the point about which we are finding torque.

Therefore the mass should be bring closer to the point about which we are finding torque.

8 0
3 years ago
Two people, one of mass 85 kg and the other of mass 50 kg, sit in a rowboat of mass 90 kg. With the boat initially at rest, the
sergeinik [125]

Answer:

0.11m

Explanation:

let's assume the boat is of uniform construction

Ignoring friction losses

Also assume the origin is at the end of the boat originally with the heavier person

the center of mass of the whole system will not change relative to the water when the two swap ends

Originally, the center of mass is

85[0] + 90[3.5/2] + 50[3.5] / (85 + 90 + 50) = 1.14m from the origin

after the swap, the center of mass is

50[0] + 90[3.5/2] + 85[3.5] / (85 + 90+ 50) = 1.030m from the origin

The center of mass has shifted

1.14-1.030 = 0.11m

as no external force acted on the system, the center of mass relative to the water will not change. The boat will therefore shift towards the end where the heavier person originally sat

4 0
3 years ago
Un auto de 2,300 kg está estacionado sobre una rampa inclinada 26.0°. Obtenga la tensión en la cadena.
Natasha_Volkova [10]

Answer:

T = mgsinθ = 2300(9.8)sin26.0 = 9880.88 ≈ 9900 N

Explanation:

8 0
2 years ago
A subway train is traveling at 22.2 m/s when it approaches a slower train 50m ahead traveling in the same direction at 6.94 m/s.
Amiraneli [1.4K]

Answer:

Time that they collide = 4.99s

Relative speed of the trains when they collide: The relative speed of The first train relative to the second, slower train at collision = 4.781 m/s

Explanation:

We will use the equations of motion to obtain the solution required

At time t = 0

speed of first train = 22.2 m/s

Initial space between the two trains = 50 m

Speed of second train = 6.94 m/s

For the first car, distance covered by the first train = y

y = distance covered between the beginning of the deceleration and the point where the the two trains hit one another.

u = initial velocity = 22.2 m/s

t = time taken for all this to happen

a = deceleration = - 2.1 m/s²

y = ut + (1/2)at²

y = 22.2t - 1.05t² (eqn 1)

For the second train,

At t = 0, y = 50 m

Let the new distance moved by the second train before collision = (y - 50)

u = initial velocity = 6.94 m/s

t = time taken = t

a = acceleration of the second train = 0 m/s² (constant velocity)

(y - 50) = ut + (1/2)at²

y - 50 = 6.94t

y = 6.94t + 50 (eqn 2)

substituting for y in eqn 2 using the expression obtained in eqn 1

y = 22.2t - 1.05t²

y = 6.94t + 50

22.2t - 1.05t² = 6.94t + 50

1.05t² - 15.26t + 50 = 0

Solving this quadratic equation

t = 4.99 s or 9.54 s

The position of the two trains are the same at those two times, but the first time is when they hit each other.

t = 4.99 s

At 4.99 s, the the velocity of the first train

v = u + at

v = 22.2 + (-2.1×4.99) = 11.721 m/s in the same direction as the second train.

Relative velocity at this point will be

= 11.721 - 6.94 = 4.781 m/s

Relative speed of the trains when they collide: The relative speed of The first train relative to the second, slower train at collision = 4.781 m/s

Hope this Helps!!!

4 0
2 years ago
If izzy mass is 0.3kg he applide 657.9n force what will be the accelration​
Sholpan [36]

Answer:

The acceleration of the body, a = 2193 m/s²

Explanation:

Given,

The mass of the body, m = 0.3 kg

The force acting on the body, F = 657.9 N

The force acting on an object is proportional to the product of mass and acceleration of the body.

                         F = m x a

Therefore, the acceleration of the body is

                           a = F / m

                              = 657.9 N / 0.3 kg

                              = 2193 m/s²

Hence, the acceleration of the body, a = 2193 m/s²

4 0
3 years ago
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