All categories

3 years ago

How do you find the water pressure at the bottom of the 55-m-high water tower?

1 answer:

5 0

-- What's the volume of a cylinder with radius=1m and height=55m ?

   ( Volume of a cylinder = π R² h )

-- How much does that volume of water weigh ?

    1 liter of water = 1 kilogram of mass
Weight = (mass) x (acceleration of gravity)

-- What's the area of the bottom of that 1m-radius cylinder ?

   Pressure = (force) / (area)

You might be interested in

A car moving at 30 m/s slows uniformly to a speed of 10 m/s in a time of 5 s. Determine 1. The acceleration of the car. 2. The d

ValentinkaMS [17]

Answer:

Explanation:

Initial velocity , u = 30 m/s

final velocity , v = 10 m/s

time , t = 5 seconds

1. Acceleration = v - u / t

= 10 - 30 / 5

= -20 / 5

= <u><em>- 4 m/s</em></u>

8 0

3 years ago

Fusion occurs with atoms of elements less massive than

Nimfa-mama [501]

The answer is A. Iron.

4 0

2 years ago

Two blocks, A and B, are connected by a rope. A second rope is connected to block B and a steady, horizontal tension force of T

Oksanka [162]

Answer:

40 N

Explanation:

We are given that

Speed of system is constant

Therefore, acceleration=a=0

Tension applied on block B=T=50 N

Friction force=f=10 N

We have to find the friction force acting on block A.

Let T' be the tension in string connecting block A and block B and friction force on block A be f'.

For Block B

$T-f-T'=m_Ba$

Where $m_B$ =Mass of block B

Substitute the values

$50-10-T'=m_B\times 0=0$

$T'==40 N$

For block A

$T'-f'=m_Aa$

Where $m_A=$ Mass of block A

Substitute the values

$40-f'=m_A\times 0=0$

$f'=40 N$

Hence, the friction force acting on block A=40 N

3 0

2 years ago

Two identical cars, one on the moon and one on the earth, have the same speed and are rounding banked turns that have the same r

Naily [24]

Answer:

The value of the centripetal forces are same.

Explanation:

Given:

The masses of the cars are same. The radii of the banked paths are same. The weight of an object on the moon is about one sixth of its weight on earth.

The expression for centripetal force is given by,

$F_{c} = \dfrac{mv^{2}}{r}$

where, $m$ is the mass of the object, $v$ is the velocity of the object and $r$ is the radius of the path.

The value of the centripetal force depends on the mass of the object, not on its weight.

As both on moon and earth the velocity of the cars and the radii of the paths are same, so the centripetal forces are the same.

3 0

2 years ago

Free runners jump long distances and land on the ground or a wall. How do they apply Newton’s second law to lessen the force of

Veseljchak [2.6K]

As we know that as per Newton's II law we have

$F = \frac{dP}{dt}$

here we will have

$dP$ = change in momentum

$dt$ = time interval in which momentum is changed

now in order to have least injury during jumping we need to have least force on the jumper

so in order to have least force we can say that the momentum must have to change in maximum time so that amount of force must be least

So we need to increase the time in which momentum of the system is changed

5 0

2 years ago