Ask question

Ask question

All categories

4 years ago

15

a circular cylinder and isused to maintain a water depth of 4 m. That is, when the water depth exceeds 4 m, thegate opens slight

ly and lets the water flow under it. Determine the weight of the gate permeter of length

1 answer:

stiv31 [10]4 years ago

3 0

Answer:

W / A = 39200 kg / m²

Explanation:

For this problem let's use the equilibrium equation of / newton

F = W

Where F is the force of the door and W the weight of water

W = mg

We use the concept of density

ρ = m / V

m = ρ V

The volume of the water column is

V = A h

We replace

W = ρ A h g

On the other side the cylinder cover has a pressure

P = F / A

F = P A

We match the two equations

P A = ρ A h g

P = ρ g h

P = 39200 Pa

The weight of the water column is

W = 1000 9.8 4 A

W / A = 39200 kg / m²

You might be interested in

A small, rigid object carries positive and negative 3.00 nC charges. It is oriented so that the positive charge has coordinates

enot [183]

Answer:

Umax = 105.8nJ

Umin =-105.8nJ

Umax-Umin = 211.6nJ

Explanation:

8 0

3 years ago

What does the rotation of the Earth cause?

lord [1]

This rotation causes days. The relative position of us to the sun, and the Earth's rotation will cause the sun to appear to rise above the horizon and below, creating day and night.

7 0

3 years ago

Read 2 more answers

A 0.50-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.9 m on a frictionless horizontal su

Sedaia [141]

Answer:

$\boxed {\boxed {\sf 18 \ m/s}}$

Explanation:

The ball is moving in a circle, so the force is centripetal.

One formula for calculating centripetal force is:

$F_c= \frac{mv^2}r}$

The mass of the ball is 0.5 kilograms. The radius is 1.9 meters. The centripetal force is 85 Newtons or 85 kg*m/s².

$F_c$ = 85 kg*m/s²
m= 0.5 kg
r= 1.9 m

Substitute the values into the formula.

$85 \ kg*m/s^2 = \frac{0.5 \ kg *v^2}{1.9 \ m}$

Isolate the variable v. First, multiply both sides by 1.9 meters.

$(1.9 \ m)(85 \ kg*m/s^2) = \frac{0.5 \ kg *v^2}{1.9 \ m}*1.9 \ m$

$(1.9 \ m)(85 \ kg*m/s^2) = {0.5 \ kg *v^2}$

$161.5 \ kg*m^2/s^2 = 0.5 \ kg*v^2$

Divide both sides by 0.5 kilograms.

$\frac {161.5 \ kg*m^2/s^2}{0.5 \ kg} = \frac{0.5 \ kg*v^2}{0.5 \ kg}$

$\frac {161.5 \ kg*m^2/s^2}{0.5 \ kg} =v^2$

$323 \ m^2/s^2 = v^2$

Take the square root of both sides of the equation.

$\sqrt {323 \ m^2/s^2} =\sqrt{ v^2$

$\sqrt {323 \ m^2/s^2} =v$

$17.9722007556 \ m/s =v$

The original measurements have 2 significant figures, so our answer must have the same.

For the number we found, 2 sig fig is the ones place. The 9 in the tenth place tells us to round the 7 to an 8.

$18 \ m/s =v$

The maximum speed is approximately <u>18 meters per second.</u>

8 0

3 years ago

20.0 Ω resistor and a 2.50 μF capacitor are connected in parallel with signal generator. The signal generator produces a sinusoi

Free_Kalibri [48]

Answer:

0.15A

Explanation:

The parameters given are;

R=20.0 Ω

C= 2.50 μF

V= 3.00 V

f= 2.48×10^-3 Hz

Xc= 1/2πFc

Xc= 1/2×3.142 × 2.48×10^-3 × 2.5 ×10^-6

Xc= 25666824.1

Z= 1/√(1/R)^2 +(1/Xc)^2

Z= 1/√[(1/20)^2 +(1/25666824.1)^2]

Z= 1/√(2.5×10^-3) + (1.5×10^-15)

Z= 20 Ω

But

V=IZ

Where;

V= voltage

I= current

Z= impedance

I= V/Z

I= 3.00/20

I= 0.15A

6 0

3 years ago

A 1.2-kg mass is projected from ground level with a velocity of 31.3 m/s at some unknown angle above the horizontal. A short tim

My name is Ann [436]

Answer:

The kinetic energy will be "399.65 J".

Explanation:

Given:

Mass,

m = 1.2 kg

Velocity,

v = 31.3 m/s

The total energy of mass will be:

⇒ $E=K+U$

or,

⇒ $E=.5mv^2+mgh$

By putting the values, we get

$=.5(1.2)(31.3)^2+0$

$=0.6\times 979.69+0$

$=587.81 \ J$

Since,

The system's total energy is unchanged, then

⇒ $E=K+U$

or,

⇒ $E=K+mgh$

$587.81=K+1.2(9.8)(16)$

$587.81=K+188.16$

$K=587.81-188.16$

$=399.65 \ J$

8 0

3 years ago