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3 years ago

Equation for pressure at a depth H inside a fluid PLSS URGENTT

Physics

1 answer:

Blababa [14]3 years ago

4 0

Answer:

We begin by solving the equation P = hρg for depth h: h=Pρg h = P ρ g . Then we take P to be 1.00 atm and ρ to be the density of the water that creates the pressure.

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At a distance of 0.220 cm from the axis of a very long charged conducting cylinder with radius 0.100cm, the electric field is 49

Scorpion4ik [409]

Answer:

At the distance of 0.220cm from the axis.

r = 0.220cm = 0.0022m, E = 490N/C, e0 = 8.854 x 10^-12F/m

Linear charge density = 2*π*e0*r*E = 2 x 3.142 x 8.854x10^-12 x 0.0022 x 490 = 5.998 x 10^-11C/m

Thus, To Calculate the Electric field at the distance r = 0.616cm from the cylinder axis, we substitute the calculated linear change density in the equation

E = (linear charge density)/2*π*e0*r

Here, r = 0.616cm = 0.00616m

E = [(5.998 x 10^-11)/(2 x 3.142 x 8.854 x 10-12 x 0.00616)]

E = 175N/C

Explanation:

The Electric field of a charged conducting cylinder obey the Gauss Law.

Therefore, the Electric field is given as:

E = (linear charge density)/4πe0r,

Where e0 is the permittivity of free space with constant value of 8.854 x 10^-12F/m, r is the radial distance from the axis.

3 0

4 years ago

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau

Thepotemich [5.8K]

Answer:

$T=6.75s$

Explanation:

We must separate the motion into two parts, the first when the rocket's engines is on and the second when the rocket's engines is off. So, we need to know the height ( $h_1$ ) that the rocket reaches while its engine is on and we need to know the distance ( $h_2$ ) that it travels while its engine is off.

For solving this we use the kinematic equations:

In the first part we have:

$h_1=v_0T+\frac{1}{2}aT^2\\h_1=0*T+\frac{1}{2}(16\frac{m}{s^2})T^2\\h_1=8\frac{m}{s^2}T^2\\$

and the final speed is:

$v_f=v_0+aT\\v_f=0+16\frac{m}{s^2}T\\v_f=16\frac{m}{s^2}T$

In the second part, the final speed of the first part it will be the initial speed, and the final speed is zero, since gravity slows it down the rocket.

So, we have:

$v_f^2=v_0^2+2gh_2\\2gh_2=v_f^2-v_0^2\\h_2=\frac{v_f^2-v_0^2}{2g}\\h_2=\frac{0^2-(16\frac{m}{s^2}T)^2}{2(-9.8\frac{m}{s^2})}\\h_2=\frac{-256\frac{m^2}{s^4}T^2}{-19.6\frac{m}{s^2}}\\h_2=13.06\frac{m}{s^2}T^2$

The sum of these heights will give us the total height, which is known:

$h=h_1+h_2\\960m=8\frac{m}{s^2}T^2+13.06\frac{m}{s^2}T^2\\960m=21.06\frac{m}{s^2}T^2\\T^2=\frac{960m}{21.06\frac{m}{s^2}}\\T^2=45.58s^2\\T=\sqrt{45.58s^2}\\T=6.75s$

This is the time that its needed in order for the rocket to reach the required altitude.

5 0

3 years ago

Select the situation for which the torque is the smallest.

alina1380 [7]

Answer:

e. The torque is the same for all cases.

Explanation:

The formula for torque is:

τ = Fr

where,

τ = Torque

F = Force = Weight (in this case) = mg

r = perpendicular distance between force an axis of rotation

Therefore,

τ = mgr

Here,

m = 200 kg

r = 2.5 m

Therefore,

τ = (200 kg)(9.8 m/s²)(2.5 m)

τ = 4900 N.m

Here,

m = 20 kg

r = 25 m

Therefore,

τ = (20 kg)(9.8 m/s²)(25 m)

τ = 4900 N.m

Here,

m = 8 kg

r = 62.5 m

Therefore,

τ = (8 kg)(9.8 m/s²)(62.5 m)

τ = 4900 N.m

Hence, the correct answer will be:

e. The torque is the same for all cases.

6 0

4 years ago

Which of the following types of energy are present at some point in the energy transfer process in a nuclear power plan? Select

Alexeev081 [22]

Solar Energy is the answer to the question tell me if i`m wrong

7 0

3 years ago

IF YOU GOOD AT SCIENCE PLEASE ANSWER THIS ASAP

xz_007 [3.2K]

Answer:

I would believe that it would be the last option

Explanation:

Physical science is a type of science that mainly focuses on natural objects that are not alive, such as minerals and rocks.

4 0

3 years ago

Read 2 more answers