This table shows four examples of experiments.

Hey can anyone please help me with this it’s due in few hours and I’m stuck with ittt

Ray Of Light [21]

Answer:

Check body of the explanation

Explanation:

Ooook, quick theory rushdown. if you're at a depth of h in a tank of a fluid, the pressure is the sum of the atmosferic pressure (if the tank is open on top) plus a term which is the product of acceleration of gravity - about $10 ms^-^2$ , the density of whatever you're sinking in, and the depth at which you are. In formula, $p(h) = p_0 + \rho g h$ , and the pressure is the same for every point of the tank at the same depth.

At this point, we can start answering!

1a. The pressure at A is - not counting atmosferic pressure - $1000 * 10 * 1 = 10^4 Pa$ , while in B is $1000*10*2 = 2*10^4 Pa$ , so it's half of it.

1b. The two points are at the same depth, so the pressure is the same - they would be even if the two cilinders weren't linked!

1c. Ditto. Same depth? same pressure!

1d. Usual equation, this time density is 800. Pressure is $800*10*2 = 1,6*10^4 Pa$ : Since the density is 4/5 of water, the pressure is also 4/5 of the one exerted by water

2a. The volume is simply the product, so $4m*3m*2m = 24m^3$

2b. Density is defined as mass over volume, so you simply multiply the volume you found earlier by the density of paraffine: $800* 24 = 1,92 *10^4kg$

2c. Weight is defined as the mass of something times the acceleration due to gravity, in our case it's $1.92 *10^4 kg * 10 ms^{-2} = 1.92 * 10^5 N$

2d. $\rho gh$ again, what a surprise! $800 {kg \over m^3} * 10 {N \over kg}} * 2 m = 1,6* 10^4 {N\over m^2} =1.6*10^4 Pa$

3. Yet again, $\rho gh$ . $1000 {kg \over m^3} * 10 {N \over kg}} * 2 m = 2* 10^4 {N\over m^2} =2*10^4 Pa$

4 0

2 years ago

Using Gauss's law, calculate the electric field at a point distance s from a long wire bearing uniform charge density. i need he

11111nata11111 [884]

Answer:

E = 2k $\frac{\lambda}{ r}$

Explanation:

Gauss's law states that the electric flux equals the wax charge between the dielectric permeability.

We must define a Gaussian surface that takes advantage of the symmetry of the problem, let's use a cylinder with the faces perpendicular to the line of charge. Therefore the angle between the cylinder side area has the same direction of the electric field which is radial.

Ф = ∫ E . dA = E ∫ dA = q_{int} /ε₀

tells us that the linear charge density is

λ = q_ {int} /l

q_ {int} = l λ

we substitute

E A = l λ /ε₀

is area of cylinder is

A = 2π r l

we substitute

E = $\frac{ l \ \lambda}{ \epsilon_o \ 2\pi \ r \ l }$

E = $\frac{\lambda}{ 2\pi \epsilon_o \ r}$

the amount

k = 1 / 4πε₀

E = 2k $\frac{\lambda}{ r}$

5 0

3 years ago

Please help ASAP. In what direction do you need to apply force to move an object vertically?

PilotLPTM [1.2K]

The answer is up . tylrhscjwizn

5 0

3 years ago

a new planet is discovered that has twice the earth’s mass and twice the earth’s radius. on the surface of this new planet, a pe

Irina18 [472]

Answer: 250n

Explanation:

The formula for gravitational force is: F = (gMm)/r^2

There are two factors at play here:

1) The mass of the planet 'M'

2) The radius 'r'

We can ignore the small M and the g, they are constants that do not alter the outcome of this question.

You can see that both M and r are double that of earth. So lets say earth has M=1 and r=1. Then, new planet would have M=2 and r=2. Let's sub these two sets into the equation:

Earth. F = M/r^2 = 1/1

New planet. F = M/r^2 = 2/4 = 1/2

So you can see that the force on the new planet is half of that felt on Earth.

The question tells us that the force on earth is 500n for this person, so then on the new planet it would be half! So, 250n!

8 0

2 years ago

The experimental apparatus shown in the figure above contains a pendulum consisting of a 0.66 kg ball attached to a string of le

lara31 [8.8K]

The problem is solved and the questions are answered below.

Explanation:

a. To calculate the speed of the 0.66 kg ball just before the collision

V₀ + K₀ = V₁ + K₁

= mgh₀ = 1/2 mv₁²

where, h= r - r cosθ

V = $\sqrt{2gh}$

V = 2.42 m/s

b. Calculate the speed of the 0.22 kg ball immediately after the collision

y = y₀ + Vy₀t - 1/2 gt²

0 = 1.2 - 1/2 gt²

t = 0.495 s

x = x₀ + Vx₀t

1.4 = 0 + vx₀ (0.495)

Vx₀ = 2.83 m/s

C. To Calculate the speed of the 0.66 kg ball immediately after the collision

m₁ v₁ = m₁ v₃ + m₂ v₄

(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)

V₃ = 1.48 m/s

D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.

E. To Calculate the height to which the 0.66 kg ball rises after the collision

V₀ + k₀ = V₁ + k₁

1/2 mv₀² = mgh₁

h₁ = v₀²/2 g

= 0.112 m

F. Based on your data, No the collision is not elastic.

Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²

= 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²

= - 0.329 J

Hence, kinetic energy is not conserved.

8 0

3 years ago