What is heat? I need help

The gauge pressure at the bottom of a cylinder of liquid is 0.30atm. The liquid is poured into another cylinder with twice the r

likoan [24]

Answer:

$P_g' = 0.075 atm$

Explanation:

Gauge pressure at the bottom of the cylinder depends on the height of water in the cylinder

So here we can say that

$P_g = \rho g h$

now when liquid is filled to height "h" in base area "A" then gauge pressure of the liquid at the bottom is given as

$P_g = 0.30 atm$

now we put the whole liquid into another cylinder with twice radius of the first cylinder

So area becomes 4 times

now by volume conservation we can say that if area is increased by 4 times then height of liquid will decrease by 4 times

so we have

$h' = \frac{h}{4}$

so gauge pressure is given as

$P_g' = \frac{0.30}{4} = 0.075 atm$

5 0

3 years ago

Before the big bang, the universe was much ____ and _____ than it is now.

Dahasolnce [82]

The answer to the question is that before the big bang, the universe was much hotter and more dense than it is now. Letter B.
It is because after the big bag occurred, the universe became cooler and less dense.
a. - does not correspond in the answer because the universe became less dense after the big bang.
c - the universe became cool and less dense after the big bang so being cool and less dense does not correspond to the question.
d - cooler does not answer the question because it only became cooler after the big bang.

5 0

3 years ago

Read 2 more answers

mass of the planet is 12 times that of earth and its radius is thrice that of earth , then find the escape velocity on that plan

Over [174]

Answer:

The escape velocity on the planet is approximately 178.976 km/s

Explanation:

The escape velocity for Earth is therefore given as follows

The formula for escape velocity, $v_e$ , for the planet is $v_e = \sqrt{\dfrac{2 \cdot G \cdot m}{r} }$

Where;

$v_e$ = The escape velocity on the planet

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m = The mass of the planet = 12 × The mass of Earth, $M_E$

r = The radius of the planet = 3 × The radius of Earth, $R_E$

The escape velocity for Earth, $v_e_E$ , is therefore given as follows;

$v_e_E = \sqrt{\dfrac{2 \cdot G \cdot M_E}{R_E} }$

$\therefore v_e = \sqrt{\dfrac{2 \times G \times 12 \times M}{3 \times R} } = \sqrt{\dfrac{2 \times G \times 4 \times M}{R} } = 16 \times \sqrt{\dfrac{2 \times G \times M}{R} } = 16 \times v_e_E$

$v_e$ = 16 × $v_e_E$

Given that the escape velocity for Earth, $v_e_E$ ≈ 11,186 m/s, we have;

The escape velocity on the planet = $v_e$ ≈ 16 × 11,186 ≈ 178976 m/s ≈ 178.976 km/s.

3 0

3 years ago

When two hydrogen atoms bond, the positive nucleus of one atom attracts the a. negative nucleus of the other atom. c. negative e

Aleksandr-060686 [28]

Answer:

c

Explanation:

4 0

3 years ago

The rectangular region of the xy plane shown has nonuniform surface charge density σ = σ0 (1+ y b), where σ0 is a constant. Find

sergiy2304 [10]

Answer:

This is net charge on the surface is Q = σ₀ x (y + 2by²)

Explanation:

The surface charge density is defined as the amount of charge Q per unit area A

σ = dq / dA

dq = σ dA

Since the surface is a rectangular region we use an xy coordinate system so the area difference

dA = dxdy

dq = σ dx dy

We replace, evaluate the integral

∫ dq = ∫ σ₀ (1 + yb) dxdy

realizamos laintegral de dx

Q -0 =σ₀ ∫ (1 + yb) (x-0) dy

Where we evaluate We must recognize that the charge Q must be zero by the time X = 0 and Y = 0. At the starting point Q = 0 for x = 0

We perform the other integral (dy)

Q = σ₀ x (y + 2y² b)

Evaluated between Y = 0 and Y = y

Q = σ₀ x (y + 2by²)

This is net charge on the surface

8 0

3 years ago