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3 years ago

Please help me!

Physics

2 answers:

mihalych1998 [28]3 years ago

7 0

The answer would be mass movement

Kay [80]3 years ago

5 0

Runoff because the mud is a liquid and moves on an amount of water that is in it like with quicksand

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What emerging technologies will make hydroelectric energy source safer, more usable, more efficient, cleaner, etc?

vitfil [10]

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3 years ago

What is question 22?

Fynjy0 [20]

the answer is a!! its pretty simple I just read the graph.

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3 years ago

Read 2 more answers

g If the interaction of a particle with its environment restricts the particle to a finite region of space, the result is the qu

Travka [436]

Answer:

the result is the quantization of __Energy__ of the particle

Explanation:

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4 years ago

Three equal point charges, each with charge 1.45 μCμC , are placed at the vertices of an equilateral triangle whose sides are of

LUCKY_DIMON [66]

Answer:

U = 80.91 J

Explanation:

In order to calculate the electric potential energy between the three charges you use the following formula:

$U=k\frac{q_1q_2}{r_{1,2}}$ (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q1: q2 charge

r1,2: distance between charges 1 and 2.

For the three charges you have:

$U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}$ (2)

You use the fact that q1=q2=q3=q and that the distance between charges are equal. Then, in the equation (2) you have:

q = 1.45μC = 1.45*10^-6C

r = 0.700mm = 0.700*10^-3m

$U_T=3k\frac{q^2}{r}=3(8.98*10^9Nm^2/C^2)\frac{(1.45*10^{-6}C)}{0.700*10^{-3}m}\\\\U_T=80.91J$

The electric potential energy between the three charges is 80.91 J

7 0

3 years ago

1. A hydraulic lift is to be used to lift a truck masses 5000 kg. If the diameter of the

zheka24 [161]

Answer:

a) F = 4.9 10⁴ N, b) F₁ = 122.5 N

Explanation:

To solve this problem we use that the pressure is transmitted throughout the entire fluid, being the same for the same height

1) pressure is defined by the relation

P = F / A

to lift the weight of the truck the force of the piston must be equal to the weight of the truck

∑F = 0

F-W = 0

F = W = mg

F = 5000 9.8

F = 4.9 10⁴ N

the area of the pisto is

A = pi r²

A = pi d² / 4

A = pi 1 ^ 2/4

A = 0.7854 m²

pressure is

P = 4.9 104 / 0.7854

P = 3.85 104 Pa

2) Let's find a point with the same height on the two pistons, the pressure is the same

$\frac{F_1}{A_1} = \frac{F_2}{A_2}$

where subscript 1 is for the small piston and subscript 2 is for the large piston

F₁ = $\frac{A_1}{A_2} \ F_2$

the force applied must be equal to the weight of the truck

F₁ = $( \frac{d_1}{d_2} )^2\ m g$

F₁ = (0.05 / 1) ² 5000 9.8

F₁ = 122.5 N

7 0

3 years ago