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3 years ago

A rock at rest has weight 138 n. what is the weight of the rock when it is accelerating downward at 3.5 m/s2?

1 answer:

7 0

<span>The weight of the rock is 138n when accelerating downward at 3.5m/ s2.Weight is a measurement of the force of gravity on an objects mass. Therefore as long as the force of gravity is the same and the mass of the rock is the same the weight should also be the same.</span>

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For a fixed amount of gas at a fixed temperature, what will happen if the volume is doubled?.

mash [69]

Answer:

the pressure will decrease by 1/2

Explanation:

PV=nRT

P=(nRT)/(V)

nRT are all constant so they will equal 1

V is 2

P=1/2

4 0

2 years ago

A fishbowl has a circular opening with a diameter of 13 cm. The fishbowl sits upright on a table in a magnetic field of 0.00110

azamat

Answer:

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Explanation:

5 0

2 years ago

A simple and common technique for accelerating electrons is shown in the figure, where there is a uniform electric field between

Sunny_sXe [5.5K]

Answer : $4.483\times 10^{15}\ m/s^2$ .

Explanation:

It is given that,

Electric field strength, $E=2.55\times 10^{4}\ N/C$

We know that,

Charge of electron, $q=1.6\times 10^{-19}\ C$

Mass of electron, $m=9.1\times 10^{-31}\ kg$

From the definition of electric field, $F=qE$ ...............(1)

According to Newton's second law, F = ma..........(2)

From equation (1) and (2)

$ma=qE$

$a=\dfrac{qE}{m}$

$a=\dfrac{1.6\times 10^{-19}\ C\times 2.55\times 10^{4}\ N/C }{9.1\times 10^{-31}\ kg}$

$a=0.4483\times 10^{16}\ m/s^2$

$a=4.483\times 10^{15}\ m/s^2$

So, the horizontal component of acceleration of an electron is $4.483\times 10^{15}\ m/s^2$ .

Hence, it is the required solution.

7 0

3 years ago

Four point charges have equal magnitudes. Three are positive, and one is negative, as the drawing shows. They are fixed in place

Semmy [17]

Answer:

Ans= 9

See attached picture for clearer solution.

Explanation:

The net electrostatic force acting on charge A = 2/ 2 + 2 /(2) 2 − 2 /(3) 2 = 2 / 2 (1 + 1/4 – 1/9 ) = 41/36 2/2 .

The net electrostatic force acting on charge B = 2/2 + 2/(2)2 − 2/2 = 1/4 2/d2 .

The net electrostatic force acting on charge C = 2/2 + 2/(2)2 + 2/2 = 2/2 (1 + 1 4 + 1) = 9/4 2/2 .

The net electrostatic force acting on charge D = 2/2+ 2 /(2)2 + 2/(3)2 = 2 /2 (1 + 1/4 + 1/9 ) = 49/36 2/ 2 .

The ratio of the largest to the smallest net force = 9/4*2/2 / 1/4 2/2 . = 9

5 0

2 years ago

jane pushed a box with 15 N force (an action) what force does the box have on Jane? which law is this?

lord [1]

Answer:

This is Newton’s third law (forces due to friction). There would be 15N being pushed back at Jane.

Explanation:

Newton’s third law states that for every action there is an equal and opposite reaction. So if Jane pushed with a force of 15N, 15N would be pushed back at her. Hope this helps! :)

5 0

2 years ago