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mr Goodwill [35]

3 years ago

11

The electric field at point P due to a point charge Q a distance R away has magnitude E. In order to double the magnitude of the

field at P, you could:__________.
a. double the distance to 2.
b. double the charge to 2 and at the same time reduce the distance to /2.
c. reduce the distance to /2.
d. reduce the distance to /4.
e. double the charge to 2.

1 answer:

Zinaida [17]3 years ago

4 0

Answer:

Option E, double the charge to 2, is the right answer.

Explanation:

Given the electric field at point = P

The point charge = Q

Distance =R

Magnitude = E

Due to a certain charge, the magnitude of the electric field at a point is defined as:

$E = \frac{kQ}{r^2}$

Here, we can see that E is proportionate to Q and 1/r^2

Hence, double the charge to 2 and at the same time reduce the distance to /2 will half the E

Therefore, doubling the charge will doubling E

So the answer is (e) is correct.

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Please help with these physics problems 1. A soccer ball is dropped from the top of a building. It takes 5.8 seconds to fall to

Tatiana [17]

Answer:

1.) h = 164.8 m

2.) U = 49.1 m/s

3.) t = 1.43 seconds

Explanation:

1.) A soccer ball is dropped from the top of a building. It takes 5.8 seconds to fall to the ground. The height of the building is...?

Since the soccer ball is dropped from the building, the initial velocity U will be equal to zero

Using second equation of motion

h = Ut + 1/2gt^2

Substitutes the time into the formula

h = 1/2 × 9.8 × 5.8^2

h = 164.8 m

2. The Falcon 9 launches to a height of 123 meters. What is its vertical initial velocity?

At maximum height final velocity = 0

Using the third law of motion

V^2 = U^2 - 2gH

0 = U^2 - 2 × 9.8 × 123

U^2 = 2410.8

U = 49.1 m/s

3. An apple falls from rest off a 10.m m tree. How long will it take before it hits the ground?

Since the apple fall from rest, the initial velocity U will be equal to zero

Using the second equation of motion,

h = Ut + 1/2gt^2

substitute all the parameters into the formula

10 = 1/2 × 9.8 × t^2

10 = 4.9t^2

t^2 = 10/4.9

t^2 = 2.04

t = 1.43 seconds

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3 years ago

Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rock

pashok25 [27]

Answer:

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

c) $H=294300\ m$

d) $t_T=544.95\ s$

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, $a=2g=2\times 9.81=19.62\ m.s^{-2}$

time for which the rocket accelerates, $t_a=100\ s$

<u>For the course of upward acceleration:</u>

using eq. of motion,

$s_a=ut+\frac{1}{2}at_a^2$

where:

$u=$ initial velocity of the rocket at the launch $=0$

$s_a=$ height the rocket travels just before its fuel finishes off

so,

$s_a=0+\frac{1}{2}\times 19.62\times 100^2$

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

<u>Now the velocity of the rocket just after the fuel is finished:</u>

$v_a=u+at_a$

$v_a=0+19.62\times 100$

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

$v^2=v_a^2-2gh$

where:

$g=$ acceleration due to gravity

$v=$ final velocity of the rocket at the top height

$0^2=1962^2-2\times 9.81\times h$

$h=196200\ m$

c) So the total height at which the rocket gets:

$H=h+s$

$H=196200+98100$

$H=294300\ m$

d)

Time taken by the rocket to reach the top height after the fuel is over:

$v=v_a+g.t$

$0=1962-9.81t$

$t=200\ s$

Now the time taken to fall from the total height:

$H=v.t'+\frac{1}{2}\times gt'^2$

$294300=0+0.5\times 9.81\times t'^2$

$t'=244.95\ s$

Hence the total time taken by the rocket to strike back on the earth:

$t_T=t_a+t+t'$

$t_T=100+200+244.95$

$t_T=544.95\ s$

e)

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

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3 years ago

A container in the shape of a cube 10.0 cm on each edge contains air (with equivalent molar mass 28.9 g/mol) at atmospheric pres

Vikentia [17]

Answer:

a) m = 1.174 grams

b) F_g = 0.01151 N

c) F_c = 1013 N

Explanation:

Given:

- The length of a cube L = 10.0 cm

- The molar mass of air M = 28.9 g/mol

- Pressure of air P = 101.3 KPa

- Temperature of air T = 300 K

- Universal Gas constant R = 8.314 J/kgK

Find:

(a) the mass of the gas

(b) the gravitational force exerted on it

(c) the force it exerts on each face of the cube

(d) Why does such a small sample exert such a great force? (6%)

Solution:

- Compute the volume of the cube:

V = L^3 = 0.1^3 = 0.001 m^3

- Use Ideal gas law equation and compute number of moles of air n:

P*V = n*R*T

n = P*V / R*T

n = 101.3*10^3 * 0.001 / 8.314*300

n = 0.04061 moles

- Compute the mass of the gas:

m = n*M

m = 0.04061*28.9

m = 1.174 grams

- The gravitational force exerted on the mass of gas is due to its weight:

F_g = m*g

F_g = 1.174*9.81*10^-3

F_g = 0.01151 N

- The force exerted on each face of cube is due its surface area:

F_c = P*A

F_c = (101.3*10^3)*(0.1)^2

F_c = 1013 N

- The molecules of a gas have high kinetic energy; hence, high momentum. When they collide with the walls they transfer momentum per unit time as force. Higher the velocity of the particles higher the momentum higher the force exerted.

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slavikrds [6]

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4 0

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If the pressure exerted on a 300.0 mL sample of hydrogen gas at constant temperature is increased from 0.500 kPa to 0.750 kPa, w

uranmaximum [27]

Answer:

200 mL

Explanation:

Given that,

Initial volume, V₁ = 300 mL

Initial pressure, P₁ = 0.5 kPa

Final pressure, P₂ = 0.75 kPa

We need to find the final volume of the sample if pressure is increased at constant temperature. It is based on Boyle's law. Its mathematical form is given by :

$V\propto \dfrac{1}{P}\\\\P_1V_1=P_2V_2$

V₂ is the final volume

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So, the final volume of the sample is 200 mL.

5 0

3 years ago