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3 years ago

A point charge is placed 3 m from a 4 uC charge. What is the strength of the electric field on the point charge at this

Physics

2 answers:

brilliants [131]3 years ago

4 0

Answer:

$4,000N/C$

Explanation:

the information we have is:

distance: $r=3m$

charge: $q= 4\mu C$

since the prefix $\mu$ is equal to $1x10^{-6}$

the charge is: $q=4x10^{-6}C$

------------------

To calculate the strength of the electric field we use the following formula:

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

where $q$ is the charge, $r$ is the distance and $k$ is Coulomb's constant

$k=9x10^9Nm^2/C^2$

substituting all the known values into the formula:

$E=\frac{(9x10^9Nm^2/C^2)(4x10^{-6}C)}{(3m)^2} \\E=\frac{36,000Nm^2/C}{9m^2} \\E=4,00N/C$

The strength of the electric field on the point charge at the distance of 3m is 4,000 N/C.

mars1129 [50]3 years ago

3 0

Answer:4000v/m

Explanation:

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Determine W (or fuel energy) required to launch a satellite of mass m at rest from a launching pad placed at the surface earth,

jeka57 [31]

Answer:

5. 9GmM/(10R)

Explanation:

m is the mass of the satellite

M is the mass of the earth

W is the energy required to launch the satellite

Energy at earth surface = Potential energy (PE) + W

W = Energy at earth surface - Potential energy (PE)

But PE = $-\frac{GMm}{R}$

Therefore: W = Energy at earth surface - $\frac{GMm}{R}$

Energy at earth surface (E) at an altitude of 5R = $-\frac{GMm}{5r} +\frac{1}{2}mV^2$

But $V=\sqrt{\frac{GM}{5R} }$

Therefore: $E=-\frac{GMm}{5R}+\frac{1}{2}m(\sqrt{\frac{GM}{5R} } )^2= -\frac{GMm}{5R}+\frac{GMm}{10R} = -\frac{GMm}{10R}$

W = E - PE

$W=-\frac{GMm}{10R}-(-\frac{GMm}{R})=-\frac{GMm}{10R}+\frac{GMm}{R}=\frac{9GMm}{10R} \\W=\frac{9GMm}{10R}$

7 0

3 years ago

A scientist spends most of his time observing the night time sky and tracking the orbits of planets. Which branch of physical sc

gulaghasi [49]

Astronomy!!?
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4 0

3 years ago

Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?

Alexandra [31]

11) $1.04\cdot 10^7 J$

12) $1.04\cdot 10^7 J$

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

$PE=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

$g=9.8 m/s^2$ (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

$PE=(8325)(9.8)(127)=1.04\cdot 10^7 J$

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

$KE_t + PE_t = KE_b + PE_b$

where

$KE_t$ is the kinetic energy at the top

$PE_t$ is the potential energy at the top

$KE_b$ is the kinetic energy at the bottom

$PE_b$ is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

$KE_t=0$

Also, at the bottom the height is zero, so the potential energy is zero

$PE_b=0$

Therefore, we find:

$KE_b=PE_t=1.04\cdot 10^7 J$

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

$KE=1.04\cdot 10^7 J$

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

$v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s$

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

$KE_1 + PE_1 = KE_2 + PE_2$

where

$KE_1$ is the kinetic energy at the top of the 1st hill

$PE_1$ is the potential energy at the top of the 1st hill

$KE_2$ is the kinetic energy at the top of the 2nd hill

$PE_2$ is the potential energy at the top of the 2nd hill

From the previous questions, we know that

$KE_1=0$

and

$PE_1=1.04\cdot 10^7 J$

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

$PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J$

So, the kinetic energy at the second hill is

$KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J$

And so, the speed is

$v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s$

4 0

3 years ago

A professional cyclist rides a bicycle that is 92 percent efficient. For every 100 joules of energy he exerts as input work on t

emmainna [20.7K]

Efficiency = Work Output / Work Input

92% = Work Output / 100

0.92 = Work Output / 100

Work Output = 0.92 * 100

Work Output = 92 joules.

8 0

3 years ago

What are possible formulas for impulse? Check all that apply. J = Fdeltat J = StartFraction force over change in time EndFractio

Alex

<u>The possible formulas for impulse are as follows:</u>

J = FΔt

J = mΔv

J = Δp

Answer: Option A, E and F

<u>Explanation:</u>

The quantity which explains the consequences of a overall force acting on an object (moving force) is known as impulse. It is symbolised as J. When the average overall force acting on an object than such products are formed and in given duration than the start fraction force over change in time end fraction J = FΔt.

The impulse-momentum theorem explains that the variation in momentum of an object is same as the impulse applied to it: J = Δp J = mΔv if mass is constant J = m dv + v dm if mass changes. Logically, the impulse-momentum theorem is equivalent to Newton second laws of motion which is also called as force law.

6 0

3 years ago