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sweet [91]
3 years ago
12

A charged object is immersed in a 17 N/C E-field. Then the total charge on the object increases by a factor of 22 over 10 s. Dur

ing this time, the electrostatic force on the object remains constant. Determine the magnitude of the E-field after the total charge has changed.
Physics
2 answers:
Marysya12 [62]3 years ago
4 0

Answer:

Eu = 0.65 N/C (Approx)

Explanation:

Given:

Time taken = 10 sec

E-field (Initial)(Ei) = 17 N/C

Find:

E-field (Final)

Assume;

Initial charge = Qi

Initial immersed force (Fi) = QiE

Fi = Qi(17)

After 10 sec

Fu = (26)(Qi)(Eu)

So

Fi = Fu

So,

Qi(17) = (26)(Qi)(Eu)

Eu = 17 / 26

Eu = 0.65 N/C (Approx)

vova2212 [387]3 years ago
4 0

Given :

Initial electric field , E_i=17\ N/C .

The total charge on the object increases by a factor of 22 .

To Find :

The magnitude of the E-field after the total charge has changed .

Solution :

Let , initial charge is q .

Therefore , final charge is 22q .

Electrostatic force F is given by :

F=qE

It is also given that the force remains constant .

Therefore ,

F_i=F_f\\\\qE_i=22qE_f\\\\E_f=\dfrac{E_i}{22}\\\\E_f=\dfrac{17}{22}\ N/C\\\\E_f=0.78\ N/C

Hence , this is the required solution .

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g A small object of mass 2.5 g and charge 18 uC is suspended motionless above the ground when immersed in a uniform electric fie
kvasek [131]

Answer: 1361.11\ N/C,\text{upward}

Explanation:

Given

Mass of particle is m=2.5\ gm

Charge of particle is q=18\ \mu C

Electrostatic force must balance the weight of the particle

\lim_{n \to \infty} a_n \Rightarrow mg=qE\\\\\Rightarrow E=\dfrac{2.5\times 9.8\times 10^{-3}}{18\times 10^{-6}}\\\\\Rightarrow E=1361.1\ N/C

Direction of the electric field is in upward direction such that it opposes the gravity force.

4 0
4 years ago
If2.0J of work is done in raising a 180g apple how far is it lifted?
Alex
We know, W = F * s
W = mg * s

Here, w = 2 J
m = 180 g = 0.180 Kg
g = 9.8 m/s

Substitute their values into the expression:
2 = 0.180*9.8 * s
1.764s = 2
s = 2 / 1.764
s = 1.13 meter

In short, Your Final Answer is 1.13 m

Hope this helps!

7 0
3 years ago
I'M GIVING 95POINTS PLZ HELP!!
Travka [436]

#1

initial height = 25000 ft

as we know that

1 ft = 0.3048 m

so we have

H = 25000(0.3048) = 7620 m

#2

If mass of the man = 90 kg

now the initial potential energy is given as

U = mgh

U = (90)(9.8)(7620)

U = 6.72 \times 10^6 J

#3

Gravitational force that is acting on it is given by

F = mg

F = 90(9.8) = 882 N

so this is the force by which earth is attracting him towards it as we can see due to this force the man is accelerating towards the earth

#4

As we know that gravitational force is given by

F = \frac{GMm}{r^2}

here we know that

r = R + h

R = radius of earth (6.37 \times 10^6 m)

h = 7620 m

now we have acceleration at that point is

a = \frac{F}{m}

a = \frac{GM}{r^2}

a = \frac{(6.67\times 10^{-11})(5.98 \times 10^{24})}{(6.37 \times 10^6 + 7620)^}

a = 9.806 m/s^2

#5

by energy conservation we have

KE = PE

\frac{1}{2}mv^2 = mgh

v^2 = 2gh

v^2 = 2(9.8)(7620)

v = 386.5 m/s

#6

final speed due to wind resistance is 150 mph

now we know that

1 mile = 1609 meter

so this speed in m/s is given as

150 mph = 150(\frac{1609 m}{3600 s})

v = 67 m/s

#7

No Luke is not accelerating when his speed is 150 mph

because at this speed his velocity will become constant

and since there is no change in velocity so his acceleration will become zero

#8

since at 150 mph the acceleration of Luke is zero

so net force on him must be zero

so we will have

wind force = weight or force due to earth

F_{wind} = 882 N

#9

If there is no wind resistance then there will be energy conservation

so KE = PE

KE = 6.72 \times 10^6 J

#10

Net will reduce the velocity of luke to be zero by taking long time

As we know that force on a system is given by

F = \frac{\Delta P}{\Delta t}

so if we increase the time interval here then we will have less force

so Luke will unhurt due to more time to decrease his speed to be zero

5 0
3 years ago
Which function represents g(x), a reflection of f(x) = 6(one-third) Superscript x across the y-axis?
Alisiya [41]

Answer:

g(x)=6(3)°x

g(x) = 6(3){x}

6 0
3 years ago
Which has a higher acceleration a 10-kg object acted upon with a net force of 20 N or an 18-kg object acted on by a net force of
Ulleksa [173]
First object:
m=10 kg
F=20 N

F=ma, then:
a=F/m
a=20/10=2 m/s²

Second object:
m=18 kg
F=30 N

a=F/m
a=30/18=1.6 m/s²

The first object has a higher acceleration.
6 0
4 years ago
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