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Mashcka [7]
3 years ago
15

Three individual point charges are placed at the following positions in the x-y plane:Q3= 5.0 nC at (x, y) = (0,0);Q2= -3.0 nC a

t (x, y) = (4 cm, 0); and Q1= ?nC at (x, y) = (2 cm,0);What isthe magnitude, and sign, ofcharge Q1such that the net force exerted on charge Q3, exerted bycharges Q1and Q2, is zero?
Physics
1 answer:
ozzi3 years ago
8 0

Answer:

Explanation:

net force exerted on charge Q₃, exerted by charges Q₁and Q₂, will be  zero

if net  electric field due to charges Q₁ and Q₂  at origin is zero .

electric field due to Q₂

= 9 X 10⁹ X 3 x10⁹ / .04²

electric field due to Q₁

= 9 X 10⁹ X Q₁ / .02²

For equilibrium

9 X 10⁹ X Q₁ / .02² = 9 X 10⁹ X 3 x10⁻⁹ / .04²

Q₁  = 3 X10⁻⁹ x .02² / .04²

= 3 / 4 x 10⁻⁹

.75 x 10⁻⁹  C

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As his socks tumbled in the dryer, they became charged. If a small piece of lint with a charge of +1.25 E -19 C is attracted to
qwelly [4]

Answer:

2.4\cdot 10^{10}N

Explanation:

When an electric charge is immersed in an electric field, it experiences a force given by the equation

F=qE

where

q is the charge

E is the magnitude of the electric field

In this problem, we have:

q=+1.25\cdot 10^{-19}C is the charge on the small piece of lint

F=3.0\cdot 10^{-9}N is the force experienced by the  charge

Therefore, we can find the magnitude of the electric field by re-arranging the equation and solving for E:

E=\frac{F}{q}=\frac{3.0\cdot 10^{-9}}{1.25\cdot 10^{-19}}=2.4\cdot 10^{10}N

5 0
4 years ago
A peregrine falcon dives at a pigeon. The falcon starts with zero downward velocity and falls with the acceleration of gravity.
Reptile [31]

Answer:

t = 3.94 s

Explanation:

This can be modeled as a case of free fall motion. Because, the falcon is going down with acceleration, that is equal to acceleration due to gravity. To find the time taken by the falcon to intercept the pigeon, we will use second equation of motion for vertical direction:

h = v_{i}t + \frac{1}{2}gt^2

where,

h = height = 76 m

vi = initial speed of falcon = 0 m/s

t = time required = ?

g = acceleration due to gravity = 9.81 m/s²

Therefore,

76\ m = (0\ m/s)(t) + \frac{1}{2}(9.81\ m/s^2)t^2\\t^2 = \frac{(2)(76\ m)}{9.8\ m/s^2}\\\\t = \sqrt{ 15.49\ s^2}\\

<u>t = 3.94 s</u>

5 0
3 years ago
A 100 kg boat is floating in water. half of the boat is submerged under water. what is the weight of the boat?
Alex17521 [72]
W=MG
w is weight 
m is mass
g is gravity 
W=(100 kg)(9.8 m/s)
W= 980 N
hope this helps
7 0
3 years ago
4. A car accelerates at 2.5 m/s^2, covers 4 km in 0.8 min. How fast was it moving at the beginning
alukav5142 [94]

Answer:

Initial velocity, u = 23.33 m/s

Explanation:

Given the following data;

Acceleration = 2.5 m/s²

Distance = 4 km to meters = 4000 meters

Time = 0.8 mins to seconds = 0.8 * 60 = 48 seconds.

To find the initial velocity, we would use the second equation of motion;

S = ut + \frac {1}{2}at^{2}

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

4000 = u*48 + ½*2.5*48²

4000 = 48u + 1.25*2304

4000 = 48u + 2880

48u = 4000 - 2880

48u = 1120

Initial velocity, u = 1120/48

Initial velocity, u = 23.33 m/s

8 0
3 years ago
How fast is a car going if he travels 2,100 km in 21 hr?
Tcecarenko [31]

Answer:

about 62mph or 100 kph

Explanation:

3 0
3 years ago
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