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sveta [45]
1 year ago
6

You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh

at is the direction of the force on the charge you are holding
Physics
1 answer:
lara [203]1 year ago
3 0

If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.

Thus the electric force applied on the charge in my hand due to each other is,

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

Here k is the Coulomb constant, and r is the distance between the charges.

It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}

And the force on the charge due to the charge on the north is,

\vec{F_2}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{j}

As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of 45^\circ from the north or from the north direction.

Thus the net force is acting in the north-east direction.

Learn more about the direction of the force here,

brainly.com/question/2037071

#SPJ4

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Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m ea
Kipish [7]

Answer:

6.5 m/s

Explanation:

We are given that

Distance, s=100 m

Initial speed, u=1.4 m/s

Acceleration, a=0.20 m/s^2

We have to find the final velocity at the end of the 100.0 m.

We know that

v^2-u^2=2as

Using the formula

v^2-(1.4)^2=2\times 0.20\times 100

v^2-1.96=40

v^2=40+1.96

v^2=41.96

v=\sqrt{41.96}

v=6.5 m/s

Hence, her final velocity at the end of the 100.0 m=6.5 m/s

5 0
2 years ago
lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A
qaws [65]

2.6×10^6\:\text{m}

Explanation:

The acceleration due to gravity g is defined as

g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

5 0
3 years ago
QUESTION 1
Sveta_85 [38]

distance from the Sun of 2.77 astronomical units or about 414 million km 257 million miles and orbiting period of 4.62 years

7 0
3 years ago
The electron gun in a television tube accelerates electrons (mass = 9.11x10^-31 kg, charge = 1.6 x 10^-19C) from rest to 3 x10^7
stich3 [128]

Answer:

Electric field acting on the electron is  127500 N/C.

Explanation:

It is given that,

Mass of an electron, m=9.11\times 10^{-31}\ kg

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

Initial speed of electron, u = 0

Final speed of electron, v=3\times 10^7\ m/s

Distance covered, s = 2 cm = 0.02 m

We need to find the electric field required. Firstly, we will find the acceleration of the electron from third equation of motion as :

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(3\times 10^7)^2-0}{2\times 0.02}

a=2.25\times 10^{16}\ m/s^2

According to Newton's law, force acting on the electron is given by :

F = ma

F=9.1\times 10^{-31}\times 2.25\times 10^{16}

F=2.04\times 10^{-14}\ N

Electric force is given by :

F = q E, E = electric field

E=\dfrac{F}{q}

E=\dfrac{2.04\times 10^{-14}}{1.6\times 10^{-19}}

E = 127500 N/C

So, the electric field is 127500 N/C. Hence, this is the required solution.

8 0
3 years ago
Which statement bet explains the relationship between the electric force between two charged objects and the distance between th
Nataliya [291]
Coulomb's Law: Force = k x q1x q2 divided distance square
where k=9x10^9 , q1 and q2 are the charge
So if you distance is halved, your force is stronger by 4 times
and if you distance is doubled, your force is 1/4
Ask me again if you aren't clear :)

4 0
3 years ago
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