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sveta [45]
2 years 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]2 years 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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Suppose you are measuring double‑slit interference patterns using an optics kit that contains the following options that you can
svetlana [45]

Answer:

3.6 m

Explanation:

\lambda_R = 650 \ nm\\\\\lambda_R = 650*10^{-9} m\\\\L \ should \ be \ minimum \\\\i.e \  0.25 \ mm\\\\= 0.25 *10^{-3} m

\lambda_R = 700 \ nm\\\\\lambda_R = 700*10^{-9} m\\\\

Also

\beta = 1 \ mm \ fringe \  width

D_{min} = \frac{\beta d}{\lambda}\\\\D_{min} = \frac{10^{-3}*0.25*10^{-3}}{700*10^{-9}}\\\\D_{min} = 3.57 \\D_{min} =  3.6 m

Therefore, the minimum distance L  you can place a screen from the double slit that will give you an interference pattern on the screen that you can accurately measure using an ordinary 30 cm (12 in) ruler. = 3.6 m

4 0
3 years ago
A 2kg object is tied to the end of a cord and whirled in a horizontal circle of radius 2 m. If the body makes three complete rev
Travka [436]

Answer:

a) 37.70 m/s

b)710.6 m/s²

Explanation:

Given that ;

Mass of object = 2 kg

Radius of the motion = 2m

Frequency of motion = 3 rev/s

The formula to apply is;

v= 2πrf   where v is linear speed

v = 2×π×2×3 =12π = 37.70 m/s

Centripetal acceleration is given as;

a= 4×π²×r×f²  

a= 4×π²×2×3²

a=710.6 m/s²

5 0
3 years ago
If the mass of an object increases by a factor 2, kinetic energy?
emmainna [20.7K]

Answer:

A) Increases by a factor of 2

Explanation:

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

K.E = \frac{1}{2}MV^{2}

Where;

K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

Given that mass, m = 2m

Substituting into the equation, we have;

K.E = ½mv²

K.E = ½*2mv²

Cross-multiplying, we have;

2K.E = 2mv²

Hence, if the mass of an object increases by a factor 2, kinetic energy is increased by a factor of 2.

5 0
3 years ago
A bucket filled with water has a mass of 54 kg and is hanging from a rope that is wound around a 0.050 m radius stationary cylin
Lubov Fominskaja [6]

Answer:

The magnitude of the torque the bucket produces around the center of the cylinder is 26.46 N-m.

Explanation:

Given that,

Mass of bucket = 54 kg

Radius = 0.050 m

We need to calculate the magnitude of the torque the bucket produces around the center of the cylinder

Using formula of torque

\tau=F\times r

\tau=mg\times r

Where, m = mass

g = acceleration due to gravity

r = radius

Put the value into the formula

\tau=54\times9.8\times0.050

\tau=26.46\ N-m

Hence, The magnitude of the torque the bucket produces around the center of the cylinder is 26.46 N-m.

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3 years ago
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viktelen [127]

Answer:

B.

Explanation:

3 0
3 years ago
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