# Two particles, one with charge −7.97×10−6 C and the other with charge 6.91×10−6 C, are 0.0359 m apart. What is the magnitude of the force that one particle exerts on the other?

-384.22N

Explanation:

From Coulomb's law;

F= Kq1q2/r^2

Where;

K= constant of Coulomb's law = 9 ×10^9 Nm^2C-2

q1 and q2 = magnitudes of the both charges

r= distance of separation

F= 9 ×10^9 × −7.97×10^−6 × 6.91×10^−6/(0.0359)^2

F= -495.65 × 10^-3/ 1.29 × 10^-3

F= -384.22N

## Related Questions

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.55 m and the acceleration achieved during the time the jumper is extending their legs is 1.2 times the acceleration due to gravity, g .How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

1.32 m.

Explanation:

Below is an attachment containing the solution.

A boat that travels 3.00 m/s relative to the water is crossing a river that is 1.00 km wide. The destination on the far side of the river is 0.500 km downstream from the starting point. (a) If the river current is 2.00 m/s, in what direction should the boat be pointed in order to reach the destination? (b) How much time will the trip take?

a) 10.29° upstream

b) t=338.7s

Explanation:

If the river is 1km wide and the destination point is 0.5km away downstream, then the angle and distance the the boat has to travel is:

The realitve velocity of the boat respect to the water is:

where β is the angle it has to be pointed at.

From the relative mvement equations:

where

From this equation we get one equation per the x-axis and another for the y-axis. If we square each of them and add them together, we will get 2 equations:

Solving for V:

V = 3.3m/s   and   V=-1.514m/s   Replacing this value into one of our previous x or y-axis equations:

The amount of time:

You say goodbye to your friend at the intersection of two perpendicular roads. At time t=0 you drive off North at a (constant) speed v and your friend drives West at a (constant) speed ????. You badly want to know: how fast is the distance between you and your friend increasing at time t?

Explanation:

Rate of Change

When an object moves at constant speed v, the distance traveled at time t is

We know at time t=0 two friends are at the intersection of two perpendicular roads. One of them goes north at speed v and the other goes west at constant speed w (assumed). Since both directions are perpendicular, the distances make a right triangle. The vertical distance is

and the horizontal distance is

The distance between both friends is computed as the hypotenuse of the triangle

We need to find d', the rate of change of the distance between both friends.

Plugging in the above relations

Solving for d

Differentiating with respect to t

The problem is solved using Pythagoras' Theorem, representing the two travel paths forming a right triangle. The rate at which the distance increases between two points moving perpendicularly can be found by differentiating the resulting equation, which yields the expression sqrt[(v^2)+(u^2)].

### Explanation:

The question is about the rate at which the distance between you and your friend is increasing at time t. It's a typical problem in kinematics. Because the roads are perpendicular to each other, we can solve the problem using Pythagoras' Theorem which states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Let's denote the distance you've traveled as D1 = v*t (because distance = speed * time) and the distance your friend has travelled D2 = u*t. The distance between you can be computed using Pythagoras' Theorem as D = sqrt(D1^2 + D2^2). Hence, D = sqrt[(v*t)^2 + (u*t)^2]. Differentiating D with respect to t using the chain rule will give us the rate at which the distance between you is increasing, which is sqrt[(v^2)+(u^2)].

brainly.com/question/34025828

#SPJ11

A large convex lens stands on the floor. The lens is 180 cm tall, so the principal axis is 90 cm above the floor. A student holds a flashlight 120 cm off the ground, shining straight ahead (parallel to the floor) and passing through the lens. The light is bent and intersects the principal axis 60 cm behind the lens. Then the student moves the flashlight 30 cm higher (now 150 cm off the ground), also shining straight ahead through the lens. How far away from the lens will the light intersect the principal axis now?A. 30 cm
B. 60 cm
C. 75 cm
D. 90 cm

B. 60 cm

All parallel light rays are bent through the focal point of a convex lens, so the rays from the flashlight 150 cm above the floor must go through the same point on the principal axis as the rays from the flashlight 120 cm above the floor. The location of the focal point does not change when the position of the object is moved either vertically or horizontally.

A typical laboratory centrifuge rotates at 3700 rpm . Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. Part A What is the acceleration at the end of a test tube that is 10 cm from the axis of rotation

Explanation:

acceleration of test tube

= ω² R

= (2πn)² R

= 4π²n²R

n = no of rotation per second

= 3700 / 60

= 61.67

R = .10 m

acceleration

= 4π²n²R

= 4 x 3.14² x 61.67² x .10

= 14999 N Approx

A person on a rocket traveling at 0.47 c (with respect to the Earth) observes a meteor come from behind and pass her at a speed she measures as0.47 c.How fast is the meteor moving with respect to the Earth?

The concept to solve this problem is related to the relativistic physics for which the speed of the object in different frames of reference is related. This concept is called Velocity-addition formula

and can be written as,

Where,

u = Velocity of a body within a Lorentz Frame

v = Velocity of a second frame

u'= The transformed velocity of the body within the second frame

c = speed of light

Replacing we have to

Therefore the meteor moving with respect to the Earth to 230'700.000m/s