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guapka [62]
2 years ago
7

What is the formula to find initial velocity?​

Physics
1 answer:
Inga [223]2 years ago
7 0

Answer:

s = ut + 1/2at^2

Explanation:

s= distance

u= initial velocity

a= acceleration

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An object is moving in a circle. If the radius of the object is doubled, and the period remains constant, the magnitude of the v
Anton [14]

Answer:

Twice

Explanation:

From the formula for velocity in a circle

V= 2πr/T

Where V is velocity

r is raduis

T is period

We see that as r increases V increases so if r is doubled V becomes doubled

4 0
3 years ago
How are kids made? my teacher asked us this question
Kazeer [188]
A single sperm and the mother's egg cell meet in the fallopian tube. When the single sperm enters the egg, conception occurs. The combined sperm and egg is called a zygote. The zygote contains all of the genetic information (DNA) needed to become a baby.
7 0
3 years ago
Four charges with equal magnitudes of 10.6 × 10-12 C are placed at the corners of a rectangle. The lengths of the sides of the r
cricket20 [7]

Answer:

Figure a. E_net = 99.518 N/C

Figure b. E_net = 177.151 N / C

Explanation:

Given:

- Attachment for figures missing in the question.

- The dimensions for rectangle are = 7.79 x 3.99 cm

- All four charges have equal magnitude Q = 10.6*10^-12 C

Find:

Find the magnitude of the electric field at the center of the rectangle in Figures a and b.

Solution:

- The Electric field generated by an charged particle Q at a distance r is given by:

                                         E = k*Q / r^2

- Where, k is the coulomb's constant = 8.99 * 10^9

Part a)

- First we see that the charges +Q_1 and +Q_3 produce and electric field equal but opposite in nature. So the sum of Electric fields:

                                 E_1 + E_3 = 0

- For Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Hence, the net Electric Field at center of the rectangle can be given as:

                                  E_net = E_2 + E_4

                                  E_2 = E_4

                                  E_net = 2*E = 2*k*Q / r^2

- The distance r from each corner to mid-point of the rectangle is constant. It can be evaluated by Pythagoras Theorem as follows:

                                  r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )

                                  r = sqrt ( 1.9151*10^-3 ) = 0.043762 m

- Plug the values in the E_net expression developed above:

                                  E_net = 2*(8.99*10^9)*(10.6*10^-12) / 1.9151*10^-3

                                 E_net = 99.518 N/C

Part b)

- Similarly for Figure b, for Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Also, Charges -Q_1 and +Q_3, they are equal in nature but act in the same direction towards the negative charge -Q_1. These Electric fields are equal in magnitude to what we calculated in part a).

- To find the vector sum of two Electric Fields E_1,3 and E_2,4 we see the horizontal components of each cancels each other out. While the vertical components E_1,3 and E_2,4 are equal in magnitude and direction.

Hence,

                                  E_net = 2*E_part(a)*cos(Q)

- Where, Q is the angle between resultant, vertical in direction, and each of the electric field. We can calculate Q using trigonometry as follows:

                                  Q = arctan ( 3.99 / 7.79 ) = 27.12 degrees.

- Now, compute the net electric field E_net:

                                  E_net = 2*(99.518)*cos(27.12)

                                  E_net = 177.151 N / C

               

5 0
3 years ago
A car moves with a speed 72km/h, the driver uses the brakes, the car stops after 8 seconds, calculate its speed after 10 seconds
vampirchik [111]
The acceleration of the car is solved by subtracting the initial speed from the final speed then dividing the result by the elapsed time.

initial speed = 72 km/hr = 20 m/s

final speed = 0 m/s

elapsed time = 5 seconds

acceleration = (0 m/s – 20 m/s) / 5 s

acceleration = - 20m/s / 5 s

acceleration = -4 m/s^2
8 0
3 years ago
A particle that carries a net charge of -41.8 μc is held in a region of constant, uniform electric field. the electric field vec
miss Akunina [59]
The total work done by the electric field on the charge is given by the scalar product between the electric force acting on the charge and the displacement of the charge:
W=F d cos \theta
where the force is F=qE, d=0.556 and \theta=55.2^{\circ}. Using the value of q and E given by the problem, we find
W=qEdcos\theta = 6.39\cdot10^{-5}J
3 0
3 years ago
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