The electric Field is a vector quantity, in other words it has a magnitude and a direction. On the other hand, the electric field follows the law of superposition. The electric field produced by two elements is equal to the sum of the electric fields produced by each element when the other element is not present. in other words, the total electric field is solved determining the electric field due to each charge and adding them together as vectors.

7 0

3 years ago

Please help me with my Physical Science! 50 POINTS

DaniilM [7]

Answer:

1, When Jane brakes, the brakes slow the car wheels turning and the road surface exerts a backwards force on the tires, causing the car to decelerate. The pocket book tends to continue on in a straight line (Newton's first law). If she brakes hard enough that the friction between the book and the car seat is insufficient to decelerate the book as fast as the car is decelerating, the book will slide off the seat, and gravity pulls it to the floor

When the diver uses his / her force to depress the springboard, the springboard pushes him back with equal force

3.Newton's Second Law (F=ma)

4. 5 N

5. 19.5 N

65kg * 0.3 m/s^2

6.0.2 N/s

10kg divided by 2N

7.-Walking then pushing the moving forward

-Dribbling

-Basketball is pushed but bounces back

Explanation:

6 0

3 years ago

Read 2 more answers

determine the force of gravitational attraction between a 78kg boy sitting 2 meters away from a 65kg girl.

bixtya [17]

Answer:

$F=8.45\times 10^{-8}\ N$

Explanation:

Given that,

Mass of a boy is 78 kg

Mass of a girl is 65 kg

We need to find the force of gravitational attraction between them if they are 2 m away.

The formula for the gravitational force is given by :

$F=\dfrac{Gm_1m_2}{r^2}\\\\F=\dfrac{6.67\times 10^{-11}\times 78\times 65}{(2)^2}\\\\=8.45\times 10^{-8}\ N$

So, the force between them is $8.45\times 10^{-8}\ N$ .

8 0

2 years ago

In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting a

IRISSAK [1]

A. $4.64\cdot 10^{11}m$

The orbital speed of the clumps of matter around the black hole is equal to the ratio between the circumference of the orbit and the period of revolution:

$v=\frac{2\pi r}{T}$