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Molodets [167]
2 years ago
10

. A ball is thrown downward at a speed of 20 m/s. Choosing

Physics
1 answer:
a_sh-v [17]2 years ago
3 0

The final velocity is v = 20 - gt

The distance traveled by the ball at time t is y = y_o + (-20)t - \frac{1}{2} gt^2

The maximum distance traveled by the object is 0 = (20)^2 - 2g(y-y_0)

The given parameters;

initial velocity of the ball, u = 20 m/s

acceleration due to gravity, g = 9.8 m/s²

The final velocity can be calculate as;

v = 20 - gt

The distance traveled by the ball at time t;

y = y_o + (-20)t - \frac{1}{2} gt^2

The maximum distance traveled by the object is calculated as;

v^2 = u^2 - 2g(y - y_0)\\\\0 = (20)^2 - 2g(y-y_0)

Learn more here: brainly.com/question/16878713

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An object is falling from a height of 7.5 meters. At what height will it’s velocity be 7 meters/second?
Nady [450]
One of the equations of gravity is this:
{v}^{2} = {u}^{2} + 2gh
Where v = final velocity which is 7m/s
u = initial velocity which is 0 for objects falling from a height
g = acceleration due to gravity and it is approximately 10m/s^2. It's a constant so pretty much remember this number. It's positive since the work being done is caused by gravity (in other words, it's falling down). It can also be negative if the work being down is against gravity (in other words, it's going up)
h = height of object

Substitute for the values and you should have something like this
{7}^{2} = {0}^{2} + 2 \times 10 \times h
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6 0
3 years ago
8. two +1 C charges are separated by 3000m. What is the magnitude of the electric force between them?
Sidana [21]

Answer:

1000 N

Explanation:

The magnitude of the electrostatic force between two charged object is given by

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb constant

q1, q2 is the magnitude of the two charges

r is the distance between the two objects

Moreover, the force is:

- Attractive if the two forces have opposite sign

- Repulsive if the two forces have same sign

In this problem:

q_1=q_2=+1C are the two charges

r = 3000 m is their separation

Therefore, the electric force between the charges is:

F=(9\cdot 10^9)\frac{(1)(1)}{3000^2}=1000 N

8 0
3 years ago
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