Answer:
The arrow is at a height of 500 feet at time t = 2.35 seconds.
Explanation:
It is given that,
An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s
The projectile formula is given by :

We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,

On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds
So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.
The charge of the object must be 
Answer: Option C
<u>Explanation:</u>
Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.
Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

Here, given E = 4500 N/C and F = 0.05 N.
We need to find charge of the object (q)
By substituting the given values, we get

<h2>Answer:</h2>
<u>Distance covered is 6.9 meters</u>
<h2>Explanation:</h2>
Data given:
Work Done = 345 kJ = 345000 J
Force = 5 x 10 ^ 4 = 50000 N
Distance = ?
Solution:
As we know that
Work Done = Force applied x Distance covered
By arranging the equation we get
Work / Force = Distance covered
By putting the values
345000 / 50000 = 6.9
So distance covered is 6.9 meters
Answer:
yes it could deform a shape or an object
Explanation: