Answer:
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Explanation:
Answer: Magnitude of the force exerted on the egg by the ground is 9.2N
Explanation:
Given the following :
Mass of egg (m) = 150g = 0.15kg
Height(h) from which egg is dropped = 3m
velocity of egg before hitting the ground (u) = 4.4m/s
Final velocity of egg (V) = 0
Time taken (t) = 0.072s
Magnitude of the force exerted on the egg by the ground can be found by applying Newton's 2nd law:
Momentum = mass × velocity
From Newton's second law:
Force = mass × change in Velocity with time ;
That is
F = m * ΔV / t
Inputting our values
F = 0.15 * (4.4 - 0) / 0.072
F = 0.15 × (4.4 / 0.072)
F = 0.15 × 61.11
F = 9.16N
F = 9.2N
Explanation:
The Coulomb's law states that the magnitude of each of the electric forces between two point-at-rest charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:
In this case we have an electron (-e) and a proton (e), so:
In this case, the electric force is negative, therefore, the force is repulsive and its magnitude is:
Use the equation for the acceleration
A = final velocity - initial velocity divided by time final - time initial
A= 54 - 32 / 8 - 0
A= 22 / 8
A= 2.75 m/s^2
Hope this helps!
<span>On the y-axis (the bottom of the table) hope this helps</span>
