31.3m/s
Explanation:
Given parameters:
Mass of rock = 40kg
Height of cliff = 50m
Unknown:
Speed of rock when it hits ground = ?
Solution:
We are going to use the appropriate motion equation to solve this problem
The rock is falling with the aid of gravitational force. The force is causing it to accelerate with an amount of velocity.
Using;
V² = U² + 2gH
V = unknown velocity
U = initial velocity = O
g = acceleration due to gravity = 9.8m/s²
H = height of fall
since the initial velocity of the bodyg is 0
V² = 2gH
V= √2gH = √2 x 9.8 x 50 = 31.3m/s
learn more:
Velocity brainly.com/question/4460262
#learnwithBrainly
True
False
True
My answers
The Cell Membrane is what controls what enters and leaves the cell.
Answer:
94.1 m
Explanation:
From Coulombs law,
F = Gm1m2/r²................... Equation 1
where F = force, m1 = first mass, m2 = second mass, G = universal constant, r = distance of separation.
Make r the subject of the equation,
r = √(Gm1m2/F)................. Equation 2
Given: F = 7×10² N, m1 = 15×10⁷ kg, m2 = 62×10⁷ kg,
Constant: G = 6.67×10⁻¹¹ Nm²/kg²
Substitute into equation 2
r = √( 6.67×10⁻¹¹×15×10⁷×62×10⁷/7×10²)
r √(886.16×10)
r √(88.616×10²)
r = 9.41×10
r = 94.1 m.
Hence the distance of separation = 94.1 m
Answer:
1500
Explanation:
is that all the question?
