Answer:
M L1 = m L2 torques must be zero around the fulcrum
M = m L2 / L1 = .3 kg * 28 cm / 22 cm = .382 kg
Answer:
Explanation:
a ) Let the distance required in former case be d₁ .
initial velocity u = 30 m /s , final velocity v =0 , deceleration a = 7.00 m /s²
v² = u² - 2 a s
0 = 30² - 2 x 7 x d₁
d₁ = 64.28 m
b) initial velocity u = 30 m /s , final velocity v =0 , deceleration a = 5.00 m /s²
v² = u² - 2 a s
0 = 30² - 2 x 5 x d₂
d₂ = 90 m
c)
t = .5 s
s₁ = ut - .5 at²
= 30 x .5 - .5 x 7 x .5²
= 15 - .875
= 14.125 m
t = .5 s
s₂ = ut - .5 at²
= 30 x .5 - .5 x 5 x .5²
= 15 - .625
= 14.375 m
90 J / 2.2 s = 41 W. 41 Watts is the power <span>required to give a brick 90 J of potential energy in a time of 2.2 s</span>
Answer:its B
Explanation:
The suns gravity keeps the planets in their gravitational orbit.
Answer:
Use the formula
g = GM/R^2
where
g = acceleration of gravity on moon
G = universal gravitational constant
= 6.67 x 10^-11 m^3/kg-s^2
M = mass of the moon
R = radius of the moon
solving for R, we get
R = 1.74 x 10^6 kg
