To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.
If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,
So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
Answer:
2452.79432 m/s
Explanation:
m = Mass of ice
= Latent heat of steam
= Specific heat of water
= Latent heat of ice
v = Velocity of ice
= Change in temperature
Amount of heat required for steam
Heat released from water at 100 °C
Heat released from water at 0 °C
Total heat released is
The kinetic energy of the bullet will balance the heat
The velocity of the ice would be 2452.79432 m/s
65 years but anything can happen to them
I’m not really sure but I hope this helps
To solve this, we
use the formula:
y = v0 t + 0.5 a t^2
where y is distance, v0 is initial velocity, t is time
and a is acceleration
Since we know that total time is 8.5 seconds, hence going
up must be 4.25 s and going down is 4.25 s.
a = 0.379 g = 0.379 (9.8 m/s^2) = 3.7142 m/s^2
going up:
y = v0 (4.25) - 0.5 (3.7142) (4.25)^2
y = 4.25 v0 – 33.5439 -->
eqtn 1
going down:
y = 0 (4.25) + 0.5 (3.7142) (4.25)^2
y = 33.5439
y = 33.5439 m
Calculating for v0 from equation 1:
33.5439 = 4.25 v0 – 33.5439
4.25 v0 = 67.0877
v0 = 15.78535 m/s
answers:
a. y = 33.5439 m
b. v0 = 15.78535 m/s
c.
