Answer:
1793.7m
Explanation:
From the principle of conservation of energy; the kinetic energy substended by the object equals the potential energy sustain by the object when it gets to its maximum position.
Now the kinetic energy; is
K.E = 1/2 × m × v2
Where m is mass
v is velocity
Hence.
K.E = 1/2 × 2.25 × (187.5)^2
Now this should be same with the potential energy which is given as;
P.E = m× g× h
Where m is mass of object
g is acceleration of free fall due to gravity = 9.8m/S2
h is maximum height substain by the object.
Hence P.E = 2.25 × 9.8 × h
From the foregoing analysis of energy conversation it implies;
1/2 × 2.25 × (187.5)^2 =2.25 × 9.8 × h
=> 1/2 × (187.5)^2 = 9.8 × h
=>1/2 × (187.5)^2 / 9.8 = h
=> 1793.69m = h
h= 1793.69m
h =1793.7m to 1 decimal place
Answer:
Because you have a greater understanding of the mentality and conduct of individuals you care about, you are better able to understand their perspective and are more likely to recognize possible situational factors for their behavior.
Answer:
2.464 cm above the water surface
Explanation:
Recall that for the cube to float, means that the volume of water displaced weights the same as the weight of the block.
We calculate the weight of the block multiplying its density (0.78 gr/cm^3) times its volume (11.2^3 cm^3):
weight of the block = 0.78 * 11.2^3 gr
Now the displaced water will have a volume equal to the base of the cube (11.2 cm^2) times the part of the cube (x) that is under water. Recall as well that the density of water is 1 gr/cm^3.
So the weight of the volume of water displaced is:
weight of water = 1 * 11.2^2 * x
we make both weight expressions equal each other for the floating requirement:
0.78 * 11.2^3 = 11.2^2 * x
then x = 0.78 * 11.2 cm = 8.736 cm
This "x" is the portion of the cube under water. Then to estimate what is left of the cube above water, we subtract it from the cube's height (11.2 cm) as follows:
11.2 cm - 8.736 cm = 2.464 cm
Answer:’B’
Explanation:Since, the nucleus is in the middle and it carries protons which carry positive charge and neutrons which carry no charge.
One of the useful forns of the formula for electrical power is: Power = (voltage squared) / (resistance). Knowing that power is proportional to (voltage squared), we can see that if the voltage is reduced to 1/2, the power is reduced to 1/4 of its original value. The 220volt/60watt appliance, when operated on 110 volts, dissipates 60/4 = 15 watts.
