In order to give a spaceship at rest in a specific reference frame s a speed increment of 0.500c, seven increments are required. Then, in this new frame, it receives an additional 0.500c increment.
The speed of an object, also known as v in kinematics, is a scalar quantity that refers to the size of the change in that object's position over time or the size of the change in that object's position per unit of time. The distance travelled by an object in a certain period of time divided by the length of the period gives the object's average speed in that period.
The spacecraft moves at v1 = 0.5c after the initial increment.The equation becomes V2 = V+V1/1+V*V1/c after the second one. 2 V2 = 0.5c+0.50c/1+(0.50c)^2/c^ 2 = 0.80c
Likewise, V3 = 0.929c
V4 = 0.976c
V5 = 0.992c
V6 = 0.99c
V7 = 0.999c
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But even more pain on pain and then pain and pain ya feel me and even more pain okay and yes more pain
If you write down the formula for friction, you will get an answer.
Ff = u * N Where N is a push down force that an object experiences.
u (mu) is a constant and has no units
It may not be accelerating and still experience friction. A is not correct.
Color and Density will not affect the frictional force. B is not so.
Buoyant forces are a different thing altogether. Generally friction has nothing to do with them. C is incorrect.
The last one is your answer. Technically mg should be the answer and not mass, but the second part is correct.
D.to bring light together
Answer:
326149.2 KJ
Explanation:
The heat transfer toward and object that suffered an increase in temperature can be calculated using the expression:
Q = m*cv*ΔT
Where m is the mass of the object, cv is the specific heat capacity at constant volume, which basically means the amount of heat necessary for a 1kg of water to increase 1C degree in temperatur, and ΔT is the change in temperature.
A 65000 L swimming pool will have a mass of:
65000L *
= 65000 kg
The specific heat capacity at constant volume of water is equal to 4.1814 KJ/KgC.
We replace the data and get:
Q = m*cv*ΔT = 65000 kg * 4.1814 KJ/KgC * 1.2°C = 326149.2 KJ