Acceleration = force/mass
1/4
s=ut+1/2at^2
0*4+1/2*1/4*4^2
1/2*1/4*16
1/2*4
=2
Answer:
The unknown substance is Aluminum.
Explanation:
We'll begin by calculating the change in the temperature of substance. This can be obtained as follow:
Initial temperature (T₁) = 25 ⁰C
Final temperature (T₂) = 100 ⁰C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 100 – 25
ΔT = 75 ⁰C
Finally, we shall determine the specific heat capacity of the substance. This can be obtained as follow:
Change in temperature (ΔT) = 75 ⁰C
Mass of the substance (M) = 135 g
Heat (Q) gained = 9133 J
Specific heat capacity (C) of substance =?
Q = MCΔT
9133 = 135 × C × 75
9133 = 10125 × C
Divide both side by 10125
C = 9133 / 10125
C = 0.902 J/gºC
Thus, the specific heat capacity of substance is 0.902 J/gºC
Comparing the specific heat capacity (i.e 0.902 J/gºC) of substance to those given in the table above, we can see clearly that the unknown substance is aluminum.
T o a stationary observer, a man jogs east at 2.5 m/s and a woman jogs west at 1.5 m/s. from the woman's frame of reference, what is the man's velocity? it is 4m/s east
Answer:
The Statement is wrong because the reverse is the case as it is the kinetic energy that is being transformed to gravitational potential energy.
Explanation:
As your friend throws the baseball into the air the ball gains an initial velocity (u) and this makes the Kinetic energy to be equal to
Here m is the mass of the baseball
Now as this ball moves further upward the that velocity it gained reduce due to the gravitational force and this in turn reduces the kinetic energy of the ball and this kinetic energy lost is being converted to gravitational potential energy which is mathematically represented as (m×g×h)
as energy can not be destroyed but converted to a different form according to the first law of thermodynamics
Looking a the formula for gravitational potential energy we see that the higher the ball goes the grater the gravitational potential energy.
Not exactly the best way to describe it but, it is used to calculate resistance of a lever as in the use of a pry bar or pulley. Technology used to increase output with little input.
