E=energy=5.09x10^5J = 509KJ
<span>M=mass=2250g=2.25Kg </span>
<span>C=specific heat capacity of water= 4.18KJ/Kg </span>
<span>ΔT= change in temp= ? </span>
<span>E=mcΔT </span>
<span>509=(2.25)x(4.18)xΔT </span>
<span>509=9.405ΔT </span>
<span>ΔT=509/9.405=54.1degrees </span>
<span>Initial temp = 100-54 = 46 degrees </span>
<span>Hope this helps :)</span>
The horizontal speed is going to be the cosine of the given speed, therefore, the horizontal speed is 19.15 m/s. To find the time, divide the 22 m distance by the velocity. This results in 1.131 seconds, which is in between C and D.
Answer:
x = 0.4 m
Explanation:
When a spring is stretched from its equilibrium position. Some energy is stored in the spring. This energy is called the elastic potential energy of the spring. The formula used to calculate the magnitude of this stored energy is given as follows:
P.E = (1/2)kx²
where,
P.E = Elastic Potential Energy Stored in the spring = 45 J
k = Spring Constant = 540 N/m
x = amount of stretching = ?
Therefore,
45 J = (1/2)(540 N/m)x²
x² = (45 J)(2)/(540 N/m)
x = √(0.167 m²)
<u>x = 0.4 m</u>
