Answer:
Heat is removed from it.
Explanation:
A object will become cold when it loses its heat to its surroundings. I would look up the 3 Principles of Heat Transfer.
I hope this helped! :)
Answer:
E = 2.7 x 10¹⁶ J
Explanation:
The release of energy associated with the mass can be calculated by Einstein's mass-energy relation, as follows:
![E = mc^2](https://tex.z-dn.net/?f=E%20%3D%20mc%5E2)
where,
E = Energy Released = ?
m = mass of material reduced = 0.3 kg
c = speed of light = 3 x 10⁸ m/s
Therefore,
![E = (0.3\ kg)(3\ x\ 10^8\ m/s)^2](https://tex.z-dn.net/?f=E%20%3D%20%280.3%5C%20kg%29%283%5C%20x%5C%2010%5E8%5C%20m%2Fs%29%5E2)
<u>E = 2.7 x 10¹⁶ J</u>
(amount of heat)Q = ? , (Mass) m= 4 g , ΔT = T f - T i = 180 c° - 20 °c = 160 °c ,
Ce = 0.093 cal/g. °c
Q = m C ΔT
Q = 4 g × 0.093 cal/g.c° × ( 180 °c- 20 °c )
Q= 4×0.093 × 160
Q = 59.52 cal
I hope I helped you^_^
Answer:
u = 10.63 m/s
h = 1.10 m
Explanation:
For Take-off speed ..
by using the standard range equation we have
![R = u² sin2θ/g](https://tex.z-dn.net/?f=R%20%3D%20u%C2%B2%20sin2%CE%B8%2Fg)
R = 9.1 m
θ = 26º,
Initial velocity = u
solving for u
![u² = \frac{Rg}{sin2\theta}](https://tex.z-dn.net/?f=u%C2%B2%20%3D%20%5Cfrac%7BRg%7D%7Bsin2%5Ctheta%7D)
![u^2 = \frac{9.1 x 9.80}{sin26}](https://tex.z-dn.net/?f=u%5E2%20%3D%20%5Cfrac%7B9.1%20x%209.80%7D%7Bsin26%7D)
![u^2 = 113.17](https://tex.z-dn.net/?f=%20u%5E2%20%3D%20113.17)
u = 10.63 m/s
for Max height
using the standard h(max) equation ..
![v^2 = (v_osin\theta)^2 -2gh](https://tex.z-dn.net/?f=v%5E2%20%3D%20%28v_osin%5Ctheta%29%5E2%20-2gh)
![h =\frac{(v_o^2sin\theta)^2}{2g}](https://tex.z-dn.net/?f=h%20%3D%5Cfrac%7B%28v_o%5E2sin%5Ctheta%29%5E2%7D%7B2g%7D)
![h = \frac{(113.17)(sin26)^2}{(2 x 9.80)}}](https://tex.z-dn.net/?f=h%20%20%3D%20%20%5Cfrac%7B%28113.17%29%28sin26%29%5E2%7D%7B%282%20x%209.80%29%7D%7D)
h = 1.10 m