Given:
m = 555 g, the mass of water in the calorimeter
ΔT = 39.5 - 20.5 = 19 °C, temperature change
c = 4.18 J/(°C-g), specific heat of water
Assume that all generated heat goes into heating the water.
Then the energy released is
Q = mcΔT
= (555 g)*(4.18 J/(°C-g)*(19 °C)
= 44,078.1 J
= 44,100 J (approximately)
Answer: 44,100 J
Gravitational potential energy =
(mass) x (gravity) x (height)
= (5.8 kg) x (9.8 m/s²) x (2.5 m)
= 142.1 Joules (C)
Answer:
If the frequency of the motion of a simple harmonic oscillator is doubled , then maximum speed of the oscillator changes by the factor 2
Explanation:
We know that in a simple harmonic oscillator the maximum speed is given by
= 
Here A is amplitude which is constant , so from above equation we see that maximum speed is directly proportional to 
of the oscillation .
Since 
= 2
Where 
is the maximum speed when frequency is doubled .
The Hubble Space Telescope is a joint ESA/NASA project and was launched in 1990 by the Space Shuttle mission STS-31 into a low-Earth orbit 569 km above the ground. During its lifetime Hubble has become one of the most important science projects ever. Hope this helps! ~ Autumn :)
