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
Answer:
E = 31.329 N/C.
Explanation:
The differential electric field at the center of curvature of the arc is
<em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>
where is the radius of curvature.
Now
,
where is the charge per unit length, and it has the value
Thus, the electric field at the center of the curvature of the arc is:
Now, we find and . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:
and this is
radians.
Therefore,
evaluating the integral, and putting in the numerical values we get:
Given data:
* The mass of the ball is 2 kg.
* The gravitational field strength at the surface of planet X is 5 N/kg.
Solution:
The weight of the ball on the planet X is,
where m is the mass of ball, a is the gravitational field strength,
Substituting the known values,
Thus, the weight of the ball on the surface of planet X is 10 N.