The force of F=10 N produces an extension of
on the string, so the spring constant is equal to
Then the string is stretched by
. The work done to stretch the string by this distance is equal to the variation of elastic potential energy of the string with respect to its equilibrium position:
Answer:
335°C
Explanation:
Heat gained or lost is:
q = m C ΔT
where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.
Heat gained by the water = heat lost by the copper
mw Cw ΔTw = mc Cc ΔTc
The water and copper reach the same final temperature, so:
mw Cw (T - Tw) = mc Cc (Tc - T)
Given:
mw = 390 g
Cw = 4.186 J/g/°C
Tw = 22.6°C
mc = 248 g
Cc = 0.386 J/g/°C
T = 39.9°C
Find: Tc
(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)
Tc = 335
Answer:
2046.37 kPa
Explanation:
Given:
Number of moles, n = 125
Temperature, T = 20° C = 20 + 273 = 293 K
Radius of the cylinder, r = 17 cm = 0.17 m
Height of the cylinder, h = 1.64 m
thus,
volume of the cylinder, V = πr²h
= π × 0.17² × 1.64
= 0.148 m³
Now,
From the ideal gas law
we have
PV = nRT
here,
P is the pressure
R is the ideal gas constant = 8.314 J / mol. K
thus,
P × 0.148 = 125 × 8.314 × 293
or
P × 0.148 = 304500.25
or
P = 2046372.64 Pa = 2046.37 kPa
Answer:
Explanation:
Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.
For faster car on the road,
v = 2v
..........(1)
For the slower car on the road,
............(2)
Equation (1) becomes,
So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.
