V2 = 4.4579 L
Since pressure is constant, use Charle’s law.
Charles's law, a statement that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature, if the pressure remains constant.
V(olume) 1 = V(olume) 2
————— —————
T(emperature) 1 T(emperature)2
4.00 L = V2
———- ———
297 K 331 K
Cross multiply
(4.00 L x 331 K) = (297 K x V2)
Simplify
1324 L/K = 297 K x V2
Isolate V2 by dividing out 297 K
1324 L/K = V2
————
297 K
(This cancels out the kelvin and leaves you with Liters as the volume measure)
V2 = 4.4579 L
Round to significant digits if required
Had to look for the options and here is my answer. What happens when the fluid discharge of an air-operated reciprocating pump is shut, this will cause the pump to OVERSTROKE. Overstroke happens when the engine is switching in a normally-closed manner.
Answer:
Explanation:
refractive index of ember = sin of angle of incidence / sin of angle of refraction
= sin 35 / sin24
= .5735 / .4067
= 1.41
This is refractive index of ember with respect to water
refractive index of ember with respect to water
= wμe = μe / μw
μe = wμe x μw
= 1.33 x 1.41
= 1.87
refractive index of ember with respect to air = 1.87 .
Answer:
a) w = 7.27 * 10^-5 rad/s
b) v1 = 463.1 m/s
c) v1 = 440.433 m/s
Explanation:
Given:-
- The radius of the earth, R = 6.37 * 10 ^6 m
- The time period for 1 revolution T = 24 hrs
Find:
What is the earth's angular speed?
What is the speed of a point on the equator?
What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?
Solution:
- The angular speed w of the earth can be related with the Time period T of the earth revolution by:
w = 2π / T
w = 2π / 24*3600
w = 7.27 * 10^-5 rad/s
- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.
v1 = R*w
v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)
v1 = 463.1 m/s
- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.
π/2 ........... s
x ............ 1/5 s
x = π/2*5 = 18°
- The radius of the earth R' at point where θ = 18° from the equator is:
R' = R*cos(18)
R' = (6.37 * 10 ^6)*cos(18)
R' = 6058230.0088 m
- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.
v2 = R'*w
v2 = (6058230.0088)*(7.27 * 10^-5)
v2 = 440.433 m/s
Answer:
the extension recorded by the student would be smaller than the actual extension of the spring
