Answer: Option (A) and (D) are the correct statements.
Explanation:
Potential energy is defined as the energy obtained by an object due to its position. Whereas kinetic energy is defined as the energy obtained by an object due to the motion of its molecules.
When there will be only one phase present then addition or removal of energy will lead to change in kinetic energy of the substance but no change in potential energy will take place.
Whereas if change in phase is occurring then adding or removing any energy will lead to change in potential energy of the substance while kinetic energy will remain the same.
Thus, we can conclude that correct statements are as follows.
- While only one phase is present, adding or removing energy changes PE but not KE.
- During a phase change, adding or removing energy changes PE but not KE.
You can tell a lot about an object that's not moving,
and also a lot about the forces acting on it:
==> If the box is at rest on the table, then it is not accelerating.
==> Since it is not accelerating, I can say that the forces on it are balanced.
==> That means that the sum of all forces acting on the box is zero,
and the effect of all the forces acting on it is the same as if there were
no forces acting on it at all.
==> This in turn means that all of the horizontal forces are balanced,
AND all of the vertical forces are balanced.
Horizontal forces:
sliding friction, somebody pushing the box
All of the forces on this list must add up to zero. So ...
(sliding friction force) = (pushing force), in the opposite direction.
If nobody pushing the box, then sliding friction force = zero.
Vertical forces:
gravitational force (weight of the box, pulling it down)
normal force (table pushing the box up)
All of the forces on this list must add up to zero, so ...
(Gravitational force down) + (normal force up) = zero
(Gravitational force down) = -(normal force up) .
Answer:
7500 m/s
Explanation:
Centripetal acceleration = gravity
v² / r = GM / r²
v = √(GM / r)
Given:
G = 6.67×10⁻¹¹ m³/kg/s²
M = 5.98×10²⁴ kg
r = 6.8×10⁵ + 6.357×10⁶ = 7.037×10⁶ m
v = √(6.67×10⁻¹¹ (5.98×10²⁴) / (7.037×10⁶))
v = 7500
The orbital velocity is 7500 m/s.
Answer:
155.17N.
Explanation:
The magnitude of the net force is expressed as;
F = mv²/r where;
m is the mass
v is the velocity of the airplane
r is the radius of the loop
Given
m = 95kg
v = 70m/s
r = 3km = 3000m
Required
Magnitude of the net force
F = 95*70²/3000
F = 95*4900/3000
F = 95*49/30
F = 4655/30
F = 155.17N
Hence the magnitude of the net force on the 95 kg pilot at the bottom of this loop is 155.17N.
Answer:
(a) 106 kPa
(b) 0.0377 mol
(c) 17.8 cm
Explanation:
(a) There are three forces on the piston. Atmospheric pressure pushing down, weight pulling down, and pressure of the gas pushing up.
∑F = ma
PA − mg − PₐA = 0
P = (PₐA + mg) / A
P = Pₐ + (mg / A)
P = 101,300 Pa + (40.0 N) / (π (0.05 m²))
P = 106,393 Pa
P = 106 kPa
(b) Use ideal gas law.
PV = nRT
(106,393 Pa) (π (0.05 m²) (0.11 m)) = n (8.314 Pa m³/mol/K) (20 + 273.15) K
n = 0.0377 mol
(c) Use ideal gas law to find the new volume of the gas.
PV = nRT
(106,393 Pa) (π (0.05 m²) h) = (0.0377 mol) (8.314 Pa m³/mol/K) (200 + 273.15) K
h = 0.178 m
h = 17.8 cm