Answer:
V
I and II
III and IV
Explanation:
The impulse is equal to the change in momentum of the object involved, so we can calculate the change in momentum in each situation and compare them all.
Taking always east as positive direction, and labelling
u the initial velocity
v the final velocity
m = 1000 kg the mass (which is always equal)
We find:
(i)
u = 25 m/s
v = 0

(II)
u = 25 m/s
v = 0

(III)
In this case,
F = 2000 N is the force
is the time
So the magnitude of the impulse is

(IV)
F = 2000 N is the force
is the time
So the magnitude of the impulse is

(V)
u = 25 m/s
v = -25 m/s

So the ranking from largest to smallest is:
V
I and II
III and IV
Answer:
option C
Explanation:
given,
energy dissipated by the system to the surrounding = 12 J
Work done on the system = 28 J
change in internal energy of the system
Δ U = Q - W
system losses energy = - 12 J
work done = -28 J
Δ U = Q - W
Δ U = -12 -(-28)
Δ U = 16 J
hence, the correct answer is option C
Answer:
80% (Eighty percent)
Explanation:
The material has a refractive index (n) of 1.25
Speed of light in a vacuum (c) is 2.99792458 x 10⁸ m/s
We can find the speed of light in the material (v) using the relationship
n = c/v, similarly
v = c/n
therefore v = 2.99792458 x 10⁸ m/s ÷ (1.25) = 239 833 966 m/s
v = 239 833 966 m/s
Therefore the percentage of the speed of light in a vacuum that is the speed of light in the material can be calculated as
(v/c) × 100 = (1/n) × 100 = (1/1.25) × 100 = 0.8 × 100 = 80%
Therefore speed of light in the material (v) is eighty percent of the speed of light in the vacuum (c)
Answer:
the molecules have different energy and the system is not in equilibrium
Explanation:
The model developed by Jack has the same energy for each of the two objects, but as each object is made up of a different number of molecules, in the system with more molecules, object 2, each one has approximately 2.4 less energy and the molecules of Object 1 have an E / 2 energy.
So when you book them together the molecules have different energy and the system is not in equilibrium
Answer:
-0.01 mm
Explanation:
We are given that
The value of one division of vernier scale =0.5 mm
The value of one main scale division=0.49 mm
We have to find the value of least count of the instrument in mm.
We know that
Leas count of vernier caliper=1 main scale division-1 vernier scale division
Least count of vernier caliper=0.49-0.50=-0.01 mm
Hence, the least count of the instrument=-0.01 mm
Answer: -0.01 mm