Answer:
The correct answers are
(a) It decreases to 1/3 L
(ii) is (c) It is constant
Explanation:
to solve this, we list out the number of knowns and unknowns so as to determine the correct equation to solve the problem
The given variables are as follows
Initial volume V1 = 1L
V2 = Unknown
Initial Temperature T1 = 300K
let us assume that the balloon is perfectly elastic
At 300K the balloon is filled and it stretches to maintain 1 atmosphere
at 100K the content of the balloon cools reducing the excitement of the gas content which also reduces the pressure, however, the balloon being perfectly elastic, contracts to maintain the 1 atmospheric pressure, hence the answer to (ii) is (c) It is constant,
For (i) since we know that the pressure of the balloon is constant
by Charles Law V1/T1 =V2/T2
or V2 = (V1/T1)×T2 =
×
=
× L = L/3 hence the correct answer to (i) is 1/3L
Answer:
8 time increase in K.E.
Explanation:
Consider Mass of truck = m kg and speed = v m/s then
K.E. = 1/2 ×mv²
If mass and speed both are doubled i.e let m₀ = 2m and v₀ = 2v then
(K.E.)₀ = 1/2 ×2m(2v)²
(K.E.)₀ = 8 (1/2 × mv²) = 8 × K.E.
The runner has initial velocity vector

and acceleration vector

so that her velocity at time
is

She runs directly east when the vertical component of
is 0:

It's not clear what you're supposed to find at this particular time... possibly her position vector? In that case, assuming she starts at the origin, her position at time
would be

so that after 10.4 s, her position would be

which is 19.9 m away from her starting position.