
- Speed of the mobile = 250 m/s
- It starts decelerating at a rate of 3 m/s²
- Time travelled = 45s

- Velocity of mobile after 45 seconds

We can solve the above question using the three equations of motion which are:-
- v = u + at
- s = ut + 1/2 at²
- v² = u² + 2as
So, Here a is acceleration of the body, u is the initial velocity, v is the final velocity, t is the time taken and s is the displacement of the body.

We are provided with,
- u = 250 m/s
- a = -3 m/s²
- t = 45 s
By using 1st equation of motion,
⇛ v = u + at
⇛ v = 250 + (-3)45
⇛ v = 250 - 135 m/s
⇛ v = 115 m/s
✤ <u>Final</u><u> </u><u>velocity</u><u> </u><u>of</u><u> </u><u>mobile</u><u> </u><u>=</u><u> </u><u>1</u><u>1</u><u>5</u><u> </u><u>m</u><u>/</u><u>s</u>
<u>━━━━━━━━━━━━━━━━━━━━</u>
Cosmic background radiation is electromagnetic radiation from the sky with no discernible source. The origin of this radiation depends on the region of the spectrum that is observed.
Answer:
oa
Explanation:
it may be oa is the right answer for this question
but I don't know properly
Since the ladder is standing, we know that the coefficient
of friction is at least something. This [gotta be at least this] friction
coefficient can be calculated. As the man begins to climb the ladder, the
friction can even be less than the free-standing friction coefficient. However,
as the man climbs the ladder, more and more friction is required. Since he
eventually slips, we know that friction is less than what's required at the top
of the ladder.
The only "answer" to this problem is putting lower
and upper bounds on the coefficient. For the lower one, find how much friction
the ladder needs to stand by itself. For the most that friction could be, find
what friction is when the man reaches the top of the ladder.
Ff = uN1
Fx = 0 = Ff + N2
Fy = 0 = N1 – 400 – 864
N1 = 1264 N
Torque balance
T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)
N2 = 439 N
Ff = 439= u N1
U = 440 / 1264 = 0.3481