Answer:
lauda and lasan I am fine with me lit Q ans pg-14
Answer: -0.84 rad/sec (clockwise)
Explanation:
Assuming no external torques act on the system (man + turntable), total angular momentum must be conserved:
L1 = L2
L1 = It ω + mm. v . r = 81.0 kg . m2 .21 rad/s – 56.0 kg. 3.1m/s . 3.1 m
L1 = -521.15 kg.m2/sec (1)
(Considering to the man as a particle that is moving opposite to the rotation of the turntable, so the sign is negative).
Once at rest, the runner is only a point mass with a given rotational inertia respect from the axis of rotation, that can be expressed as follows:
Im = m. r2 = 56.0 kg. (3.1m)2 = 538.16 kg.m2
The total angular momentum, once the runner has come to an stop, can be written as follows:
L2= (It + Im) ωf = -521.15 kg.m2/sec
L2= (81.0 kg.m2 + 538.16 kg.m2) ωf = -521.15 kg.m2/sec
Solving for ωf, we get:
ωf = -0.84 rad/sec (clockwise)
<u>Answer:</u>
The correct answer option is: 2 km South.
<u>Explanation:</u>
We are to find out the displacement for the given vectors:
- 4 km south,
- 2 km north,
- 5 km south; and
- 5 km north
So starting off with the vector going 4 km south we have a displacement of 4 km south. Then it goes 2 km so we are left with 4 - 2 = 2 km south.
Next, it goes 5 km south so we will have a displacement of 2 + 5 = 7 km south. Then again it goes 5 km north so we are left with a displacement of 7 - 5 = 2 km South.
Answer:
8.756 rad/s²
Explanation:
Given that:
A motorcycle accelerates uniformly from rest, then initial velocity v_i = 0 m/s
It final velocity v_f = 24.8 m/s
time (t) = 9.87 s
radius (r) of each tire = 0.287 m
Firstly; the linear acceleration of the motor cycle is determined as follows:
=(V_f - v_i)/t
=(24.8-0)/9.87
=2.513 m/s²
Then; the magnitude of angular acceleration
α = /r
=2.513/0.287
=8.756 rad/s²
It holds more weight in the regular water.