Answer:
B
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Explanation:
I did it and this was the correct answer
Your answer is C)
a)t=2.78 sec
b)R=835.03 m
c)
Explanation:
Given that
h= 38 m
u=300 m/s
here given that
The finally y=0
So
t=2.78 sec
The horizontal distance,R
R= u x t
R=300 x 2.78
R=835.03 m
The vertical component of velocity before the strike
Answer:
4.36 rad/s
Explanation:
Radius of platform r = 2.97 m
rotational inertia I = 358 kg·m^2
Initial angular speed w = 1.96 rad/s
Mass of student m = 69.5 kg
Rotational inertia of student at the rim = mr^2 = 69.5 x 2.97^2 = 613.05 kg.m^2
Therefore initial rotational momentum of system = w( Ip + Is)
= 1.96 x (358 + 613.05)
= 1903.258 kg.rad.m^2/s
When she walks to a radius of 1.06 m
I = mr^2 = 69.5 x 1.06^2 = 78.09 kg·m^2
Rotational momentuem of system = w(358 + 78.09) = 436.09w
Due to conservation of momentum, we equate both momenta
436.09w = 1903.258
w = 4.36 rad/s
<span>
A= 5 i − 3 J and B = −i − 6 j
A+ B
=5 i − 3 j −i − 6 j = 4 i - 9 j ---answer
A - B
= 5 i − 3 j + i + 6 j = 6 i + 3j ---answer
|A+B|
= sq root [ 16 + 81] = 9.848 ---answer
|A - B|
= sq root [ 36 + 9] = 6.708 ---answer
directions of A+ B
= tan (-1) -9 /4 = - 66 deg --answer
direction of A- B
= tan (-1) 3 /6 = 26.56 deg --answer</span>
Answer:
= 2.49 × 10³J
Explanation:
Total rotational kinetic energy K = nRT
R = 8.134 J/mol.K
n = 1 mol
T = 300 K
K = 1 × 8.314 × 300
= 2.49 × 10³J
