1.1 x 10-1J
5.5 x 10-2J
1.1 x 10-4J
5.5 x 10-8J
Correct answer is B
e = \(\frac{4FI}{Y \pi d^2} = \frac{4 \times 11 \times 10 \times 2}{7 \times 10^3 \times 3.14(2 \times 10^{-2})^2}\)
= 0.001m
E = \(\frac{1}{2} Fe = \frac{1}{2} \times 110 \times 0.001\)
= 5.5 x 10-2J
From Newton's first law of motion,
a body can only undergo translational motion
once a body remains at rest no force acts on it
the net force acting in uniform linear motion is zero
a body's inertia is its weight
Correct answer is C
The key point of this law is that if there is no net force resulting from unbalanced forces acting on an object. If all the external forces cancel each other out, then the object will maintain a constant velocity.
period
amplitude
wavelength
frequency
Correct answer is B
No explanation has been provided for this answer.
0.36J
0.72J
360.00J
720.00J
Correct answer is A
P.E at max = K.E on projection
= \(\frac{1}{2}\)mv2 = \(\frac{1}{2} \times 0.02 \times 6^2 = 0.36J\)
1.2ms-1
6.0ms-1
12.0ms-1
0.6ms-1
Correct answer is C
V = u - gt
u = v + gt
= 0 + 10 x 1.2
= 12ms\(^{-1}\)