Answer:
<em>The number 0.0217 has 3 significant digits</em>
Explanation:
<u>Significant Digits
</u>
These are digits that contribute to the significance of the number. Some rules apply to discard the non-significant digits like:
- Leading zeros
- Trailing zeros (with exceptions)
Our number is 0.0217 has two leading zeros before the 2 because they only occupy space to indicate the order of magnitude of the number. Only the 2,1,7 are significant digits, thus
The number 0.0217 has 3 significant digits
Answer:
Torque is 93 Nm anticlockwise.
Explanation:
We have value of torque is cross product of position vector and force vector.
A force of 38 N, directed 30° above the x axis in the x-y plane.
Force, F = 38 cos 30 i + 38 sin 30 j = 32.91 i + 19 j
A particle is located on the x axis 4.9 m and we have to find torque about the origin on the particle.
Position vector, r = 4.9 i
Torque, T = r x F = 4.9 i x (32.91 i + 19 j) = 4.9 x 19 k = 93.1 k Nm
So Torque is 93 Nm anticlockwise.
Answer:
u/2 √(1 + 3 cos² θ)
Explanation:
The object is thrown at an angle θ, so the velocity has two components, vertical and horizontal.
Initially, the vertical component is u sin θ and the horizontal component is u cos θ.
At the maximum height, the vertical component is 0 and the horizontal component is u cos θ.
The mean vertical velocity is:
(u sin θ + 0) / 2 = u/2 sin θ
The mean horizontal velocity is:
(u cos θ + u cos θ) / 2 = u cos θ
The net mean velocity can be found with Pythagorean theorem:
v² = (u/2 sin θ)² + (u cos θ)²
v² = u²/4 sin² θ + u² cos² θ
v² = u²/4 (1 − cos² θ) + u² cos² θ
v² = u²/4 (1 − cos² θ) + u²/4 (4 cos² θ)
v² = u²/4 (1 − cos² θ + 4 cos² θ)
v² = u²/4 (1 + 3 cos² θ)
v = u/2 √(1 + 3 cos² θ)
Explanation:
Below is an attachment containing the solution.
Answer:
No it can not
Explanation: Kinetic energy is the energy of motion so it can not be negative the kinetic energy can only be at a point of "0" which is when its not moving. (I hope this helped) :))
