Below are the choices that can be found elsewhere:
A. (4.9 × 10-14 newtons) · tan(30°)
<span>B. (4.9 × 10-14 newtons) · sin(30°) </span>
<span>C. (4.9 × 10-14 newtons) · cos(30°) </span>
<span>D. (4.9 × 10-14 newtons) · arctan(30°) </span>
<span>E. (4.9 × 10-14 newtons) · arccos(30°)
</span>
<span>Force is proportional to the angle made by the velocity with respect to the magnetic field. It is maximum when velocity is perpendicular to the magnetic field and minimum when the velocity is parallel to the magnetic field. It is proportional to sin of the angle. In this problem it will be proportional to sin(30)</span>
Answer:
a) v = 21.34 m/s
b) v = 21.34 m/s
c) v = 21.34 m/s
Explanation:
Mass of the snowball, m = 0.560 kg
Height of the cliff, h = 14.2 m
Initial velocity of the ball, u = 13.3 m/s
θ = 26°
The speed of the slow ball as it reaches the ground, v = ?
The initial Kinetic energy of the snow ball,
Potential energy of the snow ball at the given height, PE = mgh
Final Kinetic energy of the ball as it reaches the ground,
a) Using the principle of energy conservation,
b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch
c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass
v = 21. 34 m/s
To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,
Here,
= Magnification
= Focal length eyepieces
= Focal length of the Objective
Rearranging to find the focal length of the objective
Replacing with our values
Therefore the focal length of th eobjective lenses is 27.75cm
The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.
<h3>How to determine the friction factor</h3>
Using the formula
μ = viscosity = 0. 06 Pas
d = diameter = 120mm = 0. 12m
V = velocity = 1m/s and 3m/s
ρ = density = 0.9
a. Velocity = 1m/s
friction factor = 0. 52 ×
friction factor = 0. 52 ×
friction factor = 0. 52 × 0. 55
friction factor
b. When V = 3mls
Friction factor = 0. 52 ×
Friction factor = 0. 52 ×
Friction factor = 0. 52 × 0. 185
Friction factor
Loss When V = 1m/s
Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter
Head loss = 0. 289 × × ×
Head loss = 1. 80 × 10^8
Head loss When V = 3m/s
Head loss = × × ×
Head loss = 5. 3× 10^8
Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.
Learn more about friction here:
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