Answer:
0.0231 m/s
Explanation:
Given the question :
A small toddler is playing in his yard. Over the course of 29.01 s, he walks 2.099 m S before turning N and walking 1.429 m. What is his average velocity?
Distance walked south = 2.099 metre
Time taken = 29.01s
Distance walked north = 1.429 metre
Average Velocity = Displacement / time
The Displacement is hence,
(2.099 m - 1.429m) = 0.67m
Average Velocity = 0.67m / 29.01s
Average Velocity = 0.02309 m/s
Average Velocity = 0.0231 m/s
Answer:
The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N
(b) the force the light beam exerts is much too small to be felt by the man.
Explanation:
Given;
cross-sectional area of the man, A = 0.500m²
intensity of light, I = 35.5kW/m²
If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;
P = I/c
where;
P is the pressure of the incident light
I is the intensity of the incident light
c is the speed of light
F = PA
where;
F is the force of the incident light on the man
P is the pressure of the incident light on the man
A is the cross-sectional area of the man
F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N
The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N
Therefore, the force the light beam exerts is much too small to be felt by the man.
Answer:
<h2><u>Rheostat:</u></h2>=> Rheostat is an electric device used as a variable resistance, like the regulator of the fan. It is used to change the electric resistance in the electric circuit.
<h2><u>Principle / Rule:</u></h2>=> Rheostat works on the principle of Ohm’s law. Ohm’s law states that current in a circuit is inversely proportional to the resistance at the given temperature.
Answer:
the cannonball’s velocity parallel to the ground is 86.6m/S
Explanation:
Hello! To solve this problem remember that in a parabolic movement the horizontal component X of the velocity of the cannonball is constant while the vertical one varies with constant acceleration.
For this case we must draw the velocity triangle and find the component in X(see atached image).
V= Initial velocity=100M/S
V= Initial velocity=100M/S
Vx=cannonball’s velocity parallel to the ground
Solving for Vx
Vx=Vcos30
Vx=(100m/S)(cos30)=86.6m/s
the cannonball’s velocity parallel to the ground is 86.6m/S
