1st Law: Objects that are in motion tend to stay in motion. This motion can change with external forces.
<span>If you were to stop pedaling on bike while in motion, you will notice that you will keep moving. This is because a moving body (you) has inertia. If there wasn't any friction between the tires and the ground, between the axles and wheel, any air resistance, or any other force that acts against you, then you could be coasting indefinitely! </span>
<span>2nd Law: Force is equal to the mass times acceleration. </span>
<span>When you pedal, you are applying a force onto the pedal. This force is then translated through tension to apply torque onto the wheel. Turning the wheel will make you accelerate in the lateral direction. </span>
<span>3rd Law: For every action, there is an equal and opposite reaction. </span>
<span>Without this, you could pedal and pedal, but you will be not go anywhere! It is essentially the friction between the tires and the ground that propels you forward. If the ground did not apply to the tire the same amount of force that the tire was applying to the ground, the tire would not "catch" and no friction would be applied. And if there was no third law, the weight of you and your bike would "sink" into the ground because the ground would not be applying a normal force back onto you.
hope this helps and if you have any questions just hmu and ask :)</span>
Answer:
Explained
Explanation:
A student who runs every day is becoming bored with her physical fitness routine and considers quitting. Now in order maintain her fitness routine she needs to break the monotony by changing her exercise regime. She do other activities like play sports two days a week, she do aerobics, she even try some cardio vascular exercises which as good as running. She also try changing her route of running as the new view might inspire her.
Answer:
m = 0.25
Explanation:
Given that,
Object distance, u = -15cm
Height of the object, h = 48
Focal length, f = cm
We need to find the magnification of the image.
Let v is the image distance. Using mirror's equation.
Magnification,
Hence, the magnification of the image is 0.25.
Answer:
Explanation:
Given data:
v = 220 rms
power factor = 0.65
P = 1250 W
New power factor is 0.9 lag
we knwo that
s = 1923.09 < 49.65^o
s = [1250 + 1461 j] vA
![P.F new = cos [tan^{-1} \frac{Q_{new}}{P}]](https://tex.z-dn.net/?f=P.F%20new%20%3D%20cos%20%5Btan%5E%7B-1%7D%20%5Cfrac%7BQ_%7Bnew%7D%7D%7BP%7D%5D)
solving for 
![Q_{new} = P tan [cos^{-1} P.F new]](https://tex.z-dn.net/?f=Q_%7Bnew%7D%20%3D%20P%20tan%20%5Bcos%5E%7B-1%7D%20P.F%20new%5D)
![Q_{new} = 1250 [tan[cos^{-1}0.9]]](https://tex.z-dn.net/?f=Q_%7Bnew%7D%20%3D%201250%20%5Btan%5Bcos%5E%7B-1%7D0.9%5D%5D)
Faraday
