All categories

3 years ago

Can someone plz help me with this

Physics

1 answer:

Elena-2011 [213]3 years ago

3 0

1st Law: Objects that are in motion tend to stay in motion. This motion can change with external forces.

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! 

2nd Law: Force is equal to the mass times acceleration. 

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. 

3rd Law: For every action, there is an equal and opposite reaction. 

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 :)

You might be interested in

The image below The image below The image below The image below The image below The image below The image below The image below

larisa [96]

Answer:

wow i have know idea

Explanation:

5 0

3 years ago

A man is 40 kg . calculate his weight.(g=10m/s^2)

Dafna1 [17]

${ \large{ \bf{Weight = mg}}}$

${ \large{ \bf{Mass = 40 \: kg \: \: \: \: (given)}}}$

${ \large{ \bf{Weight = 40 \times 10}}}$

${ \large{ \therefore{ \bf{ \red{ Weight = 400N}}}}}$

5 0

3 years ago

A disk of radius a has a total charge Q uniformly distributed over its surface. The disk has negligible thickness and lies in th

sleet_krkn [62]

The electric potential V(z) on the z-axis is : V = $(\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z$

The magnitude of the electric field on the z axis is : E = kб 2 $\pi$ ( 1 - [z / √(z² + a² ) ] )

Given data :

V(z) =2kQ / a²(v(a² + z²) ) -z

<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>

Considering a disk with radius R

Charge = dq

Also the distance from the edge to the point on the z-axis = √ [R² + z²].

The surface charge density of the disk ( б ) = dq / dA

Small element charge dq = б( 2πR ) dr

dV $\frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }$ ----- ( 1 )

Integrating equation ( 1 ) over for full radius of a

∫dv = $\int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,$

V = $\pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ]$

= $\pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} } -z ]$

Therefore the electric potential V(z) = $(\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z$

Also

The magnitude of the electric field on the z axis is : E = kб 2 $\pi$ ( 1 - [z / √(z² + a² ) ] )

Hence we can conclude that the answers to your question are as listed above.

Learn more about electric potential : brainly.com/question/25923373

7 0

2 years ago

What is the acceleration of a body moving with uniform velocity?

AfilCa [17]

Acceleration is the rate of change of velocity, a body moving with uniform velocity does not possess acceleration at all i.e. acceleration is zero

3 0

3 years ago

A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed

fgiga [73]

Answer:

Explanation:

Time taken to accelerate to 28 m /s

= 28 / 2 = 14 s

a ) Total length of time in motion

= 14 + 41 + 5

= 60 s .

b )

Distance covered while accelerating

s = ut + 1/2 at²

= 0 + .5 x 2 x 14²

= 196 m .

Distance covered while moving in uniform motion

= 28 x 41

= 1148 m

distance covered while decelerating

v = u - at

0 = 28 - a x 5

a = 5.6 m / s²

v² = u² - 2 a s

0 = 28² - 2 x 5.6 x s

s = 28² / 2 x 5.6

= 70 m .

Total distance covered

= 196 + 1148 + 70

= 1414 m

total time taken = 60 s

average velocity

= 1414 / 60

= 23.56 m /s .

8 0

3 years ago