If the velocity is constant then the acceleration of the object is zero.
![a=0 (m/s^2)](https://tex.z-dn.net/?f=a%3D0%20%28m%2Fs%5E2%29)
Thus when we apply the equation
![\Delta X=vt+(at^2/2)](https://tex.z-dn.net/?f=%5CDelta%20X%3Dvt%2B%28at%5E2%2F2%29%20)
It remains
![\Delta X =vt](https://tex.z-dn.net/?f=%5CDelta%20X%20%3Dvt%20)
or equivalent
Answer:
The minimum possible coefficient of static friction between the tires and the ground is 0.64.
Explanation:
if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :
Fc = f
m×(v^2)/(R) = μ×m×g
(v^2)/(R) = g×μ
μ = (v^2)/(R×g)
= ((25)^2)/((100)×(9.8))
= 0.64
Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.
Answer:
C) Use a battery with more voltage.
Explanation:
The equation for the magnetic field around a coil is given by,
B = μ₀NI
where,
B = Magnetic flux density
μ₀ = permeability
N = number of turns per meter
I = Current in the wire
So when using a higher voltage battery, more current passes through the battery as resistance of the wire remains the same.
Answer:
Newton's third law of motion states that for every action, there is equal and opposite reaction.
While space walking, when the astronaut gets detached from the space ship, she floats in space holding a wrench. In order to get back to the spaceship, she should throw the wrench in the opposite direction of the spaceship. This action would cause a reaction on her own body and she would be pushed away from the wrench and towards the spaceship. Thus, she can return back to the spaceship in this way.
The driver is tooling along in his snowmobile, pointed north,
at 8.5 m/s.
He's carrying the flares with him, so the flares are also moving north
at 8.5 m/s.
When he fires the flare straight up, it has a vertical velocity of 4.3 m/s
straight up, and a horizontal velocity of 8.5 m/s towards the north.
The magnitude of the net velocity is √(4.3² + 8.5²) .
That's about 9.53 m/s, at some angle between straight up
and straight north.
The angle above horizontal is the angle that has a tangent of 4.3/8.5 .
I'll let you work out the angle.