The coefficient of friction must be 0.196
Explanation:
For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:
where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:
is the coefficient of friction between the tires and the road
m is the mass of the car
is the acceleration of gravity
v is the speed of the car
r is the radius of the curve
In this problem,
r = 750 m is the radius
is the speed
And solving for , we find the coefficient of friction required to keep the car in circular motion:
Learn more about circular motion:
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What happens to end a of the rod when the ball approaches it closely this first time is; It is strongly attracted.
<h3>Electrostatics</h3>
I have attached the image of the rod.
We are told that the ball is much closer to the end of the rod than the length of the rod. Thus, if we point down the rod several times, the distance of approach will experience no electric field and as such the charge on end point A of the rod must be comparable in magnitude to the charge on the ball.
This means that their fields will cancel.
Finally, we can conclude that when a charge is brought close to a conductor, the opposite charges will all navigate to the point that is closest to the charge and as a result, a strong attraction will be created.
This also applies to a strong conducting rod and therefore it is strongly attracted.
Read more about Electrostatics at; brainly.com/question/18108470
Answer:
1.25 m/s^2
Explanation:
m = 76 kg, R = 840 N
Let a be the acceleration of teh elevator in upward direction.
By use of Newton's second law
R - mg = m a
R = m ( g + a)
840 = 76 ( 9.8 + a)
a = 1.25 m/s^2
Frequency of the wave " the loudness"
Answer:
322 kJ
Explanation:
The work is the energy that a force produces when realizes a displacement. So, for a gas, it occurs when it expands or when it compress.
When the gas expands it realizes work, so the work is positive, when it compress, it's suffering work, so the work is negative.
For a constant pressure, the work can be calcutated by:
W = pxΔV, where W is the work, p is the pressure, and ΔV is the volume variation. To find the work in Joules, the pressure must be in Pascal (1 atm = 101325 Pa), and the volume in m³ (1 L = 0.001 m³), so:
p = 60 atm = 6.08x10⁶ Pa
ΔV = 82.0 - 29.0 = 53 L = 0.053 m³
W = 6.08x10⁶x0.053
W = 322x10³ J
W = 322 kJ
