The tension in the string with friction would be the biggest because of the involvement of the force of gravity. This would result in that the friction force that is acting on the system. There is no friction in the frictionless system, and only the force of gravity is relevant.
Answer:
C is halved
Explanation:
The frequency and the wavelength of a wave are related by the equation:

where
v is the speed of the wave
f is the frequency
is the wavelength
From the equation above, we see that for a given wave, if the wave is travelling in the same medium (and so, its speed is not changing), then the frequency and the wavelength are inversely proportional to each other.
Therefore, if the frequency doubles, the wavelength will halve in order to keep the speed constant:

Answer:
25 meters
Explanation:
1 sec is 2.5, 1x10 so 2.5x10=25
Answer:
The maximum speed at which the car can safety travel around the track is 18.6m/s.
Explanation:
Since the car is in circular motion, there has to be a centripetal force
. In this case, the only force that applies for that is the static frictional force
between the tires and the track. Then, we can write that:

And since
and
, we have:

Now, if we write the vertical equation of motion of the car (in which there are only the weight and the normal force), we obtain:

Substituting this expression for
and solving for
, we get:

Finally, plugging in the given values for the coefficient of friction and the radius of the track, we have:

It means that in its maximum value, the speed of the car is equal to 18.6m/s.
Answer:
It is possible because, the TV broadcast audio and video signals in radio frequency which travels at the speed of light while the audio signals travel to those present in the stadium at the speed of sound which is over eight hundred thousand times slower than the speed of light
Explanation:
It is possible because of the following;
1. TV signals from the camera (including the captured sound) very close to the field of play are transmitted through the radio frequency bands and as such are a form of electromagnetic radiation that travels at the speed of light which is about 300,000 km/second
It will therefore, take 1 second for a sound of the game to reach someone located at 300,000,000 meters watching a live televised game
2. The speed of sound is about 343 m/second and it therefore takes up to 2 seconds for a sound to reach someone 686 meters away from the ball in the stadium.