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:
The separation of the 2 points should be 50.0 meters.
Explanation:
According to Rayleigh's scattering criteria the angular separation between 2 points to be resolved equals
Applying the given values we get
thus the linear separation equals 
Applying the given values we get
Answer:
12 cm and 0.4
Explanation:
f = - 20 cm, u = - 30 cm
Let v be the position of image and m be the magnification.
Use lens equation
1 / f = 1 / v - 1 / u
- 1 / 20 = 1 / v + 1 / 30
1 / v = - 5 / 60
v = - 12 cm
m = v / u = - 12 / (-30) = 0.4
Answer:
v = 3.27 m/s
Explanation:
KE = 1/2 mv^2
695 J = 1/2 (130kg)(v^2)
695 J / (1/2 x 130kg) = v^2
v^2 = square root of 10.69
v = 3.27 m/s
Answer:flexible
Explanation:In the trachea, or windpipe there are tracheal rings, also known as tracheal cartilage. cartilage is strong with flushable tissue. The tracheal cartilages help support the trachea while still allowing it to move and flex during breathing.
