Answer: Add an incline or grade to the road track.
Explanation:
Refer to the figure shown below.
When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
where
m = mass of the vehicle
v = linear (tangential) velocity to the circular path.
The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
μ = dynamic coefficient of friction.
At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.
When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
Therefore the force that opposes the centrifugal force becomes
μN + Wsinθ = W(sinθ + μ cosθ).
Answer:
60words/minute
Explanation:
If Sunitha can type 1800 words in half an hour, this can be expressed as;
1800 words = 30 minutes
To get her typing speed per minute, we will use the formula
Speed = Number of words/Time used
Typing speed = 1800/30
Typing speed = 60words/minute
Hence her typing speed in words per minute is 60words/minute
Answer:
The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.
Explanation:
Given that,
The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.
A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80 degrees.
We need to find the angular separation between the refracted red and refracted blue beams while they are in the glass.
Using Snell's law for red light as :

Again using Snell's law for blue light as :

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.
here's the first part but for the 2nd one all I know is that the word "compression" goes on the spirals that are closer together.
hope this helps!
<span>In this problem, we need to solve for Bubba’s mass. To do this, we let A be the area of the raft and set the weight of the displaced fluid with the raft alone as ρwAd1g and ρwAd2g with the person on the raft, </span>where ρw is the density of water, d1 = 7cm, and d2= 8.4 cm. Set the weight of displaced fluid equal to the weight of the floating objects to eliminate A and ρw then solve for m.
<span>ρwAd1g = Mg</span>
ρw<span>Ad2g = (M + m) g</span>
<span>d2∕d1 = (M + m)/g</span>
m = [(d2<span>∕d1)-1] M = [(8.4 cm/7.0 cm) - 1] (600 kg) =120 kg</span>
This means that Bubba’s mass is 120 kg.