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Schach [20]
2 years ago
6

________ skids occur when the brakes are applied so hard that the front or rear wheels lost traction. Cornering

Physics
2 answers:
pochemuha2 years ago
8 0

Answer:

Braking skid

Explanation:

When you apply brake very hard so that the rear wheel lost traction leads to the occurrence of Braking skid.

Skid occurs when the Tyre looses it grip from the road due to sudden brake or road being smooth and many more reasons.

Skid can be due to the worn off Tyre.

some time skidding can cause accident also so  one should use good Tyre for their vehicle.

mihalych1998 [28]2 years ago
7 0

Answer:

Braking skid

Explanation:

A skid occurs when a vehicle is moving on the road and the tires of the vehicle lose their hold on the roadway.

Braking skid occurs when the brakes are applied suddenly and the braking is very hard that it results in the lose of hold of the rear wheels and the rear wheels lose grasp over the road resulting in the skid of the vehicle.

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What is 0 point energy times negative 0 point energy ????????
elixir [45]
1 point energy should be the answer
8 0
3 years ago
The reflective surface of a CD consists of spirals of equally spaced grooves. If you shine a laser pointer on a CD, each groove
Ipatiy [6.2K]

Answer:

d = 1.55 * 10⁻⁶ m

Explanation:

To calculate the distance between the adjacent grooves of the CD, use the formula, d = \frac{m \lambda}{sin(A_{m}) }..........(1)

The fringe number, m = 1 since it is a first order maximum

The wavelength of the green laser pointer, \lambda = 532 nm = 532 * 10⁻⁹ m

Distance between the central maximum and the first order maximum = 1.1 m

Distance between the screen and the CD = 3 m

A_{m} = Angle between the incident light and the diffracted light

From the setup shown in the attachment, it is a right angled triangle in which

sin(A_{m}) = \frac{opposite}{Hypotenuse} \\sin(A_{m}) =\frac{1.1}{\sqrt{1.1^{2}+3^{2}}}

sin(A_{m} ) = 0.344\\A_{m} = sin^{-1} 0.344\\A_{m} = 20.14^{0}

Putting all appropriate values into equation (1)

d = \frac{1* 532*10^{-9} }{0.344 }\\d = 0.00000155 m\\d = 1.55 * 10^{-6} m

3 0
3 years ago
4. As Juan is going to take a shower, the soap falls out of the soap dish on to the
LiRa [457]

The coefficient of friction between the soap and the floor is 0.081

If Juan steps on the soap with a force of 493 N, this is her weight, W. This weight also equals the normal reaction on the floor, N.

We know that frictional force F = μN where μ = coefficient of friction between soap and floor.

So, μ = F/N

Since F = 40 N and N = W = 493 N,

μ = F/N

μ = 40 N/493 N

μ = 0.081

So, the coefficient of friction between the soap and the floor is 0.081

Learn more about coefficient of friction here:

brainly.com/question/13923375

5 0
3 years ago
What is the relationship between the normal force and weight
GalinKa [24]

Answer: they have the same magnitude.

Explanation:

normal force = mg

weight = mg

4 0
2 years ago
A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In
atroni [7]

Answer:

The average induced emf around the border of the circular region is 8.48\times 10^{-5}\ V.

Explanation:

Given that,

Radius of circular region, r = 1.5 mm

Initial magnetic field, B = 0

Final magnetic field, B' = 1.5 T

The magnetic field is pointing upward when viewed from above, perpendicular to the circular plane in a time of 125 ms. We need to find the average induced emf around the border of the circular region. It is given by the rate of change of magnetic flux as :

\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=A\dfrac{-d(B'-B)}{dt}\\\\\epsilon=\pi (1.5\times 10^{-3})^2\times \dfrac{1.5}{0.125}\\\\\epsilon=8.48\times 10^{-5}\ V

So, the average induced emf around the border of the circular region is 8.48\times 10^{-5}\ V.

6 0
3 years ago
Read 2 more answers
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