8. (-/2 Points]

describe the forces acting on a car as it movies along a level highway in still air at a constant speed

Travka [436]

Answers

The car's forward motion is opposed by the friction between the road and the tires and by the resistance of the air.

3 0

2 years ago

To counter the effects of centrifugal force and reduce vehicle traction it is important to to counter the effects of centrifugal

tatiyna

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).

5 0

3 years ago

Find the mass of a body if the acceleration the body used to move is given as 5 m/s-2 and the force of the body is 30N take g=10

Aleonysh [2.5K]

The mass of a body if the acceleration the body used to move is given as 5 m/s-2 will be 3 kg.

<h3>What is force?</h3>

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Force is defined as the product of mass and acceleration. Its unit is Newton.

Given data;

Force,F = 30 N

Mass,m = kg

Acceleration,a = 5 m/s²

The force is found as;

F=ma

30 N =m kg × 5 m/s²

m=3 kg

Hence the mass of a body will be 3 kg.

To learn more about the force refer to the link;

brainly.com/question/26115859#SPJ1

#SPJ1

7 0

2 years ago

The diffraction limit is a limit on: The diffraction limit is a limit on: A telescope's size. A telescope's angular resolution.

Mnenie [13.5K]

Answer:

A telescope's angular resolution.

Explanation:

Diffraction limit is a minimum angular separation of two sources and it can be distinguished by the telescope. This angle is known as the diffraction limit. It is proportional to the wavelength of light and it has an inverse relation with the diameter of the telescope. Mathematically it is defined as

θ = 1.22λ/d

where θ is the angle, λ wavelength and d is the diameter of the objective mirror (lenz).

7 0

3 years ago

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest

shutvik [7]

This is note the complete question, the complete question is:

One of the lousy things about getting old (prepare yourself!) is that you can be both near-sighted and farsighted at once. Some original defect in the lens of your eye may cause you to only be able to focus on some objects a limited distance away (near-sighted). At the same time, as you age, the lens of your eye becomes more rigid and less able to change its shape. This will stop you from being able to focus on objects that are too close to your eye (far-sighted). Correcting both of these problems at once can be done by using bi-focals, or by placing two lenses in the same set of frames. An old physicist instructor can only focus on objects that lie at distance between 0.47 meters and 5.4 meters.

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness?

Answer: 3.04 D

Explanation:

when an object is held 21 cm away from the instructor's eyes, the spectacle lens must produce 0.47m ( the near point) away.

An image of 0.47m from the eye will be ( 47 - 2 )

i.e 45 cm from the spectacle lens since the spectacle lens is 2cm away from the eye.

Also, the image distance will become negative

gap between lense and eye = 2cm

Therefore;

image distance d₁ = - 45cm = - 0.45m

object distance d₀ = 21 - 2 = 19cm = 0.19m

P = 1/f = 1/ d = 1/d₀ + 1/d₁ = 1/0.19 + (-1/0.45)

P = 1/f = 5.26315789 - 2.22222222

P = 1/f = 3.04093567 ≈ 3.04 D

5 0

3 years ago