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2 years ago

11

If the road becomes wet or crowded, you should ____. slow down and increase your following distance All choices are incorrect. m

aintain your speed and following distance speed up and decrease your following distance Submit answer

1 answer:

Vaselesa [24]2 years ago

7 0

Answer:

The first one.

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According to the law of conservation of matter, we know that the total number of atoms does not change in a chemical reaction an

iogann1982 [59]

Answer:

In a chemical reaction the total mass of all the substances taking part in the reaction remains the same. Also, the number of atoms in a reaction remains the same. Mass cannot be created or destroyed in a chemical reaction.

Explanation:

7 0

3 years ago

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. how much time does it ta

Yuri [45]

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. The time taken by the car will be 4 seconds. the correct answer is option(c).

When acceleration is constant, the rate of change in velocity is also constant. In the absence of any acceleration, velocity remains constant. When acceleration is positive, velocity becomes more significant.

Let a denote acceleration, u denote initial velocity, v denote final velocity, and t denote time.

The equation of motion is stated as,

v = u + at

v² = u² + 2as

A car travels across a distance of 60.0 m while accelerating constantly from 12.0 m/s to 18.0 m/s.

Then the time taken by the car will be,

u = 12 m/s

v = 18 m/s

s = 60 m

Put these in the equation v² = u² + 2as.

18² = 12² + 2 x a x 60

a = 1.5

Then the time will be

18 = 12 + 1.5t

1.5t = 6

t = 4 seconds

Hence, the time taken is 4 s.

The complete question is:

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. How much time does it take?

1.00 s

2.50 s

4.00 s

4.50 s

To know more about acceleration refer to: brainly.com/question/12550364

#SPJ4

8 0

1 year ago

Rays of light coming from the sun (a very distant object) are near and parallel to the principal axis of a concave mirror. After

Setler [38]

A concave mirror is an example of curved mirrors. So that the appropriate answer to the given question is option D. The rays will cross at the focal point.

A concave mirror is a type of mirror in which its inner part is the reflecting surface, while its outer part is the back of the mirror. This mirror reflects all parallel rays close to the principal axis to a point of convergence. It can also be referred to as the converging mirror.

In this type of mirror, all rays of light parallel to the principal axis of the mirror after reflection will cross at the focal point.

Therefore, the required answer to the given question is option D. i.e The rays will cross at the focal point.

For reference: brainly.com/question/20380620

3 0

3 years ago

You want to slide a 0.39 kg book across a table. If the coefficient of kinetic friction is .21, what force is required to move t

uranmaximum [27]

Find the force that would be required in the absence of friction first, then calculate the force of friction and add them together. This is done because the friction force is going to have to be compensated for. We will need that much more force than we otherwise would to achieve the desired acceleration:

$F_{NoFric}=ma=0.39kg \times0.18 \frac{m}{s^2} =0.0702N$

The friction force will be given by the normal force times the coefficient of friction. Here the normal force is just its weight, mg

$F_{Fric}=0.39kg \times 9.8 \frac{m}{s^2} \times 0.21=0.803N$

Now the total force required is:

0.0702N+0.803N=0.873N

5 0

3 years ago

What are the seven types of electromagnetic waves?

Sunny_sXe [5.5K]

Radio waves. Giant satellite-dish antennas pick up long-wavelength, high-frequency radio waves. ...
Microwaves. Because cosmic microwaves can't get through the whole of Earth's atmosphere, we have to study them from space. ...
Infrared. ...
Visible light. ...
Ultraviolet light. ...
X rays. ...
Gamma rays.

5 0

3 years ago