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Tatiana [17]
3 years ago
7

Making quick glances to the roadway in front of your vehicle is called?

Physics
2 answers:
umka2103 [35]3 years ago
5 0

Answer:

safety check.

Explanation:

Elodia [21]3 years ago
5 0

Answer:

The answer is Ground Viewing. Hope this helps!

Explanation:

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21. If the Sun's rays were at 45° to a vertical pillar, how would
diamong [38]

Answer:

Let the height of the pole AB = x m. ∴ Length of shadow OB ol the pole AB = x m. Let the angle of elevation be ө, i.e. Hence, the angle of elevation of the Sun's altitude is 45°.

Explanation:

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3 years ago
Which of the following is an example of a conductor?
nordsb [41]
Answer: B. a gold chain
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2 years ago
What is the mass of a 200 kg object on Venus?
saw5 [17]
A-200 kg
I hope this helped xx
8 0
3 years ago
Read 2 more answers
Can a body have zero velocity and finite acceleration?Explain​
sergejj [24]

Answer:

Kinda? Depends what the question is fully asking

Explanation:

Acceleration is a change in velocity. So I guess if the velocity of something is -2 m/s and its positively accelerating at a value of +1 m/s, then that means every second its velocity changes by +1m/s.

So that -2 m/s thing after one second will be going -1 m/s.

After another second it'll be going 0 m/s.

After another itll be going +1 m/s and so on.

So at one point for a brief moment, it can have an acceleration but be at 0 m/s velocity.

5 0
3 years ago
A fully loaded cart with a mass of 2200 kg starts from the top of a 12-meter hill on a roller coaster.
Salsk061 [2.6K]

Answer:

A. potential energy is 258720 Joule

Explanation:

A.Gravitational potential energy is: PE = m × g × h

velocity =  15.33 m/s when the car reaches the bottom of the hill.

where, m = mass

            g = acceleration due to gravity

            h = height from the bottom of hill.

The potential energy is : m×g×h

                                     =(2200×9.8×12)

                                     =258720 Joule

B. at the bottom of the hill, the potential energy is converted into kinetic energy so PE at top = KE at bottom

                    kinetic energy= \frac{1}{2}(m*v^{2})

where v = velocity

          m= mass

therefore,               v=\sqrt\frac{2*K.E}{m} {}

                         or,  v=\sqrt{\frac{2*258720}{2200} }

                         or,   v=15.33 m/s

7 0
2 years ago
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