All categories

3 years ago

You should increase the distance between your car and the vehicle ahead when you:

1 answer:

8 0

You should increase the distance between your car and the vehicle ahead when you are being tailgated by another driver.
Tailgated is a condition when the vehicle behind you is very close to you that there is no distance between your vehicle and the vehicle behind you. In that case you should increase distance from the vehicle ahead to avoid collision in any case. These days people were mostly in hurry and everyone want to go ahead, so tailgating is very common.

You might be interested in

A boy got out of a boat and as he

emmainna [20.7K]

Answer:

third law of motion

Explanation:

6 0

3 years ago

17. Which of the following is the closest weight in newtons of a 7.0-kilogram

marusya05 [52]

Weight = (mass) x (gravity)

Weight = (7.0 kg) x (gravity)

On Earth, where (gravity) is roughly 10 N/kg . . .

Weight = (7.0 kg) x (roughly 10 N/kg)

Weight = roughly 70 Newtons

That's B on Earth.

It would be some other number on other bodies.

4 0

3 years ago

Please answer as soon as possible. A Physics question about electricity and circuits.

garik1379 [7]

Answer:

Explanation:

Electrical energy is:

E = IVt

E = 210 * 12 * 10\\\\\\E = 25200 J

5 0

3 years ago

A lamp in a child's Halloween costume flashes based on an RC discharge of a capacitor through its resistance. The effective dura

pshichka [43]

Answer:

A. 0.199 J

B. 0.0663 C

C = 0.0221 F

D. 12.68 ohms

Explanation:

From the question:

time duration, t = 0.28 seconds

Average power, P = 0.71 W

Average voltage, V = 3 V

A) Energy is given as:

E = P * t

=> E = 0.71 * 0.28 = 0.199 J

B) Electrical energy is also given as:

E = qV

where q = charge

=> q = E / V

∴ q = 0.199 / 3 = 0.0663 C

C) Capacitance is given as charge over voltage:

C = q / V

=> C = 0.0663 / 3 = 0.0221 F

D) Electrical power, P, can also be given as:

P = $V^2 / R$

where R = resistance

=> R = $V^2 / P$

R = $3^2 / 0.71 = 9 / 0.71 = 12.68 ohms$

8 0

3 years ago

THE RELATIVE ANGLE AT THE KNEE CHANGES FROM 0O TO 85O DURING THE KNEE FLEXION PHASE OF A SQUAT EXERCISE. IF 10 COMPLETE SQUATS A

Mice21 [21]

Answer: angular distance = 1700° and 29.7 rad

also the angular displacement = 0

Explanation:

To explain this, i will give a breakdown of this works.

we are asked to find both the angular distance and displacement the knee undergo.

Ok to get the distance of the knee, we would first take note that for one to squat down and get back up, the knee would travel through 85° of flexion to gpo down, and also through another 85° of extension to return standing (upright). So, the actual angular distance of the squat is 170°.

taking ten squats, the knee would have to go through 170° motion times 10 i.e;

10 * 170° = 1700°

Therefore the angulaar distance is 1700°

now converting this distance to radians since we will be required to have our answer in both degree and rad.

Given that 2pi = 360°, it means that one degree will give 57.3°;

∅ (rad) = ∅ (deg) * 2π/360°

∅ (rad) = 1170° * 2π/360° = 29.7 rad

∅ (rad) = 29.7 rad

b. For the other part, let us remember that angular displacement is equal to angular distance divided by time, so the angular displacement displacement of the knee will be zero, because the knee's position at the final third will be the same as the initial position.

cheers i hope this helps!!!!

5 0

3 years ago