A ray of light travelling obliquely from a rarer medium to a denser medium goes ______

A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is

AveGali [126]

Answer:

Centripetal acceleration = 0.79 m/s²

Explanation:

Given the following data;

Radius, r = 2.6 km

Time = 360 seconds

Conversion:

2.6 km to meters = 2.6 * 1000 = 2600 meters

To find the magnitude of centripetal acceleration;

First of all, we would determine the circular speed of the car using the formula;

$Circular \; speed (V) = \frac {2 \pi r}{t}$

Where;

r represents the radius and t is the time.

Substituting into the formula, we have;

$Circular \; speed (V) = \frac {2*3.142*2600}{360}$

$Circular \; speed (V) = \frac {16338.4}{360}$

Circular speed, V = 45.38 m/s

Next, we find the centripetal acceleration;

Mathematically, centripetal acceleration is given by the formula;

$Centripetal \; acceleration = \frac {V^{2}}{r}$

Where;

V is the circular speed (velocity) of an object.
r is the radius of circular path.

Substituting into the formula, we have;

$Centripetal \; acceleration = \frac {45.38^{2}}{2.6}$

$Centripetal \; acceleration = \frac {2059.34}{2600}$

Centripetal acceleration = 0.79 m/s²

3 0

3 years ago

A box is being dragged with a horizontal force of 65 N for 12 meters. If there is a force of friction acting on it

asambeis [7]

Answer:

A. 780 J

B. 120 J

C. 660 J

Explanation:

From the question given above the following data were obtained:

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

A. Determination of the work done by the dragging force.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Workdone (Wd) by dragging force =?

Wd = Fₔ × s

Wd = 65 × 12

Wd = 780 J

Therefore, the work done by the dragging force is 780 J

B. Determination of the work done by friction.

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Workdone (Wd) by friction =?

Wd = Fբ × s

Wd = 10 × 12

Wd = 120 J

Therefore, the work done by friction is 120 J

C. Determination of the net work done on the box.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Net work done (Wd) =?

Next, we shall determine the net force acting on the box. This can be obtained as follow:

Dragging force (Fₔ) = 65 N

Force of friction (Fբ) = 10 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fբ

Fₙ = 65 – 10

Fₙ = 55 N

Thus, the net force acting on the box is 55 N

Finally, we shall determine the net work done on the box as follow:

Distance (s) = 12 m

Net force (Fₙ) = 55 N

Net work done (Wd) =?

Wd = Fₙ × s

Wd = 55 × 12

Wd = 660 J

Therefore, the net work done on the box is 660 J

4 0

3 years ago

Why are earbud wires for an mp3 player coated with plastic

saw5 [17]

2) they add the insulation to better the durability or the earbuds

6 0

3 years ago

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Anon25 [30]

In several of the questions you've posted during the past day, we've already said that a wave with larger amplitude carries more energy. That idea is easy to apply to this question.

8 0

4 years ago

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Why is streak generally a more useful property than color in identifying a mineral?

poizon [28]

You may confuse two minerals by their colors so using streak is another way for showing their "true colors "

3 0

3 years ago

A ray of light travelling obliquely from a rarer medium to a denser medium goes _______​

A ray of light travelling obliquely from a rarer medium to a denser medium goes _______