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3 years ago

IMPORTANT 3 QUESTIONS!

Physics

1 answer:

Brums [2.3K]3 years ago

6 0

Answer:

7. 20,000,000 mL.

8. 8,000 m.

9. 120,000 secs.

10. 4

Explanation:

7. Determination of the volume in millilitres (mL)

Volume in litre (L) = 20,000 L

Volume in millilitres (mL) =..?

The volume in mL can be obtained as follow:

1 L = 1,000 mL

Therefore,

20,000 L = 20,000 x 1,000 = 20,000,000 mL.

Therefore, 20,000 L is equivalent to 20,000,000 mL.

8. Determination of the distance in metre (m)

Distance in mile = 5 mile

Distance in metre =?

First, we shall convert from mile to kilometre.

This can be done as follow:

1 mile = 1.6 km

Therefore,

5 mile = 5 x 1.6 = 8 km

Finally, we shall convert 8 km to metre (m).

This is illustrated below:

1 km = 1,000 m

Therefore,

8 km = 8 x 1,000 = 8,000 m

Therefore, 5 miles is equivalent to 8,000 m.

9. Determination of the time in seconds.

Time = 400 minutes for 5 days.

First, we shall convert 400 mins to hour.

This is illustrated below:

60 minutes = 1 hour

Therefore,

400 mins = 400/60 = 20/3 hours

The time (hours) is 20/3 hours in 1 day.

Therefore, the time (hours) in 5 days will be = 20/3 x 5 = 100/3 hours.

Next, we shall convert 100/3 hours to minutes.

This is illustrated below:

1 hour = 60 minutes

Therefore,

100/3 hours = 100/3 x 60 = 2000 mins

Finally, we shall convert 2000 mins to seconds.

This is illustrated below:

1 mins = 60 secs

2000 mins = 2000 x 60 = 120,000 secs.

Therefore, the time is 120,000 secs.

10. Determination of the number of significant figures.

To obtain the significant figures of a number, we simply count all the numbers available.

Therefore, the number of significant figures for 9876 is 4.

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A 5.20-N force is applied to a 1.05-kg object to accelerate it rightwards. The object encounters 3.29-N of friction. Determine t

tatiyna

Answer:

Explanation:

We will use the equation F - f = ma, which is just a fancy way of stating Newton's 2nd Law. For us:

F = 5.20 to the right (+)

f = 3.29 to the left (-)

m = 1.05 kg. Therefore,

5.20 - 3.29 = 1.05a and

1.91 = 1.05a so

a = 1.82 m/s/s to the right

8 0

3 years ago

A light spring with a force coefficient 11.85 N/m is compressed by 14 cm as it is held between a 0.27 kg block on the left and a

lilavasa [31]

Answer:

A) Left = 6.14 m/s2 Right=2.765 m/s2 B) Left = 4.59 m/s2 Right= 1.215 m/s2

Explanation:

In the question we are given the spring constant which is 11.85 N/m and the compression of the spring is 14 cm. There are two blocks in front of the springs which are 0.27 kg and 0.6 kg respectively.

First, we need to calculate how much force the spring are going to exert on the masses when they are released. Since this force is not going to change for both cases, we only need to do it once.

The formula for calculation the force is F = k.x where k is the spring constant or force coefficient 11.85 N/m and the x is the compression which is 14 cm or 0.14 meters. If we put them in the equation we can find that F = 1.659 N

A) In the first case scenario, where the friction is equal to 0, we can use the formula F=m.a where F is the force applied by the release of the spring, m is the mass of the block and a is the acceleration.

For the first block, when we put 1.659 N and 0.27kg in the equation, a is calculated to be 6.14 m/s2.

For the second block, the same force of 1.659 N and 0.6 kg, a is calculated to be 2.765 m/s2.

B) In the second case scenario, where the friction is equal to 0.158, we first need to calculate its effect on each block. We need to use the formula Fk = μ.N where μ is the friction constant and N is the normal force of the block which is m.g (where g = 9,81 m/s2).

The friction force for the first block is calculated to be 0.4184 N and the friction force for the second block is calculated to be 0.9299 N.

The total force for the first block is 1.659 N - 0.4184 N = 1.2405 N.

The total force for the second block is 1.659 N - 0.9299 N = 0.7290 N.

When we insert these numbers into the same equation we used to find the acceleration in the first case scenario, we can calculate the acceleration of the first block as 1.2405 = 0.27 x a and a = 4.59 m/s2.

When we insert these numbers into the same equation we used to find the acceleration in the first case scenario, we can calculate the acceleration of the second block as 0.7290 = 0.6 x a and a = 1.215 m/s2.

4 0

3 years ago

you nose out another runner to win 100.000 m dash. if your total time for the race was 13.800 s and you aced out the other runne

sammy [17]

Your average speed was

(100 m) / (13.8 s) = 7.25 m/s .

If you finished 0.001s ahead of him, then at your average speed, that corresponds to

(7.25 m/s) x (0.001 s) = 0.00725 m

That's 7.25 millimeters ... about 0.28 of an inch !

NOTE:. I think this is only valid if your speed was a constant ~7.25 m/s all the way.

3 0

3 years ago