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3 years ago

you drive in a straight line at 20.0m/s for 10.0 min, then 30.0m/s for another 10.0 min. Find the average speed.

Physics

1 answer:

Juliette [100K]3 years ago

6 0

Answer is 25 m/s.

Explanation;

Step 1 : Convert the units into same unit.

Here speeds are given in m/s while the time is in minutes. So, let's convert minutes into seconds.

1 min = 60 s

Then,

10 minutes = 10 x 60 s = 600 s

Step 2 : Find the total distance that you have travelled.

To find that you need to first find distances separately.

We know the formula, s = vt, where s is the distance, v is the speed or velocity and t is the time.

First you have travelled 10 minutes (600 s) at 20.0 m/s.

Then distance, s = 20.0 m/s x 600 s = 12 000 m

Secondly you have travelled again 10 minutes at 30.0 m/s

Then distance = 30.0 m/s x 600 s = 18 000 m

Then, the total distance = 12 000 m + 18 000 m = 30 000 m

Step 3 : Find the total time taken to travel.

Total time = 600 s + 600 s = 1200 s

Step 4 : Calculate the average speed.

Average speed = total distance / total time

= 30 000 m / 1200 s

= 25 m/s

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The given data is as follows.

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Therefore, the concentration of caffeine in one can is calculated as follows.

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Since, it is given that lethal dose is 10.0 g. Hence, number of cans are calculated as follows.

No. of cans = $\frac{\text{Lethal dose}}{\text{concentration in one can}}$

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3 years ago

Three packing crates of masses, M1 = 6 kg, M2 = 2 kg and M3 = 8 kg are connected by a light string of negligible mass that passe

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Answer:

39.81 N

Explanation:

I attached an image of the free body diagrams I drew of crate #1 and #2.

Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.

∑Fₓ = maₓ

∑Fᵧ = maᵧ

Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:

∑Fᵧ = maᵧ
T₁ - m₂g = m₂aᵧ

Note that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.

Let's solve for T in the equation...