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olga55 [171]
2 years ago
15

Place the gears in order from highest to lowest torque?

Physics
1 answer:
Trava [24]2 years ago
5 0
60, 12, 24,48- ddfjjvdd
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Make the following conversion. 34.9 cL = _____ hL. 0.349 0.0349 0.00349 349,000
AnnyKZ [126]
Hey there!

Here is your answer:

<u><em>The proper answer to this question is option C "</em></u><span><u><em>0.00349".</em></u>

Reason:

</span><span><u><em>1 L = 100 cL. Or 1 cL = 0.01 L</em></u>

</span><span><u><em>34.9 cL = 34.9 / 100 L = 0.349 L</em></u>

</span><span><u><em> 1 hL = 100 L. 0.349 L = 0.349 / 100 hL = 0.00349 hL</em></u>

<em>Therefore the answer is option C!</em>

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit</span>
4 0
3 years ago
8. At what position does the mass have the greatest acceleration?
gulaghasi [49]

Answer:

Option (e)

Explanation:

If a mass attached to a spring is stretched and released, it follows a simple harmonic motion.

In simple harmonic motion, velocity of the mass will be maximum, kinetic energy is maximum and acceleration is 0 at equilibrium position (at 0 position).

At position +A, mass will have the minimum kinetic energy, zero velocity and maximum acceleration.

Therefore, Option (e) will be the answer.

6 0
2 years ago
A forward horizontal force of 3 lb is used to pull a 60 lb sled at constant velocity on a frozen pond. the coefficent of (kineti
puteri [66]
It’s either 0.05 or 20. Assuming that the coefficient friction is a damping factor, I feel like 0.05 would be correct m
7 0
3 years ago
The gasoline in a car does 40,000 J of work on a car and generates a constant force of 20 N. How far did the car go?
AnnyKZ [126]

L=F•d=>d=L/F=40,000/20=2,000 m

7 0
2 years ago
Strontium 3890Sr has a half-life of 28.5 yr. It is chemically similar to calcium, enters the body through the food chain, and co
patriot [66]

Answer:

Thus the time taken is calculated as 387.69 years

Solution:

As per the question:

Half life of ^{3890}Sr\, t_{\frac{1}{2}} = 28.5 yrs

Now,

To calculate the time, t in which the 99.99% of the release in the reactor:

By using the formula:

\frac{N}{N_{o}} = (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}

where

N = No. of nuclei left after time t

N_{o} = No. of nuclei initially started with

\frac{N}{N_{o}} = 1\times 10^{- 4}

(Since, 100% - 99.99% = 0.01%)

Thus

1\times 10^{- 4} = (\frac{1}{2})^{\frac{t}{28.5}}}

Taking log on both the sides:

- 4 = \frac{t}{28.5}log\frac{1}{2}

t = \frac{-4\times 28.5}{log\frac{1}{2}}

t = 387.69 yrs

5 0
3 years ago
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