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kondor19780726 [428]
4 years ago
7

What is the difference between charles law and boyle's law?

Physics
1 answer:
Leokris [45]4 years ago
3 0
Boyle's law<span> talks about the relationship </span>between<span> pressure and volume (high pressure = low volume, and vice-versa), while </span>Charles's law<span> talks about the relationship </span>between<span> volume and temperature (high temperature = high volume, and vice-versa).</span>
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What is the speed of the wave above of the frequency is 7.0 hertz
alexandr1967 [171]

Answer:

Umm

Explanation:

3 0
3 years ago
If you sleep 33% in a day, how many hours is that?
lesantik [10]

In this question, one has to carefully understand that the total number of hours in the day can never be more that 24 hours. based on this important fact the answer to the question can be very easily deduced. The only requirement is calculating perfectly.
Number of hours in a day = 24 hours
Percentage of hours of sleep in a day = 33%
Amount of sleep in the day = (33/100) * 24
                                             = 7.92 hours
So 33% of sleep in a day is equal to 7.92 hours. I hope this answer has helped you. In future you can keep the procedure in mind for solving such problems.
5 0
4 years ago
Can anyone help, i need help with this as well.
VashaNatasha [74]

Answer:

B

Explanation:

I think this one is correct

5 0
3 years ago
Read 2 more answers
16. A pug puppy is sitting on a skateboard. The pug has a mass of 2.4 kg, and the skateboard has a mass
kramer

Answer:

12 N

Explanation:

Use Newton's second law:

∑F = ma

F = (2.4 kg + 1.3 kg) (3.2 m/s²)

F = 11.84 N

Rounded to two significant figures, the force is 12 N.

4 0
3 years ago
How do you solve this problem?
Katarina [22]

The particle has acceleration vector

\vec a=\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\vec\imath+\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\vec\jmath

We're told that it starts off at the origin, so that its position vector at t=0 is

\vec r_0=\vec0

and that it has an initial velocity of 12 m/s in the positive x direction, or equivalently its initial velocity vector is

\vec v_0=\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)\,\vec\imath

To find the velocity vector for the particle at time t, we integrate the acceleration vector:

\vec v=\vec v_0+\displaystyle\int_0^t\vec a\,\mathrm d\tau

\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\displaystyle\int_0^t\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\jmath

\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right]\,\vec\imath+\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\vec\jmath

Then we integrate this to find the position vector at time t:

\vec r=\vec r_0+\displaystyle\int_0^t\vec v\,\mathrm d\tau

\vec r=\left[\displaystyle\int_0^t\left(12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\mathrm d\tau\right]\,\vec\jmath

\vec r=\left[\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)t+\left(-1.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\right]\,\vec\imath+\left(2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\,\vec\jmath

Solve for the time when the y coordinate is 18 m:

18\,\mathrm m=\left(2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\implies t=3.0\,\mathrm s

At this point, the x coordinate is

\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)(3.0\,\mathrm s)+\left(-1.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)(3.0\,\mathrm s)^2=27\,\mathrm m

so the answer is C.

7 0
3 years ago
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