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3 years ago

To Calculate Velocity and Average Velocity you can use the same formula as both.

Physics

1 answer:

Elanso [62]3 years ago

4 0

Answer:

I believe it's True! Brainliest??

Explanation: Hope you have a great day :)

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Which law of motion describes squeal and oppisit forces of action and reaction

Rom4ik [11]

These two forces are called action and reaction forces and are the subject of Newton's third law of motion. Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects.

Hope this helps! :)

4 0

3 years ago

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,

viktelen [127]

Part A)
First of all, let's convert the radii of the inner and the outer sphere:
$r_A = 11.0 cm = 0.110 m$
$r_B = 16.5 cm=0.165 m$
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
$C=4 \pi \epsilon _0 \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=$
$=3.67\cdot 10^{-11}F$

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
$Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C$

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
$E\cdot (4 \pi r^2) = \frac{Q}{\epsilon _0}$
from which
$E(r) = \frac{Q}{4 \pi \epsilon_0 r^2}$
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
$E(0.111m)=2680 N/C$

And so, the energy density at r=0.111 m is
$U= \frac{1}{2} \epsilon _0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3$

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: $Q=3.67 \cdot 10^{-9}C$ . We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
$E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C$

And therefore, the energy density at this distance from the center is
$U= \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3$

8 0

3 years ago

How do I convert 200 yards into meters using Dimensional Analysis? Explain thoroughly. Thanks!

shtirl [24]

Explanation:

When doing this calculation manually, the formula of conversion that you need to use is yd / 1.0936 = m and this is the same formula used by the online calculator. For instance, 2yd = 1.83m and 15yd = 13.72m.

So 200 yards is (200 / 1.0936)

which is 182.88 Meters

Hope it helps..

8 0

3 years ago

Measurements of acceleration are given in units of ?

Lady_Fox [76]

Meters per second squared: $\frac{m}{s^{2}}$

If you think about it, acceleration is about how fast speed changes. Speed is measured in meters per second: $\frac{m}{s}$

So if you take that and just measure it over time, you get meters per second squared.

3 0

3 years ago

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What was the largest group in the colonies in the year leading up to the American revoulation

Degger [83]

Answer: The English origin was the largest

7 0

2 years ago

Read 2 more answers