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3 years ago

Is 5g greater than 20.43 mg

Physics

2 answers:

tatyana61 [14]3 years ago

7 0

Answer

Yes.

Explanation:

20.43mg is equal to 0.02043g.

worty [1.4K]3 years ago

4 0

yes that's true 6g is larger

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What is the Kinetic Energy of the glass when it is

poizon [28]

Answer: 197

Explanation:

Because mechanical energy and mass it speeds up

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3 years ago

Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with a w

amm1812

Answer:

1) f= 8.6 GHz

2) t= 0.2 ms

Explanation:

Since microwaves are electromagnetic waves, they move at the same speed as the light in vacuum, i.e. 3*10⁸ m/s.
There exists a fixed relationship between the frequency (f) , the wavelength (λ) and the propagation speed in any wave, as follows:

$v = \lambda * f (1)$

Replacing by the givens, and solving for f, we get:

$f =\frac{c}{\lambda} =\frac{3e8m/s}{0.035m} = 8.57e9 Hz (2)$

⇒ f = 8.6 Ghz (with two significative figures)

Assuming that the microwaves travel at a constant speed in a straight line (behaving like rays) , we can apply the definition of average velocity, as follows:

$v =\frac{d}{t} (3)$

where v= c= speed of light in vacuum = 3*10⁸ m/s

d= distance between mountaintops = 52 km = 52*10³ m

Solving for t, we get:

$t = \frac{d}{c} = \frac{52e3m}{3e8m/s} = 17.3e-5 sec = 0.173e-3 sec = 0.173 ms (4)$

⇒ t = 0.2 ms (with two significative figures)

6 0

2 years ago

Help please I do not understand

ahrayia [7]

What don’t you understand? If you haven’t uploaded anything

3 0

3 years ago

If two point masses 1kg & 4kg are seperated by a distance of 2m. Magnitude of gravitational force exerted by 1kg on 4kg is ?

Aliun [14]

Answer:

F = G Newtons

Explanation:

Given:

Mass of 1st body = $1\:kg$
Mass of 2nd body = $4\:kg$

To Find:

Magnitude of gravitational force

Solution:

Here, we have a formula

$F=\dfrac{G.M_{1}.M_{2}}{r^{2}}$

<u>Substituting the values</u>

$\implies\:\:F = \dfrac{G(1)(4)}{2^{2}}$

$\implies\:\:F = \dfrac{4G}{4}$

$\implies\:\:F = \dfrac{\cancel{4}G}{\cancel{4}}$

$\implies\:\:\red{F = G}$

Know More:

The applied formula for the above solution is

${\boxed{F_{G}=\dfrac{G.M_{1}.M_{2}}{r^{2}}}}$

where,

F $_{G}$ = Gravitational force
G = Gravitational constant
M $_{1}$ = mass of 1st body
M $_{2}$ = mass of 2nd body
r = distance between two bodies

6 0

3 years ago

A very long insulating cylinder has radius R and carries positive charge distributed throughout its volume. The charge distribut

blsea [12.9K]

Answer:

1. $E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. $E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3.The results from part 1 and 2 agree when r = R.

Explanation:

The volume charge density is given as

$\rho (r) = \alpha (1-\frac{r}{R})$

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.

1. Since the cylinder is very long, Gauss’ Law can be applied.

$\int {\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

$\int\, da = 2\pi r h$

where ‘h’ is the length of the imaginary Gaussian surface.

$Q_{enc} = \int\limits^r_0 {\rho(r)h} \, dr = \alpha h \int\limits^r_0 {(1-r/R)} \, dr = \alpha h (r - \frac{r^2}{2R})\left \{ {{r=r} \atop {r=0}} \right. = \alpha h (\frac{2Rr - r^2}{2R})\\E2\pi rh = \alpha h \frac{2Rr - r^2}{2R\epsilon_0}\\E(r) = \alpha \frac{2R - r}{4\pi \epsilon_0 R}\\E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

$Q_{enc} = \int\limits^R_0 {\rho(r)h} \, dr = \alpha \int\limits^R_0 {(1-r/R)h} \, dr = \alpha h(r - \frac{r^2}{2R})\left \{ {{r=R} \atop {r=0}} \right. = \alpha h(R - \frac{R^2}{2R}) = \alpha h\frac{R}{2} \\E2\pi rh = \frac{\alpha Rh}{2\epsilon_0}\\E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3. At the boundary where r = R:

$E(r=R) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R}) = \frac{\alpha}{4\pi \epsilon_0}\\E(r=R) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r} = \frac{\alpha}{4\pi \epsilon_0}$

As can be seen from above, two E-field values are equal as predicted.

4 0

3 years ago