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4 years ago

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1 answer:

3 0

On the Newtonian theory of gravity, gravitation affects anything with mass. Assuming that none of the answer choices is the only thing that exists in the universe, all of the answer choices are subject to the law of universal gravitation (hence “universal”).

Satellites, water, frogs, and stars all have mass as they are all composed of matter. Thus, all four answer choices should be circled.

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Nya has a cup of hot tea that she lets cool down for several minutes. What effect does cooling have on the tea?

Sveta_85 [38]

<span>An efect could be that sence the air us blowing on the hot tea that Nya has, an efect could be the temperatures mixing and the tea cooling down enough for Nya to drink</span>

3 0

3 years ago

Read 2 more answers

A grating that has 3900 slits per cm produces a third- order fringe at a 28.0° angle. What wavelength of light is being used? Ex

Zolol [24]

Explanation:

Given that,

Number of slits per cm, $N=3900\ lines/cm=390000\lines/m$

The third fringe is obtained at an angle of, $\theta=28$

We need to find the wavelength of light used. The grating equation is given by :

$d\ sin\theta=n\lambda$

$d=\dfrac{1}{N}=\dfrac{1}{390000}$

$d=2.56\times 10^{-6}\ m$

$\lambda=\dfrac{d\ sin\theta}{n}$ , n = 3

$\lambda=\dfrac{2.56\times 10^{-6}\times \ sin(28)}{3}$

$\lambda=4.006\times 10^{-7}\ m$

$\lambda=400\ nm$

So, the wavelength of the light is 400 nm. Hence, this is the required solution.

4 0

3 years ago

Find the x-component of the electric field at the origin due to the full arc length for a charge of 4.5 µc and a radius of 0.55

Rasek [7]

<span>Charge Q = 4.5 Âµc
Radius of the arc R= 0.55 m
Coulomb constant ke = 8.98755 Ă— 10^9 n â€˘ m 2 /c 2
Charge per unit arc length = delta q / delta theta = Q / (pi / 2) = 2Q / pi
=> Delta q = (2Q / pi) delta theta
So the x-component after subtitling the delta q is
E = 2 ke Q / pi R^2 = (2 x 8.98755 Ă— 10^9 x 4.5 x 10^-6) / 3.14 x (0.55) ^2
E = 80.89 x 10^3 / 0.94985 = 85.15 x 10^3 = 85151 N/C</span>

5 0

3 years ago

Which of these is an example of a mechanical wave?

11Alexandr11 [23.1K]

C is the answer because it it is the answer and yah great let’s hope you get it right

4 0

3 years ago

Read 2 more answers

(a) What current (in kA) is needed to transmit 180 MW of power at a voltage of 29.0 kV? kA(b) Find the power loss (in MW) in a 1

Anna11 [10]

Given:

The power is

$\begin{gathered} P=180\text{ MW} \\ =180\times10^6\text{ W} \end{gathered}$

The voltage is

$\begin{gathered} V=29.0\text{ kV} \\ =29.0\times10^3\text{ V} \end{gathered}$

The resistance is

$R=1.50\text{ }\Omega$

To find:

a) the current

b) the power loss

c) the percent loss

Explanation:

a) The current is,

$\begin{gathered} i=\frac{P}{V} \\ =\frac{180\times10^6}{29.0\times10^3} \\ =6.21\times10^3\text{ A} \\ =6.21\text{ kA} \end{gathered}$

Hence, the current is 6.21 kA.

The power loss is,

$\begin{gathered} P_{loss}=i^2R \\ =(6.21\times10^3)^2\times1.50 \\ =57.8\times10^6\text{ W} \\ =57.8\text{ MW} \end{gathered}$

Hence, the power loss is 57.8 MW.

The percentage loss is

$\begin{gathered} \frac{P_{loss}}{P}\times100\text{ \%} \\ =\frac{57.8\times10^6}{180\times10^6}\times100\text{ \%} \\ =32.1\text{ \%} \end{gathered}$

Hence, the power loss percentage is 32.1 %.

8 0

1 year ago