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3 years ago

9

Two objects are 5 kg and 10 kg respectively, and they're 10 m apart. if the distance between them is increased to 20 m, what hap

pens to the gravitational force?
a. it decreases.
b. it increases slightly.
c. it is doubled.
d. it doesn't change.

1 answer:

Sati [7]3 years ago

6 0

I think it's OPTION A, IT DECREASES.

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Which value is represented by the slope of the line?

netineya [11]

The slope of the line is

(change in ' y ' between the ends) / (change in ' x ' between the ends)

Slope = (630g - 0) / (70 cm^3 - 0)

Slope = (630 / 70) g/cm^3

<em>Slope = 9.0 g/cm^3</em>

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3 years ago

A tetrahedron has an equilateral triangle base with 25.0-cm-long edges and three equilateral triangle sides. The base is paralle

djverab [1.8K]

Answer:

a. 7.046 Nm²/C

b. 2.348 Nm²/C

Explanation:

Data given:

Base of equilateral triangle = 25.0 cm = 0.25 m

Strength of electric field = 260 N/C

In order to find the electric flux we first have to find out the area of triangle.

Area of triangle = $\frac{\sqrt{3} }{4} a^{2}$

= $\frac{\sqrt{3} }{4} (0.25)^{2}$

= 0.0271 m³

Lets find electric flux,

Electric Flux = E. A

= 260×0.0271

= 7.046 Nm²/C

Now we can find the electric flux through each of the three sides.

Electric flux through three sides = $\frac{7.046}{3}$

= 2.348 N m²/C

3 0

3 years ago

A modern compact fluorescent lamp contains 1.4 mg of mercury (Hg). If each mercury atom in the lamp were to emit a single photon

Reika [66]

Answer:

A. 1.64 J

Explanation:

First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:

$n=\frac{m}{M_m}$

where

n is the number of moles

m = 1.4 mg = 0.0014 g is the mass of mercury

Mm = 200.6 g/mol is the molar mass of mercury

Substituting, we find

$n=\frac{0.0014 g}{200.6 g/mol}=7.0\cdot 10^{-6} mol$

Now we have to find the number of atoms contained in this sample of mercury, which is given by:

$N=n N_A$

where

n is the number of moles

$N_A=6.022\cdot 10^{23} mol^{-1}$ is the Avogadro number

Substituting,

$N=(7.0\cdot 10^{-6} mol)(6.022\cdot 10^{23} mol^{-1})=4.22\cdot 10^{18}$ atoms

The energy emitted by each atom (the energy of one photon) is

$E_1 = \frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda=508 nm=5.08\cdot 10^{-7}nm$ is the wavelength

Substituting,

$E_1 = \frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{5.08\cdot 10^{-7} m}=3.92\cdot 10^{-19} J$

And so, the total energy emitted by the sample is

$E=nE_1 = (4.22\cdot 10^{18} )(3.92\cdot 10^{-19}J)=1.64 J$

4 0

3 years ago

What happens to a sheet of copper as kinetic energy of copper molecules decrease

liberstina [14]

First is melts then it expands next it gets cooler Finally it gains ener. Hope this helps you out.

8 0

3 years ago

A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne

Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

$F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz$

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0

3 years ago