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2 years ago

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Sound travels slowest through gases _____________________. Group of answer choices because the molecules of gas are close togeth

er because gases are air because the molecules of gas are far apart because gases have the lowest temperature

1 answer:

Juliette [100K]2 years ago

4 0

Molecules is the anwser

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What is the weight of a 5.00 kg object on Earth? Assume g=9.81 m/s^2.

Softa [21]

<em>weight = (mass) x (gravity)</em>

Weight = (5.00 kg) x (9.81 m/s²)

weight = (5.00 x 9.81) (kg-m/s²)

<em>Weight = 49.05 Newton</em>

7 0

3 years ago

Two protons are maintained at a separation of nm. Calculate the electric potential due to the two particles at the midpoint betw

Liono4ka [1.6K]

Answer:

The electric potential is approximately 5.8 V

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero

Explanation:

The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:

$\displaystyle{U=\frac{q}{4\pi \epsilon_0r}}$ (1)

where $q$ is the charge of the particle, $\epsilon_0$ the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and $r$ is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:

$\displaystyle{U_{midpoint}=\frac{q}{4\pi \epsilon_0r}}+\frac{q}{4\pi \epsilon_0r}}=\frac{q}{2\pi \epsilon_0r}}}$

Substituting the values $q=1.602 \cdot10^{-19}\ C$ , $\displaystyle{\frac{1}{4\pi\epsilon_0}=8.99\cdot 10^9 N\cdot m^2\cdot C^{-2}}$ and $r=0.5 \cdot 10^{-9} m$ we obtain:

$\displaystyle{U_{midpoint}=\frac{q}{2\pi \epsilon_0r}}=5.759 \approx 5.8 V}$

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.

6 0

3 years ago

What is the period of a wave if 3 waves passes every 6 seconds

Naya [18.7K]

Time period = time/no. of waves = 6/3 = 2s

4 0

2 years ago

A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image

Neko [114]

Answer:

1.Theimage will be located at -0.13m or -13 cm

2.The height of the image will be 0.052m or 5.2cm

Explanation:

Given that;

Height of object, h=20 cm = 0.2m

Object distance in front of convex mirror, o,= 50 cm =0.5m

Radius of curvature, r, =34 cm =0.34m

Let;

Image distance, i,=?

Image height, h'=?

You know that focal length,f, is half the radius of curvature,hence

f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)

f= -0.17m

Apply the relationship that involves the focal length;

$=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}$

$=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}$

Re-arrange to get i

$\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m$

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror

Apply the magnification formula

magnification, m=height of image÷height of object

$m=\frac{h'}{h} =-\frac{i}{o}$

substitute the values to get the height of image h'

$\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\h'=\frac{0.13*0.20}{0.5} \\\\\\h'=\frac{0.025}{0.5} =0.052m\\\\\\h'=5.2cm$

5 0

3 years ago

g stAn experienced spear fisherman sees a small fish swimming in a tidal pool. If the fisherman sees the fish at approximately a

Luba_88 [7]

Answer:

θ = 28.9

Explanation:

For this exercise let's use the law of refraction

n₁ sin θ₁ = n₂ sin θ₂

where we use index 1 for air and index 2 for water where the fish is

sin θ₂ = n₁ / n₂ sin θ₁

in this case the air repair index is 1 and the water 1.33

we substitute

sin θ₂ = 1 / 1.33 sin t 40

sin θ = 0.4833

θ = sin⁻¹ 0.4833

θ = 28.9

5 0

3 years ago