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3 years ago

8

1. Does a fish appear closer or farther from a person wearing swim goggles with an air pocket in front of their eyes than the fi

sh really is

2 answers:

Nutka1998 [239]3 years ago

8 0

Answer:

Closer.

Explanation:

Due to the refraction of light passing through water and air.

pychu [463]3 years ago

3 0

Answer: I believe farther. Correct me if I am wrong

Explanation:

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Two small charged spheres are located on the y-axis. One is at y = 1.00 m, the other is at y = −1.00 m, and they both have a cha

yuradex [85]

Answer:

(a) 23.946 kV

(b) -0.077 J

Explanation:

(a) The electric potential is given by the following formula:

$V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}$ (1)

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q1 = q2 = 1.60*10^{-6}C

r1 and r2 are the distance from the charges to the point in which electric potential is evaluated.

Firs, you calculate the distance r1 and r2 by taking into account the position of the charges

$r_1=\sqrt{(1.00)^2+(0.670)^2}m=1.20m\\\\r_2=\sqrt{(-1.00)^2+(0.670)^2}m=1.20m$

Next, you replace the values of the parameters to calculate V:

$V=(8.98*10^9Nm^2/C^2)\frac{1.6*10^{-6}C}{1.20m}+(8.98*10^9Nm^2/C^2)\frac{1.6*10^{-6}C}{1.20m}\\\\V=23946.66\ V=23.946\ kV$

(b) The potential electric energy is given by:

$U_T=U_{1,2}+U_{1,3}+U_{2,3}\\\\U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}\\\\r_{1,2}=2.00m\\\\r_{1,3}=1.20m\\\\r_{2,3}=1.20m\\\\U_T=(8.98*10^9)[\frac{(1.6*10^{-6})^2}{2.00m}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}]J\\\\U_T=-0.077J$

5 0

3 years ago

Paul [167]

B would be your right answer. Because Jupiter’s gravity pull is much stronger than earths

7 0

1 year ago

3. If you had parental units of RR and Rr (R being dominant round and r being recessive

yanalaym [24]

C all round

Explanation:

because the outcome of a the traits would be RR, RR, Rr, Rr so it would be all round if it was Rr and Rr than there would have RR, Rr, Rr, and rr and that would have been a

7 0

3 years ago

A 0.091-in-diameter electrical wire at 90°F is covered by 0.02-in-thick plastic insulation (k = 0.075 Btu/h·ft·°F). The wire is

olya-2409 [2.1K]

Answer:

The critical radius of the plastic insulation is 0.72 inches.

Explanation:

Given that,

Diameter = 0.091 in

Thickness = 0.02 in

Initial temperature = 90°F

Final temperature = 50°F

Heat transfer coefficient = 2.5 Btu/h.ft²°F

Material conductivity = 0.075 Btu/h.ft °F

We need to calculate the critical radius of the plastic insulation

Using formula of critical radius

$r_{cr}=\dfrac{2K}{h}$

Where, k = Material conductivity

h = Heat transfer coefficient

Put the value into the formula

$r_{cr}=\dfrac{2\times0.075}{2.5}$

$r_{cr}=0.06\ ft$

$r_{cr}=0.72\ inches$

Hence, The critical radius of the plastic insulation is 0.72 inches.

4 0

3 years ago

What is the main source of energy?

Kryger [21]

the end, all the energy that we use comes from two sources; the sun (fossil fuels, solar panels, hydro power, and wind energy), and its predecessor, some supernova that created the very heavy atoms (uranium, thorium) which are still around that provide us with geothermal heat and nuclear reactors

4 0

3 years ago