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Scorpion4ik [409]
3 years ago
12

Give the SI unit. For physics

Physics
1 answer:
IrinaVladis [17]3 years ago
6 0

Answer:

metre (m) - unit of length

kilograms (kg) - unit of mass

second (s) - unit of time

ampere (A) - unit of electrical current

kelvin (K) - unit of temperature

mole (mol) - unit of the amount of substance

Explanation:

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A sphere of radius R = 0.295 m and uniform charge density -151 nC/m^3 lies at the center of a spherical, conducting shell of inn
cupoosta [38]

Answer:

a) -1.27*10³ N/C b) 0 c) -0.21*10³ N/C d) 0.1*10³ N/C

Explanation:

a) r = 0.76R

As this distance is inside the sphere, we need to know how much charge is enclosed within this distance for the center, as follows:

Q = ρ*V(r) = ρ*\frac{4}{3} *\pi *r^{3}

where r = 0.760* R = 0.760* 0.295 m = 0.224 m, and ρ = -151 nC/m³

Q = -151e-9 *\frac{4}{3} *\pi *0.224m^{3} = -7.11e-9 C

Applying Gauss' Law to a spherical gaussian surface of r= 0.76R, as the electric field is radial, and directed inward, we can write the following equation:

E*A = Q/ε₀, where Q= -7.11 nC, A= 4*π*(0.76R)² and ε₀ =8.85*10⁻¹² C²/N*m²

We can solve for E, as follows:

E = \frac{1}{4*\pi*8.85e-12C2/N*m2 } *\frac{-7.11e-9C}{(0.76*0.295m)^{2}} =-1.27e3 N/C

⇒ E = -1.27*10³ N/C

b) r= 3.90 R

As this distance falls inside the conducting shell, and no electric field can exist within a conductor in electrostatic condition, E=0

c) r = 2.8 R

As this distance falls between the sphere and the inner radius of the shell, we can calculate the electric field, applying Gauss' law to a gaussian surface of radius equal to r= 2.80 R.

First we need to find the total charge of the sphere, as follows:

Q = ρ*V =

Q = -151e-9 *\frac{4}{3} *\pi *0.295m^{3} = -16.2e-9 C

In the same way that for a) we can write the following expression:

E*A = Q/ε₀, where Q= -16.2 nC, A= 4*π*(2.8R)² and ε₀ =8.85*10⁻¹² C²/N*m²

We can solve for E, as follows:

E = \frac{1}{4*\pi*8.85e-12C2/N*m2 } *\frac{-16.2e-9C}{(2.8*0.295m)^{2}} =-0.21e3 N/C

⇒ E = -0.21*10³ N/C

d) r= 7.30 R

In order to find the electric field at this distance, which falls beyond the outer radius of the shell, we need to find the total charge on the outer surface.

As the sphere has a charge of -16.2 nC, and the total charge of the conducting shell is 66.7nC, in order to make E=0 inside the shell, the total charge enclosed by a gaussian surface with a radius larger than the inner radius of the shell and shorter than the outer one, must be zero, which means that a charge of +16.2 nC must be distributed on the inner surface of the shell.

This leaves an excess charge on the outer surface of the shell as follows:

Qsh = 66.7 nC - 16.2 nC = 50.5 nC

Now, we can repeat the same process than for a) and c) as follows:

E*A = Q/ε₀, where Q= 50.5 nC, A= 4*π*(7.3R)² and ε₀ =8.85*10⁻¹² C²/N*m²

We can solve for E, as follows:

E = \frac{1}{4*\pi*8.85e-12C2/N*m2 } *\frac{50.5e-9C}{(7.3*0.295m)^{2}} =0.1e3 N/C

⇒ E = 0.1*10⁻³ N/C

6 0
2 years ago
PLZ WHAT I DOO<br> I BROKE A WINDOW WHAT I DO
Hatshy [7]

Hmmmm, you either tell your parent, try to fix it, or blame it on the dog with a baseball bat that always comes on your yard and tries to eat you.

o(* ̄▽ ̄*)ブ

3 0
2 years ago
One speed skater starts across a frozen lake at an average speed of 8 m/s. Ten seconds later, a second speed skater starts from
Luba_88 [7]

Answer:

40 s

Explanation:

After 10 seconds, the first skater would have a 8m/s * 10s = 80 m head start

Let t be the number of seconds after the second skater starts will the second skater overtake the first skater

The distance traveled by the first skater after t seconds is

s_1 = v_1t = 8t

Similarly the distance traveled by the 2nd skater after t seconds is

s_2 = v_2t = 10t

Since the 2nd skater catches up to the 1st one after 80 m behind, the distance traveled by the 2nd one must be 80m greater than the distance of the 1st skater

s_2 = s_1 + 80

We can substitute s_1 = 8t, s_2 = 10t

10t = 8t + 80

2t = 80

t = 80 / 2 = 40 s

7 0
3 years ago
A rock at the top of a 20 meter tall hill.The rock has a mass of 10 kg. How much potentail   energy does it have?
ch4aika [34]
P.E.=mgh
= 10x10x20
=2000J
6 0
3 years ago
What effect or effects would be most significant if the Moon's orbital plane were exactly the same as the ecliptic plane?
Lilit [14]

Answer:

c. Solar eclipses would be much more frequent.

Explanation:

The <u>ecliptic plane</u> is the apparent orbit that the sun describes around the earth (although it is the earth that orbits the sun), is the path the sun follows in earth's sky.

A <u>solar eclipse</u> occurs when the moon gets between the earth and the sun, so a shadow is cast on the earth because the light from the sun is blocked.

The reason why solar eclipses are not very frequent is because the moon's orbital plane is not in the same plane as the orbit of the earth around the sun, but rather that it is somewhat inclined with respect to it.

So <u>if both orbits were aligned, the moon would interpose between the sun and the earth more frequently, producing more solar eclipses.</u>

So, if the moon's orbital plane were exacly the same as the ecliptic plane solar eclipses would be more frequent.

the answer is: c.

8 0
3 years ago
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