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marishachu [46]
3 years ago
9

Please helppppppp!!!!!!!!!!!!!!!!!!!!!!!!

Physics
1 answer:
kykrilka [37]3 years ago
3 0

Answer:

I think the Bulb No. 2 will stop emitting light if the bulb No. 1 burns out.

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Carbon tetrachloride (CCl4) is ionic or convalent
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Covalent, they’re between two non-metals.
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Two particles having charges of 0.50~\text{nC}0.50 nC (q_1q ​1 ​​ ) and 10~\text{nC}10 nC (q_2q ​2 ​​ ) are separated by a dista
zzz [600]

Answer:

The third charge must be placed 0.548 m from q₁.

Explanation:

Let r = 3m  be the distance between charge q₁ and q₂.

Let x be the distance between charge q₁ and charge q₃ (the third positive charge)

Then r - x is the distance between charge q₂ and q₃

Let the electrostatic force between q₁ and q₃ be F = kq₁q₃/x²

Let the electrostatic force between q₂ and q₃ be F' = kq₂q₃/(r - x)²

Since F + (-F') = 0 (the signs on the forces are different since the forces are in opposite directions)which is required when the net force on q₃ is zero, then

F - F' = 0

F = F'

So, kq₁q₃/x² = kq₂q₃/(r - x)²

q₁/x² = q₂/(r - x)²

[(r - x)/x]² = q₂/q₁

taking square-root of both sides,

(r - x)/x = ±√q₂/q₁

r/x - 1 = ±√q₂/q₁

r/x = 1 ±√q₂/q₁

x = r/(1 ±√q₂/q₁)

substituting the values of the variables r = 3 m, q₁ = 0.50 nC and q₂ = 10 nC

x = 3 m/(1 ±√10 nC/0.5 nC)

x = 3 m/(1 ±√20)

x = 3 m/(1 ± 4.472)

x = 3 m/5.472 or 3 m/-3.472

x = 0.548 m or -0.864 m

So the third charge must be placed 0.548 m to the right of q₁ or 0.864 m to the left of q₁.

Since we are concerned about the line of charge that connects q₁ and q₂, the third charge must be placed 0.548 m from q₁.

7 0
3 years ago
A car moves along a horizontal road with constant velocity v0 = v0xi until it encounters a smooth inclined hill, which it climbs
Serga [27]

Answer:

a ave = ( v1x - v0x ) i/Δt  + v1y j/Δt

Explanation:

8 0
3 years ago
n order better to map the surface features of the Moon, a 361 kg361 kg imaging satellite is put into circular orbit around the M
ollegr [7]

Answer:

Explanation:

Mass of satellite

M_s = 361 kg

Distance of satellite from moon

h = 147 km = 147,000m

Radius of the moon is

R_m = 1740 km = 1740,000m

Mass of the moon is

M_m = 7.36 × 10²² kg.

The kinetic energy is equal to the potential energy of the body to the surface of the moon from the conservation of energy.

K.E = P.E = mgh

Gravity on moon is g = 1.62 m/s²

K.E = 361 × 1.62 × 147,000

K.E = 8.597 × 10^7 J.

B. The gravitational potential energy can be calculated using

U = G•M_s × M_m (1/R_s - 1 / R)

R is the total distance from the centre of the moon to the satellite

R = h + R_m = 147 + 1740 = 1887km

R = 1,887,000 m

U = 6.67 × 10^-11 × 361 × 7.36 × 10²² (1/1,740,000 - 1/1,887,000)

U = 6.67 × 10^-11 × 361 × 7.36 × 10²² × 4.48 × 10^-8

U = 7.93 × 10^7 J

Then,

The total energy becomes

E = K.E + U

E= 8.597 × 10^7 + 7.93 × 10^7 J

E = 1.653 × 10^8 J

3 0
4 years ago
In this experiment, you need to examine the idea of thermal energy transfer. Using a controlled experiment, what might a good qu
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the effect would be 2w=btyh,3

Explanation:

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