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Nady [450]
3 years ago
10

A metal rod A and a metal sphere B, on insulating stands, touch each other. They are originally neutral. A positively charged ro

d is brought near (but not touching) the far end of A. While the charged rod is still close, A and B are separated. The charged rod is then withdrawn. Is the sphere then positively charged, negatively charged, or neutral?
Physics
1 answer:
ryzh [129]3 years ago
3 0

Answer:

The sphere is positively charged

Explanation:

This is because when the positively charged rod is brought near the metal rod A, the electrons in metal rod A and sphere B are attracted towards it into metal rod A while the positive charges in the are repelled into sphere B. So, when the charged rod is withdrawn, and metal rod A and sphere B are separated, metal rod A is now negatively charged, but sphere B is positively charged.

So, sphere B is positively charged.

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3. A girl is running the 200 m dash. She starts by acceleration at 8m/s^2 for 7s. Then continues at this speed until the end of
antoniya [11.8K]

Answer:

10.6 s

Explanation:

Given that a girl is running the 200 m dash. She starts by acceleration at 8m/s^2 for 7s. Then continues at this speed until the end of the race. How long did it take for her to complete the race?

Solution.

If she accelerated for 7s, the velocity at which she accelerated will be:

Acceleration = velocity/time

8 = V/7

Make V the subject of the formula by cross multiplying.

V = 8 × 7

V = 56 m/s

She maintains the speed through out the journey.

Speed = distance/time

Make time the subject of formula

Time = distance/speed

Time = 200 / 56

Time = 3.57s

Therefore, she will complete the race by 7 + 3.6 = 10.6 s

7 0
2 years ago
A 1.2-kg ball drops vertically onto the floor, hitting with a speed of 25 m/s. Consider the impulse during this collision. Would
Ksju [112]

Answer:

3kg

Explanation:

impulse = MV

then

m1v1=m2v2

when the values are subtitude

then

m2=1.2*25/10

m2=<em>3</em><em>0</em><em>k</em><em>g</em><em>/</em><em>/</em>

7 0
3 years ago
While researching scuba diving, Pablo reads how hot a tank should get while being filled with air. Which law best explains why t
Lady_Fox [76]

Answer:

Gay-Lussac’s law, because as the pressure increases, the temperature increases

Explanation:

First of all, we can notice that the volume of the tank is fixed: this means that the volume of the air inside is also fixed.

This means that in this situation we can apply Gay-Lussac's law, which states that:

"for a gas kept at constant volume, the pressure of the gas is proportional to the absolute temperature of the gas".

Mathematically:

p\propto T

where p is the pressure in Pascal and T is the temperature in Kelvin.

In this case, the tank is filled with air: this means that the pressure of the gas inside the tank increases. And therefore, according to Gay-Lussac's law, the temperature will increase proportionally, and this explains why the tank gets hot.

5 0
2 years ago
Read 2 more answers
Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed of v=0.8c
Studentka2010 [4]

Answer:

30.96 m

Explanation:

If the particle has a lifetime of 129 ns as measured by observer A, and has a speed of 0.8c as measured by observer A, the distance between the markers will be:

d = v * Δt

v = 0.8*c = 0.8 * 3e8 = 2.4e8

Δt = ζ = 129 ns = 1.29e-7 s

d = 2.4e8 * 1.29e-7 = 30.96 m

This is the distance as measured by observer A.

3 0
3 years ago
There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C
liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

V_{tot} = V_{q1} + V_{q2}

V_q = \dfrac{k \cdot q}{r}

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

V_{tot} =  \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05}  = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0

The electric field at the point exactly midway between the plates, V_{tot} = 0

3) The electric field, 'E', between plates is given as follows;

E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, F_e = E × e

∴ F_e = 1.1294 × 10¹² N/C ×  -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, F_e ≈ 1.807 × 10⁻⁷ N

4 0
2 years ago
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