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BartSMP [9]
3 years ago
9

What will the surface charge density be if the radius of the disk is doubled but its total charge remains the same

Physics
1 answer:
zaharov [31]3 years ago
8 0

Answer:

the new surface charge density = Q/4πr²( initial surface charge density divided by 4)

Explanation:

charge density(surface) = Q/A = charge/area

let r be the initial radius of the disk

therefore, area A = πr²

charge density = Q/πr²

Now that the radius is doubled, let it be represented as R

∴ R = 2r

Recall, charge density = Q/A

A = πR = π(2r)² = 4πr²

the new surface charge density = Q/4πr²

the initial surface charge density divided by 4

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masya89 [10]

Answer: 2F

Explanation:

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3 years ago
what color does acid turn red litmus papera substance tested with blue litmus paper, is clear and colorless, and feels slippery.
Mariulka [41]
When red litmus paper comes into contact with any alkaline substance, it turns blue. Some examples of alkaline substances are ammonia gas, milk of magnesia, baking soda and limewater.
4 0
2 years ago
A 465 g block slides along a frictionless surface at a speed of 0.35 m/s. It runs into a horizontal massless spring with spring
katovenus [111]

Answer:

a) The duration, during which the block remain in contact with the spring is 0.29 s

b) The period of the simple harmonic oscillatory motion depends only on the mass and spring constant, therefore when the speed is doubled, the duration of contact remains the same as 0.29 s.

Explanation:

Mass of the block = 465 g

Surface speed = 0.35 m/s

Spring constant , k = 54 N/m

T = 2\times \pi \times \sqrt{\frac{m}{K} } = 2\times \pi \times \sqrt{\frac{0.465}{54} }  = 0.58 s

a) Since the period for the oscillatory motion is 0.58 s, then the time when the block and spring remain in contact is T/2 = 0.29 s

b) When the speed is doubled, we have

T = 2\times \pi \times \sqrt{\frac{m}{K} }

Therefore, since T is only dependent on the mass, m and the  spring constant, K, then the time it takes when the speed is doubled remain as

T /2 = 0.29 s

7 0
3 years ago
Suppose that the coefficient of kinetic friction between Zak's feet and the floor, while wearing socks, is 0.250. Knowing this,
insens350 [35]

Answer:

The distance travel before stopping is 1.84 m

Explanation:

Given :

coefficient of kinetic friction \mu_{k} = 0.250

Zak's speed v = 3 \frac{m}{s}

Gravitational acceleration g = 9.8 \frac{m}{s^{2} }

Work done by frictional force is given by,

  W = \Delta K

 \mu _{k} mg d = \frac{1}{2} m v^{2}

  d = \frac{v^{2}  }{2 g \mu _{k} }

  d = \frac{9}{2 \times 9.8 \times 0.250}

  d = 1.84 m

Therefore, the distance travel before stopping is 1.84 m

3 0
3 years ago
Ammonia enters the expansion valve of a refrigeration system at a pressure of 10 bar and a temperature of 22oC and exits at 2.0
KonstantinChe [14]

Answer:

The  the quality of the refrigerant at the exit of the expansion valve is 0.179.

Explanation:

Given that,

Initial pressure = 10 bar

Temperature = 22°C

Final pressure = 2.0 bar

We using the value of h

h = 293.4\ kJ/kg

The refrigerant during expansion undergoes a throttling process

Therefore, h_{1}=h_{2}

We need to calculate the quality of the refrigerant at the exit of the expansion valve

At 2.0 bar,

The property of ammonia

h_{f}=47.8 kJ/kg

h_{g}=1417.7 kJ/kg

Using formula

h_{2}=h_{f}+x(h_{g}-h_{f})

Put the value into the formula

293.4 =47.8+x(1417.7-47.8)

x=\dfrac{293.4-47.8}{1417.7-47.8}

x=0.179

Hence, The  the quality of the refrigerant at the exit of the expansion valve is 0.179.

6 0
3 years ago
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