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3 years ago

A sphere has a surface uniformly charged with 3.30 C. At what distance from its center is the potential 4.50 MV

Physics

1 answer:

Bogdan [553]3 years ago

6 0

Answer:

r = 6.6 x 10³ m = 6600 m

Explanation:

The potential at a distance from a charged sphere can be given as follows:

$V = \frac{kq}{r}\\\\r = \frac{kq}{V}$

where,

r = distance = ?

k = Colomb Constant = 9 x 10⁹ Nm²/C²

q = charge on sphere = 3.3 C

V = potential = 4.5 MV = 4.5 x 10⁶ V

Therefore,

$r = \frac{(9\ x\ 10^9\ Nm^2/C^2)(3.3\ C)}{4.5\ x\ 10^6\ V}$

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3 years ago

How does flowing water change the earths surface

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3 years ago

A tank has the shape of an inverted circular cone with height 16m and base radius 3m. The tank is filled with water to a height

rewona [7]

Answer:

$W=17085KJ$

Explanation:

From the question we are told that:

Height $H=16m$

Radius $R=3$

Height of water $H_w=9m$

Gravity $g=9.8m/s$

Density of water $\rho=1000kg/m^3$

Generally the equation for Volume of water is mathematically given by

$dv=\pi*r^2dy$

$dv=\frac{\piR^2}{H^2}(H-y)^2dy$

Where

y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

$dw=(pdv)g (H-y)$

Substituting dv

$dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)$

$dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy$

Therefore

$W=\int dw$

$W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy$

$W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)$

$W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0$

$W=3420.84*0.25[2401-65536]$

$W=17084965.5J$

$W=17085KJ$

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3 years ago

What is the volume of a 1.2kg and displaced 1.0g/cm3

Zinaida [17]

Mass = 1.2 kg = 1200 grams.

Volume = mass/density = 1200 cm3.

Hope this helps!

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3 years ago

Water vapor enters a turbine operating at steady state at 500°C, 40 bar, with a velocity of 200 m/s, and expands adiabatically t

faltersainse [42]

Answer:

W = 5701 KW

Explanation:

From the question let inlet be labelled as point 1 and exit as point 2, for the fluid steam, we can get the following;

Inlet (1): P1 = 40 bar ; T1 = 500°C and V1 = 200 m/s

Exit(2) : At saturated vapour; P2 = 0.8 bar and V2 = 150 m/s

Volumetric flow rate = 15 m^(3)/s

Now, to solve this question, we assume constant average values, steeady flow and adiabatic flow.

Specific volume for steam at P2 = 0.8 bar in the saturated vapour state can be gotten from saturated steam tables(find a sample of the table attached to this answer).

So from the table,

v2 = 2.087 m^(3)/kg

Now, mass flow rate (m) = (AV) /v

Where AV is the volumetric flow rate.

Thus, the mass flow rate at exit could be calculated as;

m = 15/(2.087) = 7.17 kg/s

We also know energy equation could be defined as;

Q-W = m[(h1 - h2) + {(V2(^2) - (V1(^2)} /2)} + g(Z2 - Z1)]

Since the flow is adiabatic, potential energy can be taken to be zero. Therefore, we get;

-W = m[(h2 - h1) + {(V2(^2) - (V1(^2)} /2)}

From, table 2, i attached , at P1 = 40 bar and T1 = 500°C; specific enthalpy was calculated to be h1 = 3445.3 KJ/Kg

Likewise, at P2 = 0.8 bar; from the table, we get specific enthalpy as;

h2 = 2665.8 KJ/Kg

So we now calculate power developed;

W = - 7.17 [(2665.8 - 3445.3) + {(150^(2) - 200^(2))/2000 = 5701KW

Since the sign is not negative but positive, it means that the power is developed from the system.

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3 years ago