These type of power plants are virtually pollution free

what is the resistance of a 12m long wire having radius( 2*10 to the power - 4) and resistivity (3.14*10 to the power - 8)ohmM

Arisa [49]

R = ρl/A

Where R = Resistance in Ohms, Ω, ρ = Resistivity in Ωm, l = Length in m.

Area in m²

ρ = Resistivity = 3.14 * 10⁻⁸ Ωm, Length l = 12m,

Area = πr² = π* (2*10⁻⁴)² m² ≈ 3.14 * (2*10⁻⁴)² m²

R = ρl/A

≈ 3.14 * 10⁻⁸ * 12 / (3.14 * (2*10⁻⁴)²)

≈ 3

Resistance, R ≈ 3 Ω

3 0

2 years ago

The nutritional calorie (Calorie) is equivalent to 1 kcal. One pound of body fat is equivalent to about 4.1 x 103 Calories. If a

Luba_88 [7]

Answer:

$t = 70.13 hours$

Explanation:

As we know that 1 lb fat can be used to expend 4.1 k Calorie of energy

so here we know that we have to use 7.56 lb of body fat

so here the total energy that is to be utilized is given as

$Q = (4.1 \times 10^3)(7.56)$

so we will have

$Q = 30996 Calorie$

now we know that runner expends the energy at rate of 1850 kJ/h

so we have

$E = 1850 \times 10^3 kJ/h$

so it is

$E = 0.44 \times 10^3 calorie/h$

now we know that time to burn this energy is given as

$t = \frac{Q}{E}$

$t = \frac{30996}{0.44 \times 10^3}$

$t = 70.13 hours$

5 0

2 years ago

Two children hang by their hands from the same tree branch. The branch is straight, and grows out from the tree trunk at an angl

vazorg [7]

Answer:879.29 N-m

Explanation:

Given

mass of first child $m_1=44 kg$

distance of first child from tree is $r_1=1 m$

tree is inclined at an angle of $\theta =27^{\circ}$

mass of second child $m_1=27 kg$

distance of second child from tree is $r_2=2.1 m$

Weight of first child $=m_1g=431.2 kg$

Weight of second child $=m_2g=264.6 kg$

Torque of first child weight $=m_1g\cos \theta \cdot r_1$

$T_1=44\times 9.8\times \cos 27\times 1=384.202 N-m$

Torque of second child weight $=m_2g\cos \theta \cdot r_2$

$T_2=27\times 9.8\times \cos 27\times 2.1=495.096 N-m$

Net torque $T_{net}=T_1+T_2=384.202+495.096=879.29 N-m$

6 0

2 years ago

How hot is the sun in degrees Celsius?

Vlada [557]

Answer:

The temperature of the mantle varies greatly, from 1000° Celsius near its boundary with the crust, to 3700° Celsius

hope this helps you!

8 0

2 years ago

A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, dow

shusha [124]

Answer:

150000000

$\dfrac{F_e}{F_g}=0.00000203873598369$

49050000 N/C

Explanation:

q = Charge = 24 pC

m = Mass of honeybee = 0.12 g

E = Electric field = 100 N/C

g = Acceleration due to gravity = 9.81 m/s²

$1\ C=6.25\times 10^{18}\ electrons$

Number electrons is

$n=24\times 10^{-12}\times 6.25\times 10^{18}\\\Rightarrow n=150000000$

The number of electrons added or removed was 150000000

Force is given by

$F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N$

The ratio is

$\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369$

The ratio is $\dfrac{F_e}{F_g}=0.00000203873598369$

Balancing the forces we get

$Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C$

The electric field required is 49050000 N/C

4 0

3 years ago