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3 years ago

Which of theses substances would be a poor conductor of electricity

Physics

1 answer:

lys-0071 [83]3 years ago

8 0

Answer:

B. Water and sugar.

Explanation:

In the given options water and sugar would be the poor conductor of electricity. Other given options such as water and salt, water and Hcl and water and NaOH are better conductor of electricity because Hcl ,NaOH, salt (Nacl) can break into their ionic form whereas water and sugar will not.

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What is the motion of the particles in this kind of wave?

MatroZZZ [7]

I would say C but I’m not complete sore

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2 years ago

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How many grams of ice at 0°c will melt if 1.50 kj of heat are added? (the molar heat of fusion of water is 6.01 kj/mol.) 4.49 g

Nata [24]

The computation would be:moles = mass/ Molar Mass, but we are looking for the mass, so rearranging, will give us: mass = moles x MM
Q = moles x Hf
Q = (mass/MM) x Hf
mass = (Q x MM) / Hf
= (1.50-kJ x 18.0-g/mol) / 6.01-kJ/mol
=4.49 g H20 is the answer

7 0

3 years ago

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Single Stage 0.0/3.0 points (graded) Consider a rocket carrying 100,000 kg of propellant, 10,000 kg of structure, and 5000 kg of

Nonamiya [84]

Answer: The Propellant fraction is 0.87.

The payload fraction is 0.04.

Δv = 8991.81 m/s

Explanation: To determine the fractions, first, calculate the total mass of the rocket:

$m_{t} = m_{prop} + m_{str} + m_{pay}$

$m_{t} = 100,000 + 10,000 + 5,000$

$m_{t} = 115,000$

The Propellant Fraction will be

$m_{prop} = \frac{m_{prop}}{m_{t}}$

$m_{prop} = \frac{100,000}{115,000}$

$m_{prop} =$ 0.87

The Payload Fraction is:

$m_{pay} = \frac{m_{pay}}{m_{t}}$

$m_{pay} = \frac{5,000}{115,000}$

$m_{pay} =$ 0.04

The value of Δv is calculated by the formula:

Δv = $-V_{e}. ln(\frac{m_{final}}{m_{initial}} )$

The exhaust velocity ( $V_{e}$ ) is:

$V_{e} = g_{0}.Isp$

$V_{e} =$ 9.81*450

$V_{e} =$ 4414.5

$m_{final}$ is the total mass after the rocket consume all the propellant and $m_{initial}$ is the total mass before the action.

Δv = $-V_{e}. ln(\frac{m_{final}}{m_{initial}} )$

Δv = $-4414.5.ln(\frac{15,000}{115,000} )$

Δv = - 4414.5.ln(0.13)

Δv = 8991.81

Δv will be 8991.81 m/s.

7 0

3 years ago

Two long parallel wires placed side-by-side on a horizontal table carry identical size currents in opposite directions. The wire

N76 [4]

Answer:

D) Points down

Explanation:

We can find the direction of the magnetic field produced by each of the wire by using the right hand rule:

- the thumb must correspond to the direction of the current in the wire

- the other fingers wrap around the thumb and gives the direction of the magnetic field lines

For the wire on the right, we have:

- thumb (current): towards you

- other fingers (magnetic field): at the left of the wire, they point down

For the wire on the left, we have:

- thumb (current): away from you

- other fingers (magnetic field): at the right of the wire, they point down

So, both magnetic fields point down at the point halfway between the two wires, so the net field is also pointing down.

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3 years ago

ASK YOUR TEACHER As an astronaut, you observe a small planet to be spherical. After landing on the planet, you set off, walking

Vsevolod [243]

Given Information:

circumference of planet = C = 25.7 km

height of release = h = 1.43 m

time = t = 30 s

Required Information:

Mass of planet = M = ?

Answer:

Mass of planet = M = 7.963×10¹⁴ kg

Explanation:

The mass of the planet can be found using Newton's law of gravitation,

g = GM/R²

M = gR²/G

Where g is the gravitational acceleration at the planet, R is the radius of the planet and G is the gravitational constant.

G = 6.6743×10⁻¹¹ m³/kg⋅s²

First we need to find R and g

The relation between circumference and radius is given by

C = 2πR

Where R is the radius of planet

25.7 = 2πR

R = 25.7/2π

R = 4.09 km

From the equations of motion we have,

h = ut + ½gt²

Where u is the initial speed which is zero, t is the time and h is the height of release.

Re-arrange the above equation for g

h = ½gt²

g = 2h/t²

g = (2*1.43)/(30)²

g = 0.003177 m/s²

Finally, the mass of the planet is

M = gR²/G

M = 0.003177*(4.09×10³)²/6.6743×10⁻¹¹

M = 7.963×10¹⁴ kg

Therefore, the mass of the given planet is 7.963×10¹⁴ kg

3 0

3 years ago

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