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3 years ago

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The density of water is 1g/cm3. What is this in kg/m3? The density of water is 1g/cm3. What is this in kg/m3? Follow 3 answers 3

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1 answer:

Lelu [443]3 years ago

8 0

1 gram per cubic centimeter is equal to 1000 kilograms per cubic meter. Just multiply it with a 1000 and you have your solution.

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What's the degree to which molecules are able to pass through a membrane

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The degree is 4678% to pass theough all membranes.

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2 years ago

Which option would it be

guapka [62]

It would be 4 atm, because the way to figure out the final pressure is that (P1)(V1)=(P2)(V2)

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3 years ago

What role does team collaboration play in successfully planning a mission to Mars?

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2 years ago

Suppose the magnitude of the proton charge differs from the magnitude of the electron charge by a mere 1 part in 109

Oksana_A [137]

Answer:

A) F = 1.09 10 5 N, b) Yes

Explanation:

Part A

For this exercise we need the number of free electrons in copper, as the valence of copper +1 there is a free electron for each atom. Let's use the concept of density to find the mass of copper in the sphere

ρ = m / V

.m = ρ V = ρ 4/3 π r³

The radius is half the diameter

r = 1.9 10⁻² / 2 = 0.95 10⁻² m

ρ = 8960 kg / m3

m = 8960 4/3 π (0.95 10⁻²)³

m = 3.2179 10⁻² kg

The molecular weight of copper is 63,546 g / mol which has 6,022 10²³ atoms

With this we can use a rule of proportions to enter the number of atom is this mass

#_atom = 6.022 10²³ 3.2179 10⁻² / 63.546 10⁻³

#_atom = 3,049 10²³

Therefore there is the same number of electrons, as they indicate that the charge of the protone and the electon differs by 1/10⁹ the total charge for each spherical is

q = e / 10⁹ #_atom

q = e / 10⁹ 3,049 1023

q = 3,049 10⁴ (-1.6 10⁻¹⁹)

q = -4,878 10-5 C

Electric force is

F = k q₁q₂ / r²

F = k q² / r²

Let's calculate

F = 8.99 10⁹ (4.878 10⁻⁵)²2 / (1.4 10⁻²)²

F = 1.09 10 5 N

This is a force of repulsion.

Part B

The magnitude of this force is in very easy to detect

4 0

3 years ago

The circuits, P and Q, show two different ammeter-voltmeter methods of measuring resistance. Suppose the ammeter has a resistanc

qaws [65]

Answer:

Uncorrected values for

For circuit P

R = 2.4 ohm

For circuit Q

R = 2.4 ohm

Corrected values

for circuit P

R = 12 OHM

For circuit Q

R = 2.3 ohm

Explanation:

Given data:

Ammeter resistance 0.10 ohms

Resister resistance 3.0 ohms

Voltmeter read 6 volts

ammeter reads 2.5 amp

UNCORRECTED VALUES FOR

1) circuit P

we know that IR =V

$R = \frac{6}{2.5} - 2.4 ohm$

2) circuit Q

R = 2.4 ohm as no potential drop across ammeter

CORRECTED VALUES FOR

1) circuit p

IR = V

$\frac{3R}{R+3} \times 2.5 = 6$

R= 12 ohm

2) circuit Q

$I\times (R+0.1) =V$

$R+0.1 =\frac{6}{2.5}$

R = 2.3 ohm

5 0

3 years ago