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irina [24]
3 years ago
10

SCIENCE HELP! PLEASE HELP URGENT!

Physics
1 answer:
Nonamiya [84]3 years ago
6 0
It is also tripled, there is a rule to everything, whatever you do to one thing, you do the exact thing to the other. Hope this solves it :)
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A quantity of N2 occupies a volume of 1.4 L at 290 K and 1.0 atm. The gas expands to a volume of 3.3 L as the result of a change
lions [1.4K]

Answer:

\rho = 0.50 g/L

Explanation:

As we know that

PV = nRT

here we have

P = 1.0 atm

P = 1.013 \times 10^5 Pa

so we have

V = 1.4 \times 10^{-3} m^3

T = 290 K

now we have

(1.013 \times 10^5)(1.4 \times 10^{-3}) = n(8.31)(290)

n = 0.06

now the mass of gas is given as

m = n M

m = (0.06)(28)

m = 1.65 g

now density of gas when its volume is increased to 3.3 L

so we will have

\rho = \frac{m}{V}

\rho = \frac{1.65 g}{3.3 L}

\rho = 0.50 g/L

5 0
3 years ago
The voltage or potential difference provides the ____ required for electrons to travel through a circuit
galben [10]

Answer:

a. push

Explanation:

Voltage is the force or pressure that is responsible for pushing the charge or electrons to flow in a closed-looped electrical circuit. This flow of electrons (charge) is called the electric current. It is also defined as the difference in electric potential per unit charge between two points in an electric field.

5 0
3 years ago
How much kinetic energy does an 80 kg man have while running at 3 m/s?
Soloha48 [4]

Hello!

\large\boxed{KE = 360 J}

Use the equation KE = 1/2mv² to solve for the kinetic energy of the man.

We are given the mass and velocity, so plug these values into the equation:

KE = 1/2(80)(3²)

KE = 1/2(720)

KE = 360 J

7 0
3 years ago
How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter to p
katrin [286]

Answer:

Required charge q=2.6\times 10^{9}C.

n=1.622\times 10^{10}\ electrons

Explanation:

Given:

Diameter of the isolated plastic sphere = 25.0 cm

Magnitude of the Electric field = 1500 N/C

now

Electric field (E) is given as:

E =\frac{kq}{r^2}

where,

k = coulomb's constant = 9 × 10⁹ N

q = required charge

r = distance of the point from the charge where electric field is being measured

The value of r at the just outside of the sphere = \frac{25.0}{2}=12.5cm=0.125m

thus, according to the given data

1500N/C=\frac{9\times 10^{9}N\times q}{(0.125m)^2}

or

q=\frac{0.125^2\times 1500}{9\times 10^{9}}

or

Required charge q=2.6\times 10^{9}C.

Now,

the number of electrons (n) required will be

n=\frac{required\ charge}{charge\ of\ electron}

or

n=\frac{2.6\times 10^{-9}}{1.602\times 10^{-19}}

or

n=1.622\times 10^{10}\ electrons

6 0
4 years ago
Energy that a microwave use to heat food?
yanalaym [24]
Electromagnetic energy
6 0
3 years ago
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