If magnesium has 13 neutrons, what is its mass for the isotope

Which situation best describes the act of reducing?

yawa3891 [41]

A - i think
paying bills online?

5 0

4 years ago

Which of the following best describes the internal layering of Jupiter, from the center outward?A) Core of rock and metal; mantl

statuscvo [17]

Answer:

Option D

Explanation:

Jupiter is made almost totally from hydrogen and some hydrogen compounds. It may have a solid hydrogen core, then a liquid hydrogen layer, then a gaseous layer.

It is not known if Jupiter has a solid surface, or even a liquid surface. We measure Jupiter's diameter from the top of its gas layer.

The core is often described as rocky, but its detailed composition is unknown, as are the properties of materials at the temperatures and pressures.

The presence of a core during at least part of Jupiter's history is suggested by models of planetary formation that require the formation of a rocky or icy core massive enough to collect its bulk of hydrogen and helium from the protosolar nebula. Assuming it did exist, it may have shrunk as convection currents of hot liquid metallic hydrogen mixed with the molten core and carried its contents to higher levels in the planetary interior. A core may now be entirely absent, as gravitational measurements are not yet precise enough to rule that possibility out entirely.

8 0

3 years ago

A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o

gulaghasi [49]

A. 0.77 A

Using the relationship:

$P=\frac{V^2}{R}$

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega$

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega$

The two light bulbs are connected in series, so their equivalent resistance is

$R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega$

The two light bulbs are connected to a voltage of

V = 240 V

So we can find the current through the two bulbs by using Ohm's law:

$I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A$

B. 142.3 W

The power dissipated in the first bulb is given by:

$P_1=I^2 R_1$

where

I = 0.77 A is the current

$R_1 = 240 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_1 = (0.77 A)^2 (240 \Omega)=142.3 W$

C. 42.7 W

The power dissipated in the second bulb is given by:

$P_2=I^2 R_2$

where

I = 0.77 A is the current

$R_2 = 72 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_2 = (0.77 A)^2 (72 \Omega)=42.7 W$

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

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

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

7 0

3 years ago

The ventilation fan of the bathroom has a volumetric flow rate of 1,640 L/s. If the air density is 1.28 kg/m3, calculate the mas

liq [111]

Answer:

7557120 kg/hour

Explanation:

Given data;

Volume of air in one second = 1640 L

Density of air = 1.28 kg/L

Mass of air in 1 hour =?

Since mass = density × volume

==> Mass of air in one second = 1.28 ×1640 = 2099.2 kg

==> Mass of air in one minute = 2099.2×60=125952 kg

==> Mass of air in one hour = 125952× 60 = 7557120 kg

So rate of flow of air is 7557120 kg/hour

5 0

4 years ago

Can someone please help me with this question?

dexar [7]

The answer is B because, the farther the goes the speed starts to decrease and stops it depends on the frequency of the wave.

5 0

4 years ago