A.Substitute in the units for each one and combine like terms.

As the Sun ages, what will happen to life here on Earth? Why?

Burka [1]

Answer:

As the Sun ages it will enter another stage of stellar evolution where it's atmosphere begins to inflate.As the Sun heats up and expands, life on Earth will become increasingly difficult. Long before the Sun becomes a red giant some 4 or 5 billion years from now, our planet will be rendered uninhabitable.

~I HOPE IT HELPS♡

5 0

3 years ago

How do plants absorb light energy from sun

brilliants [131]

There are plant cells called chloroplasts, plants leave chloroplasts which have chlorophyll that capture and absorb light energy from the sun.

3 0

3 years ago

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A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri

Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = $\frac{I}{nqA}$

Where;

n = number of free electrons per cubic meter

q = electron charge

A = cross-sectional area of the wire

First let's calculate the number of free electrons per cubic meter (n)

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M) -------------(ii)

Substitute the values of Nₐ, ρ and M into equation (ii) as follows;

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

Now let's calculate the drift electron

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

Substitute these values into equation (i) as follows;

v = $\frac{I}{nqA}$

v = $\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}$

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0

3 years ago

Physics Question: Determine the work that is being done by tension in pulling the box 193.0 cm along the table. Also determine t

lozanna [386]

1) The work done by tension is 12.0 J

2) The final speed of the box is 1.57 m/s

Explanation:

The work done by a force on an object is given by:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

For the box in this problem, we have:

F = 6.2 N is the force applied (the tension in the string)

d = 193.0 cm = 1.93 m is the displacement

$\theta=0$ assuming that the string is horizontal

Substituting, we find the work done by the tension:

$W=(6.2)(1.93)(cos 0)=12.0 J$

2)

In order to determine the final speed of the box, we need to determine its acceleration first.

Beside the tension, acting forward, the other force acting horizontally on the box is the force of friction, whose magnitude is

$F_f = \mu mg$

where

$\mu=0.16$ is the coefficient of friction

m = 2.8 kg is the mass of the box

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$F_f = (0.16)(2.8)(9.8)=4.4 N$

Therefore, the net force on the box is

$F=6.2 N - 4.4 N = 1.8 N$

And the acceleration can be found by using Newton's second law:

$F=ma$

where

F = 1.8 N is the net force

m = 2.8 kg is the mass of the box

a is the acceleration

Solving for a,

$a=\frac{F}{m}=\frac{1.8}{2.8}=0.64 m/s^2$

Now we can finally find the final speed using the suvat equation:

$v^2-u^2=2as$

where

v is the final speed

u = 0 is the initial speed

$a=0.64 m/s^2$ is the acceleration

s = 1.93 m is the displacement

Solving for v,

$v=\sqrt{u^2+2as}=\sqrt{0+2(0.64)(1.93)}=1.57 m/s$

Learn more about work:

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3 0

3 years ago

Which wave behavior results from two waves colliding and the temporary combined wave results in a smaller wave than the original

mart [117]

When two waves (either mechanical or electromagnetic) combine in such a way that their peaks and troughs combine to produce a wave of large amplitude, the wave behaviour is known as constructive interference. The opposite process is where the peaks of one wave combines with the troughs of another wave in order to produce a wave of smaller amplitude. This process is known as destructive interference.

7 0

3 years ago