I will pick the brainiest! Please, and thanks! :) Please give a real answer! :)

lesantik [10]

Number two because the sun emits heat in the form of light to earth

5 0

2 years ago

Read 2 more answers

What’s the origin of a meteor shower?

Bogdan [553]

Answer: I hope this helps I read this in a book once

Explanation:

Shower meteors come from the dust released by comets as they travel through our solar system. The dust spreads out along the comet's orbit and forms an elliptical trail of debris that passes around the sun and crosses the orbits of the planets.

6 0

3 years ago

To test the performance of its tires, a car

velikii [3]

Answer:

The coefficient of static friction between the tires and the road is 1.987

Explanation:

Given:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

To Find:

The coefficient of static friction between the tires and the road = ?

Solution:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

8 0

3 years ago

Hydrogen is the second most abundant gas in the atmosphere? True or false?

Alenkinab [10]

False as oxygen is the second most abundant and nitrogen is the most abundant at 78%.

7 0

3 years ago

Calculate the average drift speed of electrons traveling through a copper wire with a crosssectional area of 80 mm2 when carryin

Vedmedyk [2.9K]

Answer:

The correct answer is $2.8*10^{-5}ms^{-1}$

Explanation:

The formula for the electron drift speed is given as follows,

$u=I/nAq$

where n is the number of of electrons per unit m³, q is the charge on an electron and A is the cross-sectional area of the copper wire and I is the current. We see that we already have A , q and I. The only thing left to calculate is the electron density n that is the number of electrons per unit volume.

Using the information provided in the question we can see that the number of moles of copper atoms in a cm³ of volume of the conductor is $8.93/63.5 molcm^{-3}$ . Converting this number to m³ using very elementary unit conversion we get $140384molm^{-3}$ . If we multiply this number by the Avagardo number which is the number of atoms per mol of any gas , we get the number of atoms per m³ which in this case is equal to the number of electron per m³ because one electron per atom of copper contribute to the current. So we get,

$n=140384*6.02*10^{23} = 8.45*10^{28}electrons.m^{-3}$

if we convert the area from mm³ to m³ we get $A=80*10^{-6}m^{2}$ .So now that we have n, we plug in all the values of A ,I ,q and n into the main equation to obtain,

$u=30/(8.45*10^{28}*80*10^{-6}*1.602*10^{-19})\\u=2.8*10^{-5}m.s^{-1}$

which is our final answer.

6 0

3 years ago