HELP please im gonna fail

What is an instrument commonly used to measure wind speed?

xeze [42]

The correct answer is letter D. Anemometer. It is a device that is used to measure wind speed. It is a very common weather station instrument and is available to use and to make. Anemos, from the greek word that means wind.

5 0

3 years ago

Read 2 more answers

#1 Not sure where to start. This is for AP Physics!

yaroslaw [1]

First,

$\rho=\dfrac mV$

where $\rho$ is density, $m$ is mass, and $V$ is volume. We can compute the volume of the roll:

$2.7\,\dfrac{\mathrm g}{\mathrm{cm}^3}=\dfrac{1275\,\mathrm g}V$

$\implies V\approx472.22\,\mathrm{cm}^3\approx4.72\,\mathrm m^3$

When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness $x$ . Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).

So we have

$V=Ax$

where $A$ is the given area, so

$4.72\,\mathrm m^3=\left(18.5\,\mathrm m^2\right)x$

$\implies x\approx0.255\,\mathrm m=255\,\mathrm{mm}$

If we're taking significant digits into account, the volume we found would have been $V=470\,\mathrm m^3$ , in turn making the thickness $x=250\,\mathrm{mm}$ .

8 0

3 years ago

The Earth and the Moon are attracted to each other by universal gravitation. The Earth is much more massive than is the Moon. Do

OverLord2011 [107]

Answer:

Earth attract the Moon with a force that is greater.

Explanation:

According to the law of gravitation, the gravitational force between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Mathematically, F1 = Gm1m2/r²... 1

Let m1 be the mass of the earth and m2 be that of the moon

If the Earth is much more massive than is the Moon, the new force of attraction between them will become;

F2= G(2m1)m2/r²

F2 = 2Gm1m2/r² ... (2)

Dividing eqn 1 by 2 we have;

F1/F2 = (Gm1m2/r²)÷(2Gm1m2/r²)

F1/F2 = Gm1m2/r²×r²/2Gm1m2

F1/F2 = 1/2

F2=2F1

This shows that that the earth will attract the moon by a force 2times the initial force of the masses(i.e a much greater force)

6 0

3 years ago

1. Which of the following transformations is an example of a β−-decay?

KatRina [158]

Answer:

1. C: 31/14 Si becomes 31/15 because a nuetron

2. A: 238 92U because the very long half-life means a very small rate of decay

3. D: Charge conservation is not satisfied

4. B: of the four nuclear decay processes only the α-decay changes the baryon number and does so in increments of four

Explanation:

I just took the quick check. Enjoy the answers I did not get to have

4 0

2 years ago

Car A starts out traveling at 35.0 km/h and accelerates at 25.0 km/h2 for 15.0 min. Car B starts out traveling at 45.0 km/h and

lawyer [7]

1 kilometre=1000 metre

1 hour = 3600 second

$1\ km/hr=\frac{1000}{3600} m/s$

$1\ km/hr=\frac{5}{18} m/s$

The initial velocity of car A is 35.0 km/hr i.e

$35.0\ km/hr=35*\frac{5}{18} m/s$

= 9.72 m/s

The initial velocity of car B is 45 km/hr =12.5 m/s

The initial velocity of car C is 32 km/hr = 8.89 m/s

The initial velocity of car D is 110 km/hr=30.56 m/s

The acceleration of car A is given as $25\ km/hr^2$

$=\ 25*\frac{1000}{3600*3600} m/s^2$

$=0.00192901234 m/s^2$

The time taken by car A = 15 min.

From equation of kinematics we know that-

v= u+at [Here v is the final velocity and a is the acceleration and t is the time]

Final velocity of A, v = 9.72 m/s +[0.00192901234×15×60]m/s

=11.456111106 m/s

The acceleration of B is given as $15\ km/hr^2$

$=0.00115740740740 m/s^2$

The time taken by car B =20 min

The final velocity of B is -

v= u+at

= u-at [Here a is negative due to deceleration]

=12.5 m/s +[0.0011574074074×20×60]

=13.8888888.....

=13.9

The acceleration of C is given as $40\ km/hr^2$

$=\ 0.003086419753 m/s^2$

The time taken by car C =30 min

The final velocity of C is-

v = u+at

=8.89 m/s+[0.003086419753×30×60] m/s

=14.4455555555..m/s

=14.45 m/s

The car C is decelerating.The deceleration is given as- $60\ km/hr^2$

$=0.0046296296296m/s^2$

The time taken by car D= 45 min.

The final velocity of the car D is-

v =u+at

=30.56 -[0.00462962962962×45×60]m/s

=18.06 m/s

Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.

3 0

3 years ago