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gayaneshka [121]
3 years ago
12

Without an atmosphere, the equatorial curve would show minimum daily values on the solstices in June when the sub-solar point is

located at 23.5°N and in December when the sub-solar point is at ________ latitude.
Physics
1 answer:
lapo4ka [179]3 years ago
3 0

Without an atmosphere, the equatorial curve would show minimum daily values on the solstices in June when the sub-solar point is located at 23.5°N and in December when the sub-solar point is at 23.5°S latitude.

Explanation:

At the sub-solar point, the sun strikes directly at the surface with an angle of 90 degrees at a given point.

Solistice refers to that point in time when the sun’s zenith is located at the farthest point from the equator.  

During summer solistice on June 21, the sun’s zenith reaches northernmost point, sub-solar point is fixed at 23.5°S Tropic of Cancer making the earth tilt 23.4 degrees

During winter soliscitse on December 21, the sub-solar point is fixed at)  Tropic of Capricorn.

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When two magnets are brought close together their magnetic
Gnoma [55]

Answer:

You know the saying "Opposites attract" well that is how you can remember that South and North Magnetic Poles connect.  

Explanation:

Hope this helps ya

6 0
3 years ago
The acceleration due to gravity on the surface of Mars is gmars = 3.7 m/s2. How much would an 11 kg bag of potatoes weigh on Mar
allsm [11]
Fnet = (mass) (acceleration)
        = 11 kg x 3.7m/s^2
        = 41 N

6 0
3 years ago
The record for the highest speed achieved in alaboratory for a
Ira Lisetskai [31]

Answer:

0.000234 seconds

Explanation:

Since the row is 0.15m, its radius of rotation must be 0.15 / 2 = 0.075 m

We can start by calculating the angular speed of the rod:

\omega = \frac{v}{r} = \frac{2010}{0.075} = 26800 rad/s

Since one revolution equals to 2π rad. The speed in revolution per second must be

26800 / 2π = 4265 revolution/s

The number of seconds per revolution, or period, is the inverse:

1/4265 = 0.000234 seconds

3 0
3 years ago
A mass of 5 kg of saturated water vapor at 150 kPa is heated at constant pressure until the temperature reaches 200°C. Calculate
yulyashka [42]

Answer:

The work done by the steam is 213 kJ.

Explanation:

Given that,

Mass = 5 kg

Pressure = 150 kPa

Temperature = 200°C

We need to calculate the specific volume

Using formula of work done

W=Pm\DeltaV

W=Pm(\dfrac{RT_{1}}{P_{atm}}-\dfrac{RT_{2}}{P_{atm}}

W=\dfrac{PmR}{P_{atm}}(T_{2}-T_{1})

Where,R = gas constant

T = temperature

P = pressure

P_{atm}=Atmosphere pressure

m = mass

Put the value into the formula

W=\dfrac{150\times10^{3}\times5\times287.05}{1.01\times10^{5}}\times(473-373)

W=213\ kJ

Hence, The work done by the steam is 213 kJ.    

6 0
3 years ago
Please someone help, I’m very confused and it’s due soon, thanks
Anit [1.1K]

Answer:

  1. 1 s
  2. 19.6 m
  3. 2 s
  4. 0.8 m/s^2
  5. 28 m/s
  6. 79 m/s
  7. 0.37 s
  8. 26 m/s
  9. 242 m/s
  10. 19,930 m

Explanation:

In physics, many of the relationships between speed, distance, and acceleration are tied up in the equations for potential and kinetic energy. For an object of mass M* at height h in a gravity field with acceleration g, the potential energy is

  PE = Mgh

At velocity v, the kinetic energy of the object is ...

  KE = 1/2Mv^2

When an object is dropped or launched from rest, the height and velocity are related by the fact that kinetic energy gets translated to potential energy, or vice versa. This gives rise to ...

  PE = KE

  Mgh = (1/2)Mv^2

The mass (M) can be factored out of this, so we have ...

  2gh = v^2

This can be solved for height:

  h = v^2/(2g) . . . . [eq1]

or for velocity:

  v = √(2gh) . . . . [eq2]

__

When acceleration is constant, as assumed here, the velocity changes linearly (to/from 0). So, over the time of travel, the average velocity is half the final velocity. That is,

  t = 2h/v

Depending on whether you start with h or with v, this resolves to two more equations:

  t = 2(v^2/(2g))/v = v/g . . . . [eq3]

  t = 2h/(√(2gh)) = √(4h^2/(2gh)) = √(2h/g) . . . . [eq4]

The last of these can be rearranged to give distance as a function of time:

  h = gt^2/2 . . . . [eq5]

or acceleration as a function of time and distance:

  g = 2h/t^2 . . . . [eq6]

__

These 6 equations can be used to solve the problems posed. Just "plug and chug." For problems in Earth's gravity, we use g=9.8 m/s^2. (You may want to keep these equations handy. Be aware of the assumptions they make.)

_____

* M is used for mass in these equations so as not to get confused with m, which is used for meters.

_____

1) Use [eq4]: t = √(2·6 m/(9.8 m/s^2)) ≈ 1.107 s ≈ 1 s

__

2) Use [eq5]: h = (9.8 m/s^2)(2 s)^2/2 = 19.6 m

__

3) Use [eq4]: t = √(25 m/(4.9 m/s^2)) ≈ 2.259 s ≈ 2 s

__

4) Use [eq6]: g = 2(10 m)/(5 s)^2 = 0.8 m/s^2

__

5) Use [eq2]: v = √(2·9.8 m/s^2·40 m) = 28 m/s

__

6) Use [eq2]: v = √(2·9.8 m/s^2·321 m) ≈ 79.32 m/s ≈ 79 m/s

__

7) Using equation [eq3], we will find the time until Tina reaches her maximum height. Her actual off-the-ground total time is double this value. Using [eq3]: t = v/g = (1.8 m/s)/(9.8 m/s^2) = 9/49 s. Tina is in the air for double this time:

  2(9/49 s) ≈ 0.37 s

__

8) Use [eq2]: v = √(2·9.8 m/s^2·33.5 m) ≈ 25.624 m/s ≈ 26 m/s

__

9) Use [eq2]: v = √(2·9.8·3000) m/s ≈ 242.49 m/s ≈ 242 m/s

(Note: the terminal velocity in air is a lot lower than this for an object like a house.)

__

10) Use [eq1]: h = (625 m/s)^2/(2·9.8 m/s^2) ≈ 19,930 m

_____

<em>Additional comment</em>

Since all these questions make use of the same equation development, I have elected to answer them. Your questions are more likely to be answered if you restrict your posts to 3 or fewer questions each.

5 0
3 years ago
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