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seraphim [82]
3 years ago
14

Where are you on Earth if you experience each of the following? (Refer to the discussion in Observing the Sky: The Birth of Astr

onomy as well as this chapter.)
The stars rise and set perpendicular to the horizon.
The stars circle the sky parallel to the horizon.
The celestial equator passes through the zenith.
In the course of a year, all stars are visible.
The Sun rises on March 21 and does not set until September 21 (ideally).
Physics
1 answer:
Aloiza [94]3 years ago
7 0

Explanation:

We know that the sky appears to us like a sphere called as celestial sphere which appears to rotate around an imaginary axis because of Earth's rotation. Since the axis cuts the celestial sphere at celestial poles all the object seems to circle around the celestial poles.

Condition 1: The stars rise and set perpendicular to the horizon

The observer is at the equator

Condition 2: The stars circle the sky parallel to the horizon

The observer is at the Pole of the Earth

Condition 3: The celestial equator passes through the zenith

The observer is at the equator

Condition 4: In the course of a year, all stars are visible

The observer is at the equator

Condition 5: The Sun rises on March 21 and does not set until September 21 (ideally)

The observer is at North Pole

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11. A 6.0‐m wire with a mass of 50 g, is under tension. A transverse wave, for which the frequency is 810 Hz, the wavelength is
Irina-Kira [14]

Answer:

Time will be 19 ms so option (a) is correct option

Explanation:

We have given that mass of wire m = 50 gram = 0.5 kg

Frequency f = 810 Hz

Wavelength = 0.4 m

Velocity is given by

v=wavelength\times frequency=810\times 0.4=324m/sec

Amplitude is given as d = 6 m

So time t=\frac{distance}{velocity}=\frac{6}{324}=0.01851=19ms

So option (a) is correct option

4 0
3 years ago
Difference between freefall and weightlessness.​
Margaret [11]

Answer:

Differences between freefall and weightlessness are as follows:

<h3><u>Freefall</u></h3>
  • When a body falls only under the influence of gravity, it is called free fall.
  • Freefall is not possible in absence of gravity.
  • A body falling in a vacuum is an example of free fall.

<h3><u>Weightlessness</u></h3>
  • Weightlessness is a condition at which the apparent weight of body becomes zero.
  • Weightlessness is possible in absence of gravity.
  • A man in a free falling lift is an example of weightlessness.

Hope this helps....

Good luck on your assignment....

6 0
3 years ago
Give an example for each of the following, where the force:
max2010maxim [7]

Explanation:

  • A cricket player hitting the ball from opposite direction.
  • A footballer kicking ball with more force.
  • Heating of a plastic bottle.
  • Applying brake of a car.
  • rolling a stopped marble on a table.

7 0
2 years ago
Read 2 more answers
To protect a material from the influence of an external magnetic field, the material should be kept in?
Evgen [1.6K]

To protect a material from the influence of an external magnetic field, the material should be kept in soft iron ring.

So the correct answer is A.

Hope this helps,

Davinia.

5 0
2 years ago
A projectile is fired at an upward angle of 45.0º from the top of a 265-m cliff with a speed of .sm 185 What will be its speed w
nignag [31]

The final velocity of the projectile when it strikes the ground below is 198.51 m/s.

<h3>Time of motion of the projectile</h3>

The time taken for the projectile to fall to the ground is calculated as follows;

h = vt + ¹/₂gt²

where;

  • h is height of the cliff
  • v is velocity
  • t is time of motion

265 = (185 x sin45)t + (0.5)(9.8)t²

265 = 130.8t + 4.9t²

4.9t² + 130.8t - 265 = 0

solve the quadratic equation using formula method,

t = 1.89 s

<h3>Final velocity of the projectile</h3>

vyf = vyi + gt

where;

  • vyf is the final vertical velocity
  • vyi is initial vertical velocity

vyf = (185 x sin45) + (9.8 x 1.89)

vyf = 149.322 m/s

vxf = vxi

where;

  • vxf is the final horizontal velocity
  • vxi is the initial horizontal velocity

vxf = 185 x cos(45)

vxf = 130.8 m/s

vf = √(vyf² + vxf²)

where;

  • vf is the speed of the projectile when it strikes the ground below

vf = √(149.322²  +  130.8²)

vf = 198.51 m/s

Learn more about final velocity here: brainly.com/question/6504879

#SPJ1  

3 0
2 years ago
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