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3 years ago

Why is polaris used as a celestial reference point?

1 answer:

8 0

At the equator, the North Celestial and South Celestial Poles would lie on the horizon where the meridian intersects the horizon. Polaris is called the "North Star." ... Polaris is special because the Earth's North Pole points almost exactly towards it.

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Is it possible to put a distance versus time graph to be universal go mine

xz_007 [3.2K]

Yes I'm pretty sure you can

3 0

3 years ago

What is the energy of an electromagnetic wave that has a frequency of

sukhopar [10]

Answer:

$Energy, \; E = 2.6504 * 10^{-34} \; Joules$

Explanation:

Given the following data;

Frequency = 4.0 x 10⁹ Hz

Planck's constant, h = 6.626 x 10-34 J·s.

To find the energy of the electromagnetic wave;

Mathematically, the energy of an electromagnetic wave is given by the formula;

E = hf

Where;

E is the energy possessed by a wave.

h represents Planck's constant.

f is the frequency of a wave.

Substituting the values into the formula, we have;

$Energy, \; E = 4.0 x 10^{9} * 6.626 x 10^{-34}$

$Energy, \; E = 2.6504 * 10^{-34} \; Joules$

8 0

3 years ago

A balloon-powered car rolls across the floor at a speed of 0.711 m/s. How long does it take to cover 8.25 m?

devlian [24]

Answer:

Time = 11.60 seconds.

Explanation:

Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.

Mathematically, speed is given by the equation;

$Speed = \frac{distance}{time}$

Given the following data;

Speed = 0.711m/s

Distance = 8.25m

To find the time;

Making time the subject of formula, we have;

$Time = \frac{distance}{speed}$

Substituting into the equation, we have;

$Time = \frac{8.25}{0.711}$

Time = 11.60 secs.

5 0

3 years ago

When you jump straight up as high as you can, what is the order of magnitude of the maximum recoil speed that you give to the Ea

Klio2033 [76]

Answer:

5.66 × 10⁻²³ m/s

Explanation:

If i assume i can jump as high as h = 2 m, my initial velocity is gotten from v² = u² + 2gh. Since my final velocity v = 0, u = √2gh = √(2 × 9.8 × 2) = √39.2 m/s = 6.26 m/s.

Since initial momentum = final momentum,

mv₁ + MV₁ = mv₂ + MV₂ where m, M, v₁, V₁, v₂ and V₂ are my mass, mass of earth, my initial velocity, earth's initial velocity, my final velocity and earth's final velocity respectively.

My mass m = 54 kg, M = 5.972 × 10²⁴ kg, v₁ = 6.26 m/s, V₁ = 0, v₂ = 0 and V₂ = ?

So mv₁ + M × 0 = m × 0 + MV₂

mv₁ = MV₂

V₂ = mv₁/M = 54kg × 6.26 m/s/5.972 × 10²⁴ kg = 338.093/5.972 × 10²⁴ = 56.61 × 10⁻²⁴ m/s = 5.661 × 10⁻²³ m/s ≅ 5.66 × 10⁻²³ m/s

5 0

3 years ago

At a hot air balloon race, a person on the ground shots a ball of confetti from a cannon to start the race. One of the balloons

Sati [7]

Answer:

a) The balloon and the ball will meet after 1.43 s.

b) The ball will reach the balloon at 15.7 m above the ground.

Explanation:

The height of the confetti ball is given by the following equation:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the ball at time t

y0 = initial height

v0 = initial velocity

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

The height of the ball is given by this equation:

y = y0 + v · t

Where v is the constant velocity.

When the ball and the ballon meet, both heights are equal. Let´s consider the ground as the origin of the frame of reference so that y0 = 0:

y balloon = y ball

y0 + v · t = y0 + v0 · t + 1/2 · g · t² (y0 = 0)

11 m/s · t = 18 m/s · t -1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 18 m/s · t - 11 m/s · t

0 = -4.9 m/s² · t² + 7 m/s · t

0 = t( -4.9 m/s² · t + 7 m/s)

t = 0 and

0 = -4.9 m/s² · t + 7 m/s

-7 m/s / - 4.9 m/s² = t

t = 1.43 s

They will meet after 1.43 s

b) Now let´s calculate the height of the balloon after 1.43 s

y = v · t

y = 11 m/s · 1.43 s = 15.7 m

The ball will reach the balloon at 15.7 m above the ground.

7 0

3 years ago