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4 years ago

What is a lunar eclipse?

Physics

2 answers:

motikmotik4 years ago

8 0

Answer: A lunar eclipse is when the Earth casts a shadow on the moon.

Explanation:Eclipses and transference are astronomical circumstances where a celestial body partially or covers the different celestial objects. A solar eclipse happens when the New Moon moves between Earth and the Sun while a lunar eclipse occurs when Earth casts a shadow on the Full Moon.

ser-zykov [4K]4 years ago

5 0

If im not wrong lunar eclipse is when the Earth, moln and sun all line up, leaving the Earth in the middle. You can recheck later.

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If the battery of your phone can provide 2 mA of current to your phone and holds a charge of 130 C, how long will it take a full

Naya [18.7K]

Answer: 65000 seconds

Explanation:

Given that,

Current (I) = 2 mA

(Since 1 mA = 1 x 10^-3A

2 mA = 2 x 10^-3A)

Charge (Q) = 130 C

Time taken for a fully charged phone to die (T) = ?

Recall that the charge is the product of current and time taken.

i.e Q = I x T

130C = 2 x 10^-3A x T

T = 130C / (2 x 10^-3A)

T = 65000 seconds (time will be in seconds because seconds is the unit of time)

Thus, it will take a fully charged phone 65000 seconds to die

5 0

4 years ago

Point A and B located at 4 meters and 9 meters from a source of the sound. If IA and IB are intensity at point A and point B, th

elena-14-01-66 [18.8K]

The intensity ratio at point A and B will be 81:16.

<u>Explanation:</u>

Sound waves are known to get faded with increase in the distance. This is because, the intensity of the sound is inversely proportional to the square of the distance of the source from the observer. So, if an observer is standing greater distance from the source of the sound, he/she will find difficulty in hearing the sound.

So, as the distance between the source and observer increases, the intensity of the sound wave decreases.

$I = \frac{1}{r^{2} }$

As here two points A and B are located at 4 m and 9 m distance from the source, then the intensity of sound at A and B will be inversely proportional to their respective square of the distance as shown below.

$I_{A} =\frac{1}{r_{A}^{2} } = \frac{1}{4 \times 4}=\frac{1}{16}$

Similarly,

$I_{B} =\frac{1}{r_{B}^{2} } = \frac{1}{9 \times 9}=\frac{1}{81}$

So, the ratio of intensity at point A and B will be

$\frac{I_{A} }{I_{B} } = \frac{81}{16} =81:16$

Thus, the intensity ratio at point A and B will be 81:16.

7 0

3 years ago

A daredevil is shot out of a cannon at 31.4 ◦ to the horizontal with an initial speed of 25.9 m/s. A net is positioned a horizon

Tanzania [10]

Answer:

The net should be placed at a height of 8.45 m

Explanation:

In order to calculate the height you have to apply the equations of <em>projectil motion.</em>

For x-axis, the position at a time t is given by:

Xf=Xo +(VoCosα)t (I)

Where Xf is the final position, Xo is the initial position, Vo is the initial speed and t is the time.

For the y-axis, the position at a time t is given by:

Yf=Yo + (VoSinα)t - 0.5g $t^{2}$ (II)

Where Yf is the final position, Yo is the initial position, Vo is the initial speed, t is the time and g is the acceleration of gravity.

For both equations, ∝ is the angle formed by the cannon and the horizontal

The height where the net should be placed in order to catch the daredevil is given by replacing the time when the daredevil will be in the horizontal position of 39.7 m in the equation of position for the y-axis.

So, you have to calculate the time using (I)

Let the system of reference be placed at x=o and y=o

Xf= 0 + 25.9Cos(31,4◦)t

Xf=39.7

39.7=22.1t

Dividing both sides by 22.1 to find the value of t:

t=1.79 seconds

Replacing the time in equation (II) to find the height:

Yf=0 +25.9Sin(31,4◦)(1.79) - 0.5(9.8) $1.79^{2}$

Yf= 8.45 m

So the height is 8.45 m

4 0

4 years ago

A ball is thrown vertically into the air with a initial velocity of 20 m/s. Find the maximum height of the ball and find the amo

REY [17]

Answer:

The maximum height of the ball is 20 m. The ball needs 2 s to reach that height.

Explanation:

The equation that describes the height and velocity of the ball are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the ball at time t

y0 = initial height

v0 = initial velocity

t = time

g = acceleration

v = velocity at time t

When the ball is at its maximum height, its velocity is 0, then, using the equation of the velocity, we can calculate the time at which the ball is at its max-height.

v = v0 + g · t

0 = 20 m/s - 9.8 m/s² · t

-20 m/s / -9.8 m/s² = t

t = 2.0 s

Then, the ball reaches its maximum height in 2 s.

Now, we can calculate the max-height obtaining the position at time t = 2.0 s:

y = y0 + v0 · t + 1/2 · g · t²

y = 0 m + 20 m/s · 2 s - 1/2 · 9,8 m/s² · (2 s)²

y = 20 m

8 0

4 years ago

Sound travels about 750 miles per hour. if you stand in a canyon and sound a horn, you will hear an echo. suppose it takes about

diamong [38]

The answer to this question would be:3850ft

To answer this question, you need to convert the speed velocity from miles/hour into feet/second. The equation would be: 750 miles/hour x 5280 foot/mile x 1 hour/3600second = 1100 ft/s
Then multiply the time with the velocity= 3.5 second x 1100 ft/s= 3850ft

5 0

4 years ago

Read 2 more answers