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Alex
1 year ago
13

Tiffany has been waiting for weeks to see one particular meteor shower that peaks around august 12. What meteor shower is tiffan

y most likely waiting to see?
Physics
1 answer:
Andru [333]1 year ago
5 0

Tiffany has been waiting for weeks to see one particular meteor shower that peaks around august 12. Tiffany is waiting to see Perseid Meteor Shower.

A very bright supermoon known as a "Sturgeon Moon" may make it difficult for photographers to capture this year's climax of the Perseid Meteor Shower tonight (August 12). The Perseid meteor shower, which peaks overnight on August 12 and 13, is one of the most powerful annual meteor showers and occurs between July 14 and September 1. Unfortunately, that falls on the night of this month's full moon, which became fully illuminated yesterday night (Aug. 11). The meteor shower is named after the constellation, the northern constellation Perseus.

Learn more about Perseid Meteor Shower here:

brainly.com/question/16517600

#SPJ4

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Answer:

depends... do you add suger to your ketchup?

Explanation:

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Why is the moon's umbra much smaller during a solar eclipse?
Stolb23 [73]

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the sun is larger than the moon so it creates a shadow

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2 years ago
Is it true or false
tester [92]

Answer:

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Explanation:

8 0
2 years ago
Calculate the RMS voltage of the following waveforms with 10 V peak-to-peak:
Deffense [45]

Answer:

a) T=0.01s

b) T=0.001s

c) T=0.00001s

Explanation:

From the question we are told that:

Given Frequencies

a. 100 Hz,

b. 1 kHz,

c. 100 kHz.

Generally the equation for Waveform Period is mathematically given by

T=\frac{1}{f}

Therefore

a)

For

T=100 Hz

T=\frac{1}{100}

T=0.01s

b)

For

F=1kHz

T=\frac{1}{1000}

T=0.001s

c)

For

F=100kHz

T=\frac{1}{100*100}

T=0.00001s

6 0
3 years ago
You shoot an arrow into the air. Two seconds later (2.00 s) the arrow has gone straight upward to a height of 35.0 m above its l
sdas [7]

This question can be solved by using the equations of motion.

a) The initial speed of the arrow is was "9.81 m/s".

b) It took the arrow "1.13 s" to reach a height of 17.5 m.

a)

We will use the second equation of motion to find out the initial speed of the arrow.

h= v_it + \frac{1}{2}gt^2\\

where,

vi = initial speed = ?

h = height = 35 m

t = time interval = 2 s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

35\ m = (v_i)(2\ s)+\frac{1}{2}(9.81\ m/s^2)(2\ s)^2\\\\v_i(2\ s)=19.62\ m\\\\v_i = \frac{19.62\ m}{2\ s}

<u>vi =  9.81 m/s</u>

b)

To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

h= v_it + \frac{1}{2}gt^2\\

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 17.5 m

vi = initial speed = 9.81 m/s

t = time = ?

Therefore,

17.5 = (9.81)t+\frac{1}{2}(9.81)t^2\\4.905t^2+9.81t-17.5=0

solving this quadratic equation using the quadratic formula, we get:

t = -3.13 s (OR) t = 1.13 s

Since time can not have a negative value.

Therefore,

<u>t = 1.13 s</u>

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

4 0
2 years ago
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