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Natasha2012 [34]
2 years ago
11

A person standing on the edge of a high cliff throws a rock straight up with an initial velocity, v0v0, of 13.0m/s13.0m/s. on th

e way back down the rock misses the edge of the cliff and continues to fall. ignoring the effects of air resistance, calculate the position and velocity of the rock a
Physics
1 answer:
Firdavs [7]2 years ago
8 0
For the position, we can use the equation below for our calculations:
H = h + v₀²/2g
where 
H is the total height from the base of the cliff to the highest point of the rock on air
h is the height of the cliff
v₀ is the initial height
g is equal to 9.81 m/s²

H = h + 13²/2(9.81)
H = h + 8.61 m

Because the height of the cliff is not known, the answer would just be h +8.61 m.

For the velocity, we use the equation: v = √2gH. Substituting the answer for H, the velocity is

v = √2(9.81)(h+8.61)
v = 4.43√(h+8.61)
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Determine
djyliett [7]

(i) The total capacitance for the circuit is 5 μF.

(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.

(iii) The charge stored in 3μF capacitor is  6 x 10⁻⁶ C.

<h3>Total capacitance of the circuit</h3>

The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.

<h3>C1 and C2 are in series </h3>

\frac{1}{C_{12}} = \frac{1}{C_1 } + \frac{1}{C_2} \\\\\frac{1}{C_{12}} = \frac{1}{4 } + \frac{1}{4} \\\\\frac{1}{C_{12}} = \frac{1}{2} \\\\C_{12} = 2 \ \mu F

<h3>C1 and C2 are parallel to C3</h3>

C_{123} = C_{12} + C_3\\\\C_{123} = 2\ \mu F + 2\ \mu F \\\\C_{123} = 4 \ \mu F

<h3>C(123) is series to C5 and C6</h3>

\frac{1}{C_{t} } = \frac{1}{C_{123}} + \frac{1}{C_5} + \frac{1}{C_6} \\\\\frac{1}{C_{t} } = \frac{1}{4} + \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{t} } = \frac{12}{24} \\\\\frac{1}{C_{t} } = \frac{1}{2} \\\\C_t = 2 \ \mu F

<h3>C7 and C8 are in series</h3>

\frac{1}{C_{78}} = \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{78}} = \frac{2}{6} \\\\\frac{1}{C_{78}} =\frac{1}{3} \\\\C_{78} = 3 \ \mu F

<h3>Total capaciatnce of the circuit</h3>

Ct + C(78) = 2 μF + 3 μF = 5 μF

<h3 /><h3>Total charge stored in the circuit</h3>

The total charge stored in the capacitor is calculated as follows;

Q = CV

Q = (5 x 10⁻⁶) x (20)

Q = 1 x 10⁻⁴ C

<h3>Charge stored in 3μF capacitor</h3>

Q =  (3 x 10⁻⁶) x (20)

Q = 6 x 10⁻⁶ C

Learn more about capacitance of capacitor here: brainly.com/question/13578522

8 0
2 years ago
A sample of argon gas (molar mass 40 g) is at four times the absolute temperature of a sample of hydrogen gas (molar mass 2 g).
qaws [65]

To solve this problem, let us recall that the formula for gases assuming ideal behaviour is given as:

rms = sqrt (3 R T / M)

where

R = gas constant = 8.314 Pa m^3 / mol K

T = temperature

M = molar mass

Now we get the ratios of rms of Argon (1) to hydrogen (2):

rms1 / rms2 = sqrt (3 R T1 / M1) / sqrt (3 R T2 / M2)

or

rms1 / rms2 = sqrt ((T1 / M1) / (T2 / M2))

rms1 / rms2 = sqrt (T1 M2 / T2 M1)

Since T1 = 4 T2

rms1 / rms2 = sqrt (4 T2 M2 / T2 M1)

rms1 / rms2 = sqrt (4 M2 / M1)

and M2 = 2 while M1 = 40

rms1 / rms2 = sqrt (4 * 2 / 40)

rms1 / rms2 = 0.447

 

Therefore the ratio of rms is:

<span>rms_Argon / rms_Hydrogen = 0.45</span>

7 0
3 years ago
When did humans learn that the earth is not the center of the universe?
Vinvika [58]

Answer:

When did humans learn that the Earth is not the center of the universe?

Answer

1

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4 Answers

Asked in 3 Spaces





Science - Next Generation

Alexander Somm

, Consultant, Investor Relations at Novelpharm AG (2015-present)

Answered Oct 16

What, it isn’t?!

Sorry, I had to.

As far as I have read and understood, the Sumerians and later the Babylonians both had astronomical calendars that already differentiated planets and stars. Earth was not the center to them, the Sun likely was. That was around 2,200 - 1,600 BC.

After that, Greek philosopher Aristarchus of Samos (310 - 230 BC) was the first (recorded) to have believed the solar system was organized around the Sun, rather than the Earth. His heliocentric model was unpopular during Aristarchus’ lifetime, although it would inspire astronomers centuries later, such as Copernicus and Galileo.

Now, there are numerous archeological findings (cave paintings) and studies, that all suggest an understanding of complex astronomy in prehistoric times dating back as far as 40,000 years. This also explains how early, prehistoric migrants may have navigated the seas.

Explanation:

hope it helps

have a good day

4 0
1 year ago
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