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Komok [63]
2 years ago
5

Trick question (really easy) just for fun! If u get this right u get brainliest!

Physics
2 answers:
Maurinko [17]2 years ago
6 0

Answer:

piper duh

Explanation:

Katarina [22]2 years ago
6 0

Answer:

Mary is the name of fourth child

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State two factors that affect the rate of diffusion of a substance
Ronch [10]

Explanation:

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  • <em>Diffusion</em><em> </em><em>of</em><em> </em><em>substance</em><em> </em><em>plays</em><em> </em><em>an</em><em> </em><em>important</em><em> </em><em>role</em><em> </em><em>on</em><em> </em><em>cellular</em><em> </em><em>transport</em><em> </em><em>in</em><em> </em><em>plants</em><em>.</em><em> </em>
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4 0
3 years ago
Land heats up and cools down quickly because...
chubhunter [2.5K]
I would say it reflects the sun easily. That’s also how we see it :)
6 0
3 years ago
A gas occupies a volume of 20 cubic meters at 9,000 pascals. If the pressure is lowered to 5,000 pascals, what volume will the g
forsale [732]
We need to consider no change in the temperature of gas (isothermal transformation)

Volume and pressure are inversely proportional magnitudes, so we can write:

P_1.V_1=P_2.V_2\\&#10;\\&#10;9.20=5.V_2\\&#10;\\&#10;V_2=\frac{180}{5}=36 \ m^3
5 0
3 years ago
A 1.90-m-long barbell has a 25.0 kg weight on its left end and a 37.0 kg weight on its right end. if you ignore the weight of th
gregori [183]

Answer: The center of gravity is 1.1338 m away from the left side of the barbell

Explanation:

Length of the barbell = 1.90 m

The distance center of gravity from left = x

Mass on the left side = 25 kg

The distance center of gravity from right = 1.90 - x

Mass on the right side = 37 kg

At the balance point: m_1x_1=m_2x_2

25 kg\times x=37 kg\times (1.90-x)

x=1.1338 m

The center of gravity is 1.1338 m away from the left side of the barbell

7 0
3 years ago
Read 2 more answers
Two trains on separate tracks move toward each other. Train 1 has a speed of 145 km/h; train 2, a speed of 72.0 km/h. Train 2 bl
tekilochka [14]

Answer:

Therefore,

The frequency heard by the engineer on train 1

f_{o}=603\ Hz

Explanation:

Given:

Two trains on separate tracks move toward each other

For Train 1 Velocity of the observer,

v_{o}=145\ km/h=145\times \dfrac{1000}{3600}=40.28\ m/s

For Train 2 Velocity of the Source,

v_{s}=90\ km/h=90\times \dfrac{1000}{3600}=25\ m/s

Frequency of Source,

f_{s}=500\ Hz

To Find:

Frequency of Observer,

f_{o}=?  (frequency heard by the engineer on train 1)

Solution:

Here we can use the Doppler effect equation to calculate both the velocity of the source v_{s} and observer v_{o}, the original frequency of the sound waves f_{s} and the observed frequency of the sound waves f_{o},

The Equation is

f_{o}=f_{s}(\dfrac{v+v_{o}}{v -v_{s}})

Where,

v = velocity of sound in air = 343 m/s

Substituting the values we get

f_{o}=500(\dfrac{343+40.28}{343 -25})=500\times 1.205=602.64\approx 603\ Hz

Therefore,

The frequency heard by the engineer on train 1

f_{o}=603\ Hz

7 0
3 years ago
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