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3 years ago

The right eye and right lung are __________.

Physics

1 answer:

PilotLPTM [1.2K]3 years ago

3 0

The correct answer is A) Ipsilateral

Explanation:

The term ipsilateral is commonly used to describe objects or structures that are on the same side of a body or structure. This term is correct to describe the right eye and the right lung because these two organs are on the same side of the body (the right side). This can also be used to describe other organs such as the left humerus and the left hand or the right ear and the right feet because these pairs are also on the same side. According to this, the correct answer is A.

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An aluminum cup contains 225 g of water and a 40-g copper stirrer, all at 27Â°C. A 470-g sample of silver at an initial temperat

vagabundo [1.1K]

Answer:

Mal = 0.232 kg = 232 g

Explanation:

mass of water (Mw) = 225 g = 0.225 kg

mass of copper stirrer (Mcu) = 40 g = 0.04 kg

initial temperature of water (Tw) = 27 degrees

mass of silver (Mag) = 470 g = 0.47 kg

initial temperature of silver (Ts) = 85 degrees

final temperature of the mixture (T) = 32 degrees

find the mass of the aluminum cup (Mal)

applying the conservation of energy

((Mal.cAl) + (Mw.cW) + (Mcu.cCu))(ΔTw) = (Mag.cAg)(ΔTag)

we require the specific heat capacities of water (cW) , aluminium (cAl) , copper (cCu) and silver (cAg) which are as follows

water (cW) =4186 J/kg.K

aluminium (cAl) = 900 J/kg.K

copper (cCu) = 387 J/kg.K

silver (cAg) = 234 J/kg.K

now we can put in our values to get the mass of the Aluminium cup (Mag)

((Mal.900) + (0.225 x 4186) + (0.04 x 387))(32-27) = (0.47 x 234)(85-32)

(900 . Mal + 957.33) x 5 = 5828.94

900 .Mal + 957.33 = 1165.79

900.Mal = 1165.79 - 957.33 = 208.5

Mal = 0.232 kg = 232 g

8 0

3 years ago

What is the greenhouse effect?

Zanzabum

A on that problem :)[email protected]

6 0

3 years ago

Read 2 more answers

A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit

vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

$\omega =\sqrt{\dfrac{K}{m}}$

The maximum speed v given as

$v=\omega A$

A=Amplitude

$v=\sqrt{\dfrac{K}{m}}\times A$

$0.32=\sqrt{\dfrac{13.5}{0.75}}\times A$

A=0.075 m

A= 0.75 cm

The speed at distance x

$v=\omega \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}$

v= 0.21 m/s

5 0

3 years ago

A 1000 kg car moving a 10 m/s collides with a stationary 2000 kg truck. The two vehicles interlock as a result of the collision.

IgorLugansk [536]

Answer:

v₃ = 3.33 [m/s]

Explanation:

This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.

In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.

$(m_{1}*v_{1})+(m_{2}*v_{2})=(m_{1}+m_{2})*v_{3}$

where:

m₁ = mass of the car = 1000 [kg]

v₁ = velocity of the car = 10 [m/s]

m₂ = mass of the truck = 2000 [kg]

v₂ = velocity of the truck = 0 (stationary)

v₃ = velocity of the two vehicles after the collision [m/s].

Now replacing:

$(1000*10)+(2000*0)=(1000+2000)*v_{3}\\v_{3}=3.33[m/s]$

7 0

3 years ago

Can someone help me with this please?

Slav-nsk [51]

Answer:

C2, C1, C4, C5 and C6 are in parallel. Therefore, we use the formula Cp = C1 + C2 + ....

Cp = C2 + C1 + C4 + C5 + C6 = ( 7 * 10 ^-3) + (18 * 10^-6) + (0.8F) + (200 * 10^-3 F) + (750 * 10^-6) = 1.008F

Now, Cp will become one capacitor and it will be aligned with C3, therefore it will now become a circuit in series.

We use the formula: 1/Cs = 1/C1 + 1/C2 + .... + ....1/Cn

Thus,

1/Cs = 1/C3 + 1/Cp

1/Cs = 1/(14 * 10^-3 F) + 1/(1.008F)

Cs = 1.4 * 10 ^-2 or if we do not round too much it will give exactly 0.0138 F

So the answer should be a)

8 0

3 years ago