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salantis [7]
3 years ago
10

Question 4

Physics
1 answer:
zmey [24]3 years ago
8 0
I think the answer is B but not too sure.
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Olegator [25]
Water goes through evaporation.
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Help please this is about the dialysis and things
AVprozaik [17]
1. B. Hemo is on a machine, peritoneal is in the abdomen
2. B. Nothing is as good as having your own kidney, and not everyone is a good candidate for all types of dialysis.
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8 0
3 years ago
As part of his job, Warren attends seminars where he must listen to presentations given by experts in his field. When he is list
jeka57 [31]

Answer:

Option (B)

Explanation:

In terms of communication, a receiver is usually referred to as a person who listens, reads as well as observes. In simple words, a receiver is an individual or it can be a group of individuals, to whom any type of message is being diverted. The other name for the receiver is 'audience'.

In the given condition, Warren is attending a seminar in which he is listening to the speaker, as a part of the audience. So, it can be concluded that Warren is a receiver who is receiving information or hearing the speaker.

Thus, the correct answer is option (B).

7 0
3 years ago
Two equally charged, 1.00 g spheres are placed with 2.00 cm between their centers. when released, each begins to accelerate at 2
Leya [2.2K]
1) Force = m*a = 1.00 g * (1kg / 1000 g) * 225 m/s^2 = 0.225 N

2) Charge

Force = K (charge)^2 /(distance)^2 => charge = √ [Force * distance^2 / k]

k = 9.00 * 10^9 N*m^2 / C^2

charge = √ [0.225 N * (0.02 m)^2 / 9.00* 10^9 N*m^2 / C^2 ]

charge = 0.0000001 C = 0.0001 mili C
3 0
3 years ago
the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​
Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

v_y(t)=vo*sin(A)-g*t

the velocity is Zero when the projectile reach in the maximum altitude:

0=vo-gt\\t=\frac{vo}{g}

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

d_x(t)=vo*cos(A)*t\\

R=Range

R=d_x(t=2*\frac{vo}{g})

R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

B=30°

7 0
3 years ago
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