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konstantin123 [22]
3 years ago
10

A white blood cell has a diameter of approximately 10 micrometers, or 0.01 um. A model represents its diameter as 5 um. what is

the ratio of model size to actual size?

Physics
2 answers:
denpristay [2]3 years ago
4 0

Answer:

Option B. 500 : 1

Explanation:

The following data were obtained from the question:

Actual size = 0.01 μm

Mode size = 5 μm

Ratio of model to actual size =.?

The ratio of the model size to the actual size can be obtained as illustrated below:

Model size : Actual size => 5 : 0.01

Model size/Actual size => 5 / 0.01

Model size/Actual size => 500/1

Model size : Actual size => 500 : 1

From the above illustration, the ratio of the model size to that of the actual size is 500 : 1

aleksandrvk [35]3 years ago
3 0

Answer:

the answer is B

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Answer:

Explanation:

A ) When gymnast is motionless , he is in equilibrium

T = mg

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B )

When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.

T = mg

= 618.03 N

C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2

Net force on it = T - mg   , acting in upward direction

T - mg = m a

T =  mg + m a

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= 63 ( 9.81 + .6)

= 655.83 N

D )  If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2

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6 0
3 years ago
When is thermal equilibrium achieved between two identical objects?
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3 0
3 years ago
Pure silicon contains approximately 1.0 X 1016 free electrons per cubic meter. (
Margarita [4]
A. The formula for mean free time is:

t = V/(4π√2 r²vN)
where
N = 1×10¹⁶ molecules (per m³)
V = 1 m³
r = 111×10⁻⁷m (atomic radius of silicon)

Let's solve for v first:
v = √(3RT/M) = √(3(8.314 m³·Pa/mol·K)(25 + 273 K)/28.1 g/mol Si)
v = 16.26 m/s

t = (1 m³)/(4π√2 (111×10⁻⁷m)²(16.26 m/s)(1×10¹⁶ molecules))
<em>t = 2.81×10⁻9 s</em>

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3 0
4 years ago
When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa
MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2

In parallel

\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}

\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0

Solving the above quadratic equation

\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}

\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

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