11) Do the past tense as in the example.

Note that the scale bar under the photo is labeled 1 μm (micrometer). The scale bar works the same way as a scale on a map, wher

AnnZ [28]

Answer:

Hello your question has some missing parts attached below is the complete question

answer : 4 μm

Explanation:

since the scale bar works the same way as a scale on a map , each bar will therefore represent 1 μm and the mature parent cell's is about 4 times the labeled value hence the Mature parent cell diameter will approximately be : 4 μm

8 0

3 years ago

g A loop circuit has a resistance of R1 and a current of 2 A. The current is reduced to 1.4 A when an additional 2.2 Ω resistor

Mrrafil [7]

Answer:

R1 = 5.13 Ω

Explanation:

From Ohm's law,

V = IR............... Equation 1

Where V = Voltage, I = current, R = resistance.

From the question,

I = 2 A, R = R1

Substitute into equation 1

V = 2R1................ Equation 2

When a resistance of 2.2Ω is added in series with R1,

assuming the voltage source remain constant

R = 2.2+R1, and I = 1.4 A

V = 1.4(2.2+R1)................. Equation 3

Substitute the value of V into equation 3

2R1 = 1.4(2.2+R1)

2R1 = 3.08+1.4R1

2R1-1.4R1 = 3.08

0.6R1 = 3.08

R1 = 3.08/0.6

R1 = 5.13 Ω

6 0

3 years ago

An unknown radioactive sample is observed to decrease in activity by a factor of two in a one hour period. What is its half-life

elena55 [62]

Answer:

The half-life is $t_{1/2} = 1.005 h$

Explanation:

Using the decay equation we have:

$A=A_{0}e^{-\lambda t}$

Where:

λ is the decay constant
A(0) the initial activity
A is the activity at time t

We know the activity decrease by a factor of two in a one hour period (t = 1 h), it means that $A = \frac{A_{0}}{2}$

$\frac{A_{0}}{2}=A_{0}e^{-\lambda*1 h}$

$0.5=e^{-\lambda*1 h}$

Taking the natural logarithm on each side we have:

$ln(0.5)=-\lambda$

$\lambda=0.69 h^{-1}$

Now, the relationship between the decay constant λ and the half-life t(1/2) is:

$\lambda = \frac{ln(2)}{t_{1/2}}$

$t_{1/2} = \frac{ln(2)}{\lambda}$

$t_{1/2} = \frac{ln(2)}{0.69}$

$t_{1/2} = 1.005 h$

I hope it helps you!

6 0

3 years ago

You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn f

Verdich [7]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. We will calculate the initial velocity of the object, and from it, we will calculate the final position. With the considerations made in the statement we will obtain the total height. Initial velocity of the acorn,

$u = 0m/s$