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jolli1 [7]
3 years ago
11

I just need help plzzzz

Physics
2 answers:
AleksAgata [21]3 years ago
6 0

Answer:

B.) Veins in a multicellular organism.

Explanation:

Ksenya-84 [330]3 years ago
3 0

Answer:

I think D it's the correct answer!! Let me know if i'm wrong

Explanation:

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What is the frequency of a wave having a period equal to 18 seconds? a. 6.6 × 10-2 hertz b. 5.5 × 10-2 hertz c. 3.3 × 10-2 hertz
irakobra [83]
Frequency = 1/period. ... 1 / 18 sec = (1/18) per sec. That's 0.056 per sec or 0.056 Hz. (rounded) (5.6 x 10^-2 Hz)
6 0
3 years ago
Read 2 more answers
He pictures show a grassy field that has been abandoned. Which best explains why this is an example of increasing entropy?
solmaris [256]
<h2>Answer:</h2>

The correct answer is option D. Which is "Over time, the lawn has naturally become disorderly".

<h3>Explanation:</h3>
  • Entropy, the measure of a system's thermal energy per unit temperature that is unavailable for doing useful work.
  • It is also a measure of the molecular disorder, or randomness, of a system.
  • According to above definitions of entropy option D is correct.
  • <u>So entropy of system is randomness/disorder that is increasing with time in case of lawn.</u>

6 0
3 years ago
What is the difference between rutherford's model of the atom and bohr's model of the atom
Amanda [17]

Answer:

Rutherford described the atom as consisting of a tiny positive mass surrounded by a cloud of negative electrons. Bohr thought that electrons orbited the nucleus in quantised orbits. Bohr built upon Rutherford's model of the atom. ... So it was not possible for electrons to occupy just any energy level.

Explanation:

7 0
3 years ago
Your grandfather clock's pendulum has a length of 0.9930 m.
posledela

We will start from the period definition to find the relationship between the perioricity and the length. After finding the relationship between the two lengths and periods we will proceed to calculate the length two with the loss of the period, that is

T = 2\pi \sqrt{\frac{l}{g}}

Therefore the period is proportional to the square root of the length

T \propto l^{1/2}

Then

\frac{T_2}{T_1} = \frac{\sqrt{L_2}}{\sqrt{L_1}}

(\frac{T_2}{T_1})^2 = \frac{L_2}{L_1}

L_2 = (\frac{T_2}{T_1})^2(L_1)

The period T_1 is equivalent to the seconds that a day has, that is 86400 seconds while period two will be the seconds that have one day less the loss of 16 consecutive announced in the statement therefore,

L_2 = (\frac{86400}{86384})(0.993)

L_2 = 0.99336

The total change in the lenght is

\Delta L = L_2 - L_1

\Delta L = 0.99336 - 0.993

\Delta L = 0.00036m = 3.6*10^{-4}m = 360 \mu m

Therefore the pendulum should be adjust in 360 \mu m

5 0
3 years ago
To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3
hodyreva [135]

Answer : The correct option is, (c) 3.7\times 10^2J/^oC

Explanation :

First we have to calculate the energy or heat.

Formula used :

E=V\times I\times t

where,

E = energy (in joules)

V = voltage (in volt)

I = current (in ampere)

t = time (in seconds)

Now put all the given values in the above formula, we get:

E=(3.6V)\times (2.6A)\times (350s)

E=3276J

Now we have to calculate the heat capacity of the calorimeter.

Formula used :

C=\frac{E}{\Delta T}=\frac{E}{T_{final}-T_{initial}}

where,

C = heat capacity of the calorimeter

T_{initial} = initial temperature = 20.3^oC

T_{final} = final temperature = 29.1^oC

Now put all the given values in this formula, we get:

C=\frac{3276J}{(29.1-20.3)^oC}

C=372.27J/^oC=3.7\times 10^2J/^oC

Therefore, the heat capacity of the calorimeter is, 3.7\times 10^2J/^oC

7 0
3 years ago
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