An athlete can exercise by making mechanical waves in ropes. What is the

A coin has a radius of 1.06 cm and a thickness of 1.2 mm. Find its<br><br> volume in m^3.

ANTONII [103]

Answer:

The volume of the coin is 4.236 x 10⁻⁷ m³

Explanation:

Given;

radius of a coin, r = 1.06 cm = 0.0106 m

thickness of the coin, h = 1.2 mm = 0.0012 m

The volume of the coin is given by;

volume = Area x thickness

Area of the coin = πr² = π (0.0106)² = 3.5304 x 10⁻⁴ m²

The volume of the coin = (3.5304 x 10⁻⁴ m²) x (0.0012 m)

The volume of the coin = 4.236 x 10⁻⁷ m³

Therefore, the volume of the coin is 4.236 x 10⁻⁷ m³

4 0

3 years ago

What happens to the orbit time as the distance increases.

Maslowich

The orbital period increases if the orbital distance is increased.

6 0

3 years ago

A house has well-insulated walls. It contains a volume of 105 m3 of air at 305 K.

REY [17]

Answer: $85.46\ kJ$

Explanation:

Given

Volume of air $V=105\ m^3$

Temperature of air $T=305\ K$

Increase in temperature $\Delta T=0.7^{\circ}C$

Specific heat for diatomic gas is $C_p=\dfrac{7R}{2}$

Energy required to increase the temperature is

$\Rightarrow Q=nC_pdT\\\\\Rightarrow Q=n\times \dfrac{7R}{2}\times \Delta T\\\\\Rightarrow Q=\dfrac{7}{2}nR\Delta T\\\\\Rightarrow Q=\dfrac{7}{2}\times \dfrac{PV}{T}\times \Delta T\quad [\text{using PV=nRT}]$

Insert the values

$\Rightarrow Q=\dfrac{7}{2}\times \dfrac{1.01325\times 10^5\times 105}{305}\times 0.7\\ \text{Assuming air pressure to be atmospheric P=}1.01325\times 10^5\ N/m^2\\\\\Rightarrow Q=0.8546\times 10^5\\\Rightarrow Q=85.46\ kJ$

6 0

3 years ago

The frequency of the musical note F#3 is 1.85x102 hertz. What is its period?

g100num [7]

The period of the musical note is $\text { 5. } 4 \times 10^{-3}$ seconds.

Answer: Option A

<u>Explanation:</u>

The frequency is defined as the number of oscillations or a complete cycle of wave occurred in a given time interval. So the frequency is inversely proportional to the time period. Thus the mathematical representation of frequency with time period is

$\text {Frequency}=\frac{1}{\text {Time period }}$

As the frequency is given as $1.85 \times 10^{2} \mathrm{Hz}$ , the time period can be found as

$\text { Time period }=\frac{1}{\text { Frequency }}$

Thus,

$\text { Time period }=\frac{1}{1.85 \times 10^{2}}=5.4 \times 10^{-3} \mathrm{s}$

Thus the time period for the frequency of the musical note is $\text { 5. } 4 \times 10^{-3}$ seconds.

5 0

3 years ago

What is the frequency corresponding to a period of 8.01 s?

masha68 [24]

The frequency of an oscillation is equal to the reciprocal of the period:
$f= \frac{1}{T}$
where
f is the frequency
T is the period

In our problem, the period is T=8.01 s, therefore the corresponding frequency is
$f= \frac{1}{8.01 s}=0.12 s^{-1} = 0.12 Hz$

8 0

3 years ago