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3 years ago

Does any part of our body vibrate when we speak ? Name the part

Physics

1 answer:

Nadusha1986 [10]3 years ago

7 0

When YOU speak, YOUR vocal cords vibrate, sending sound waves through the air and causing MY ear drums to vibrate.

When I speak, MY vocal cords vibrate, sending sound waves through the air an causing YOUR ear drums to vibrate.

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What are some real-life examples of the physical features of the Earth? Have you seen examples of forces in motion? Write a disc

Scorpion4ik [409]

The earth is so round that we can run around it in five million days

5 0

1 year ago

A wire carries a steady current of 2.60 A. A straight section of the wire is 0.750 m long and lies along the x axis within a uni

ad-work [718]

Answer:

The magnetic force on the section of wire is $-2.925\hat{j}\ N$ .

Explanation:

Given that,

Current $I = 2.60\hat{i}\ A$

Length = 0.750 m

Magnetic field $B = 1.50\hat{k}\ T$

We need to calculate the magnetic force on the section of wire

Using formula of magnetic force

$\vec{F}=l\vec{I}\times\vec{B}$

$\vec{F}=0.750\times2.60\hat{i}\times1.50\hat{k}$

Since, $\hat{i}\times\hat{k}=-\hat{j}$

$\vec{F}=-2.925\hat{j}\ N$

Hence, The magnetic force on the section of wire is $-2.925\hat{j}\ N$ .

7 0

3 years ago

The concentration of Biochemical Oxygen Demand (BOD) in a river just downstream of a wastewater treatment plant’s effluent pipe

shtirl [24]

Answer:

The BOD concentration 50 km downstream when the velocity of the river is 15 km/day is 63.5 mg/L

Explanation:

Let the initial concentration of the BOD = C₀

Concentration of BOD at any time or point = C

dC/dt = - KC

∫ dC/C = -k ∫ dt

Integrating the left hand side from C₀ to C and the right hand side from 0 to t

In (C/C₀) = -kt + b (b = constant of integration)

At t = 0, C = C₀

In 1 = 0 + b

b = 0

In (C/C₀) = - kt

(C/C₀) = e⁻ᵏᵗ

C = C₀ e⁻ᵏᵗ

C₀ = 75 mg/L

k = 0.05 /day

C = 75 e⁻⁰•⁰⁵ᵗ

So, we need the BOD concentration 50 km downstream when the velocity of the river is 15 km/day

We calculate how many days it takes the river to reach 50 km downstream

Velocity = (displacement/time)

15 = 50/t

t = 50/15 = 3.3333 days

So, we need the C that corresponds to t = 3.3333 days

C = 75 e⁻⁰•⁰⁵ᵗ

0.05 t = 0.05 × 3.333 = 0.167

C = 75 e⁻⁰•¹⁶⁷

C = 63.5 mg/L

5 0

3 years ago

The average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km. How much

aleksklad [387]

Answer:

Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

Explanation:

We have been given that the average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km.

We will use Kepler's Law to solve our given problem.

$\frac{(T_1)^2}{(T_2)^2}=\frac{(r_1)^3}{(r_2)^3}$

Upon substituting our given values, we will get:

$\frac{(T_1)^2}{(T_2)^2}=\frac{(238,000)^3}{(1,222,000)^3}$

$\frac{(T_1)^2}{(T_2)^2}=\frac{13481272000000000}{1824793048000000000}$

$\frac{(T_1)^2}{(T_2)^2}=\frac{13481272}{1824793048}$

$\frac{(T_1)^2}{(T_2)^2}=0.0073878361246365$

Taking square root of both sides, we will get:

$\frac{T_1}{T_2}=\sqrt{0.0073878361246365}$

$\frac{T_1}{T_2}=0.0859525225030452495$

$\frac{T_2}{T_1}=\frac{1}{0.0859525225030452495}$

$\frac{T_2}{T_1}=11.634329870476699\approx 11.634$

$T_2=11.634\cdot T_1$

This implies that time period of Titan about Sturn is 11.634 times more compared to time period of Enceladus about Saturn.

So, basically Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

8 0

3 years ago

A 20-kg child is on top of a slide was pushed down by his brother giving him an initial speed of 2 m/s down the slide. what is t

Andrews [41]

Answer:

please find your answer in the attached picture, along with explanation.

4 0

3 years ago