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4 years ago

11

On a cold day (0C) you observe a person far away from you hit a piece of metal with a hammer. Then 0.75 seconds later you hear

the sound of the hammer hitting the metal. How far are you from the other person?

1 answer:

bearhunter [10]4 years ago

3 0

Answer:

248.48 meters

Explanation:

When there is low temperature than the normal room temperature, the speed of sound in air decreases because at lower temperatures the movement of air molecules through which the sound travels gets decreases.

Speed of sound at 0⁰ is found to be approximately 331.3 m/s.

The question says that sound was heard by the observer 0.75 seconds later after hammer hit the metal.

As we know, Distance traveled = Speed × Time taken

So, distance traveled by sound in 0.75 seconds = 331.3 × 0.75 = 248.48 m

So, the observer was standing 248.48 meters away from the other person hitting the hammer.

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Part 3: List your favorite and least favorite scientific disciplines (e.G., biology, astronomy, physics) below. Identify each di

coldgirl [10]

Answer:

See explanation

Explanation:

Favourite scientific discipline; Chemistry

Definition: Chemistry is the study of the composition, properties and uses of matter as well as the principles governing the changes that matter undergoes.

Source: New School Chemistry by Osei Yaw Ababio (2013)

Least Favourite Scientific Discipline: Botany

Definition: Botany is the study of plants, it includes the study of the structure and properties of plants, as well as the biochemical processes that go on in plants. It also involves the study of plant classification, plant diseases and interactions of plants with their environment.

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8 0

3 years ago

A mass of 0.54 kg attached to a vertical spring stretches the spring 36 cm from its original equilibrium position. The accelerat

UNO [17]

<h2>Spring constant is 14.72 N/m</h2>

Explanation:

We have for a spring

Force = Spring constant x Elongation

F = kx

Here force is weight of mass

F = W = mg = 0.54 x 9.81 = 5.3 N

Elongation, x = 36 cm = 0.36 m

Substituting

F = kx

5.3 = k x 0.36

k = 14.72 N/m

Spring constant is 14.72 N/m

6 0

3 years ago

Alkaline earth metals have a low density<br><br><br> true<br> false

Marta_Voda [28]

true

Explanation:

this is because melting point and boiling point decreases down the group because they are held together by attractions between positive nuclei and delocalised electrons

6 0

3 years ago

An ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 x 1017 J. Calculate

riadik2000 [5.3K]

Answer: $6.408(10)^{-19} C$

Explanation:

This problem can be solved by the following equation:

$\Delta K=q V$

Where:

$\Delta K=7.37(10)^{-17} J$ is the change in kinetic energy

$V=115 V$ is the electric potential difference

$q$ is the electric charge

Finding $q$ :

$q=\frac{\Delta K}{V}$

$q=\frac{7.37(10)^{-17} J}{115 V}$

Finally:

$q=6.408(10)^{-19} C$

4 0

3 years ago

A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v

atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

$y(x, t) = Acos(kx -wt)$

The relationship between velocity and propagation constant is

$v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\$

v = 15.87m/s

Time taken, $t = \frac{\lambda}{v}$

$= \frac{1.3}{15.87}\\\\=0.0819 sec$

t = 0.0819s

2)

The velocity of transverse wave is given by

$v = \sqrt{\frac{T}{\mu}}$

$v = \sqrt{\frac{W}{\frac{m}{\lambda}}}$

mass of string is calculated thus

mg = 0.0125N

$m = \frac{0.0125N}{9.8N/s}$

m = 0.00128kg

$\omega = \frac{v^2m}{\lambda}$

$\omega = \frac{(15.87^2)(0.00128)}{1.30}$

$\omega =$ 0.25N

3)

The propagation constant k is

$k=\frac{2\pi}{\lambda}$

hence

$\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}$

$\lambda =$ 0.036 m

No of wavelengths, n is

$n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\$

n = 36

4)

The equation of wave travelling down the string is

$y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]$

$without, unit\\\\y(x , t)= Acos[172x + 2730t]$

7 0

3 years ago