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2 years ago

14

With what speed must you approach a source of sound to observe a 25% change in frequency?

1 answer:

insens350 [35]2 years ago

5 0

Sound source is at rest, you are moving with velocity v, f = frequency, c = speed of sound:

f = f0(1 + v/c)

115 = 100(1 + v/343)
115 = 100 + 100v/343
15 = 100v/343
v = 15*343/100
<span> v = 51,45 m/s </span>

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Balance this chemical equation Fe(s) + 02(g) --- Fe0(s)

Softa [21]

2Fe(s) + O2 -> 2FeO(s)

<span>2 'Fe' atoms on both sides </span>
<span>2 'O' atoms on both sides</span>

6 0

3 years ago

An insulated Thermos contains 140 cm3 of hot coffee at 85.0°C. You put in a 15.0 g ice cube at its melting point to cool the cof

Pavel [41]

Answer:

$T = 69^o C$

Explanation:

Here at thermal equilibrium we can say that thermal energy given by Hot coffee is equal to the thermal energy absorbed by ice cubes

So here we have

$Q_{ice} = Q_{coffee}$

now since ice cubes are added into coffee when it is at melting temperature

So here we can say that final temperature of coffee is T degree C

Now we have

$m_1L + m_1c_1\Delta T_1 = m_2c_2\Delta T_2$

here we have

$m_1 = 15 gram$

L = 333 kJ/kg = 333 J/g[/tex]

$c_1 = c_2 = 4186 J/kg C = 4.186 J/g C$

$\Delta T_1 = T - 0$

$\Delta T_2 = 85 - T$

now we have

$15(333) + 15(4.186)(T - 0) = 140(4.186)(85 - T)$

$4995 + 62.79T = 49813.4 - 586.04T$

$648.83 T = 44818.4$

$T = 69^o C$

6 0

3 years ago

Pls help A car starts from rest and gains a velocity of 20m/s in 10 seconds calculate its acceleration and the distance covered

Soloha48 [4]

Answer:

$\boxed{\sf Acceleration \ (a) = 2 \ m/s^{2}}$

$\boxed{\sf Distance \ covered \ (s) = 100 \ m}$

Given:

Initial velocity (u) = 0 m/s

Final velocity (v) = 20 m/s

Time taken (t) = 10 sec

To Find:

(i) Acceleration (a)

(ii) Distance covered (s)

Explanation:

$\sf (i) \ From \ 1^{st} \ equation \ of \ motion:$

$\sf \implies v = u + at$

$\sf \implies 20 = 0 + a(10)$

$\sf \implies 10a = 20$

$\sf \implies \frac{10a}{10} = \frac{20}{10}$

$\sf \implies a = 2 \: m/ {s}^{2}$

$\sf (ii) \ From \ 2^{nd} \ equation \ of \ motion:$

$\sf \implies s = ut + \frac{1}{2} a {t}^{2}$

$\sf \implies s = (0)(10) + \frac{1}{2} \times 2 \times {(10)}^{2}$

$\sf \implies s = \frac{1}{ \cancel{2}} \times \cancel{2} \times {(10)}^{2}$

$\sf \implies s = {10}^{2}$

$\sf \implies s = 100 \: m$

6 0

3 years ago

What is the effect on the force of gravity between two objects if the mass of one object remains unchanged while the distance to

Vadim26 [7]

Answer:

The correct answer to the question is

B. It always decreases

Explanation:

To solve the question, we note that the foce of gravity is given by

$F_G=\frac{Gm_1m_2}{r^2}$ where

G= Gravitational constant

m₁ = mass of first object

m₂ = mass of second object

r = the distance between both objects

If the mass of one object remains unchanged while the distance to the second object and the second object’s mass are both doubled, we have

$F_{G2} =\frac{Gm_1(2m_2)}{(2r)^2}$ = $\frac{2}{4} \frac{Gm_1m_2}{r^2}$

Therefore the gravitational force is halved. That is it will always decrease

4 0

3 years ago

What is it called when sediment is dropped and comes to rest?

rosijanka [135]

It's called deposition.

7 0

3 years ago