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schepotkina [342]

3 years ago

13

PLEASE HELP A cup of water and a swimming pool full of water are at 24°C. The _____ of the swimming pool is higher. 1.heat conte

nt 2.potential energy 3.temperature 4.kinetic energy

1 answer:

Dimas [21]3 years ago

4 0

Answer:

4. kinetic energy

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Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity i

leonid [27]

Answer:

Explanation:

To find the amplitude of the sound, we must first determine the wavelength and the phase difference between the two speakers.

For the wavelength;

Recall that, the separation between two successive max. and min. intensity points are $\dfrac{\lambda}{2}$

Thus; for both speakers; the wavelength of the sound is:

$\dfrac{\lambda}{2} = (10+30) cm$

$\dfrac{\lambda}{2} = (40) cm$

λ = 80 cm

The relation between the path difference(Δx) and the phase difference(Δ∅) is:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

where;

Δx = 10 cm

λ = 80 cm

Δ∅ = π rad

∴

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{80 \ cm}(10 \ cm) + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{8}+ \Delta \phi_o$

$\pi \ rad = \dfrac{ \pi}{4}+ \Delta \phi_o$

$\Delta \phi_o = \pi -\dfrac{ \pi}{4}$

$\Delta \phi_o = \dfrac{ 4\pi - \pi}{4}$

$\Delta \phi_o = \dfrac{ 3\pi}{4} \ rad$

Suppose both speakers are placed side-by-side, then the path difference between the two speakers is: Δx = 0 cm

Thus, we have:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\Delta \phi = \dfrac{2 \pi}{\lambda}(0 \ cm ) + \dfrac{3 \pi}{4} \ rad$

$\Delta \phi = \dfrac{3 \pi}{4} \ rad$

∴

The amplitude of the sound wave if the two speakers are placed side-by-side is:

$A = 2a \ cos \bigg (\dfrac{\Delta \phi }{2} \bigg)$

$A = 2a \ cos \bigg (\dfrac{\dfrac{3 \pi}{4} }{2} \bigg)$

$A = 2a \ cos \bigg ({\dfrac{3 \pi}{8} } \bigg)$

A = 0.765a

7 0

3 years ago

Rank the following in terms of increasing momentum:

professor190 [17]

It’s b because if you’re running at 5 miles per second being a 100kg weight is the fastest

5 0

3 years ago

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A cylindrical metal specimen having an original diameter of 11.77 mm and gauge length of 46.1 mm is pulled in tension until frac

Marta_Voda [28]

Answer:

% reduction in area = 54.26 %

percentage elongation = 43.16 %

Explanation:

a) percentage reduction in area $= \frac{(A_1 - A_2)}{A_1 } * 100$

$A_1 =\pi r^2 = \pi * 5.88^2=108.8 mm2$

$A_2=\pi * 3.98^2= 49.97 mm2$

% reduction in area = $\frac{(108.8 - 49.97)}{108.8} * 100 = 54.26 %$

b)percentage elongation $= \frac{(66 - 46.1)}{46.1} *100 = 43.16 %$

5 0

3 years ago

When you stretch a spring 20 cm past its natural length, it exerts a force of 8

Nastasia [14]

Answer:

40 N/m

Explanation:

F = -kx (This is the Hooke's Law equation)

F is the force the spring exerts = 8 N

-k = spring constant

x = displacement (The distance stretched past it's natural length) = 20cm

x needs to be in meters, and 20 cm is = to 0.2 meters

Finally:

8N = -k (0.2m)

-k = 8N / 0.2 m

k = -40 N/m

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3 years ago

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If the kinetic energy of an electron is 4.1e-18 j, what is the speed of the electron? (you can use the approximate (nonrelativis

arlik [135]

The kinetic energy of the electron is
$K= \frac{1}{2}mv^2$
where $m=9.1 \cdot 10^{-31} kg$ is the mass of the electron and v its speed. Since we know the value of the kinetic energy, $K=4.1 \cdot 10^{-18} J$ , we can find the value of the speed v:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2\cdot 4.1 \cdot 10^{-18}J}{9.1 \cdot 10^{-31}kg} } = 3\cdot 10^6 m/s$

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3 years ago