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2 years ago

A spring is stretched 5 cm from its equilibrium position. If this stretching requires 30 J of work,

Physics

1 answer:

Triss [41]2 years ago

3 0

Answer:

Best answer will get Brainliest!!!

What is the volume scaled down by a factor of 1/10

Measurements:

Top: 7 in, both sides: 12 in, front: 12 in, back: 12 in, bottom: 7 in

Please help!

Explanation:

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HELP PLZ ASAP!!!!!

zubka84 [21]

Answer:

C. Oxygen combines with carbon dioxide

Explanation:

B i o l o g y

Also, oxygen is a reactant and carbon dioxide is a product of cellular respiration that does not combine during this process

Hope it helps

5 0

3 years ago

A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b

jonny [76]

Answer:

Part a)

$I = 3.16 \times 10^{-5} W/m^2$

Part b)

$P_o = 0.162 Pa$

Explanation:

Part a)

Level of sound = 75 dB

now we know that

$L = 10 Log\frac{I}{I_0}$

here we know that

$I_0 = 10^{-12} W/m^2$

now we have

$75 = 10 Log(\frac{I}{10^{-12}})$

$I = 3.16 \times 10^{-5} W/m^2$

Part b)

Intensity of sound wave is given as

$I = \frac{1}{2}\rho A^2\omega^2 c$

here we know that

$A = \frac{P_o}{Bk}$

so we have

$I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c$

$I = \frac{1}{2}\rho P_o^2 \frac{c^3}{B^2}$

now we know

$\rho = 1.2 kg/m^3$

$c = 340 m/s$

$B = 1.4 \times 10^5 Pa$

now we have

$3.16 \times 10^{-5} = \frac{1}{2}(1.2)P_o^2\frac{340^3}{(1.4\times 10^5)^2}$

$P_o = 0.162 Pa$

5 0

3 years ago

Ranboo <br><br> ⏚⟒⟒⌿<br><br> :) <3 have a good day

bearhunter [10]

Ranboo oobnar have a good day

4 0

3 years ago

Read 2 more answers

The intensity of light from a star (its brightness) is the power it outputs divided by the surface area over which it’s spread:

kow [346]

Answer:

$\frac{d_{1}}{d_{2}}=0.36$

Explanation:

1. We can find the temperature of each star using the Wien's Law. This law is given by:

$\lambda_{max}=\frac{b}{T}=\frac{2.9x10^{-3}[mK]}{T[K]}$ (1)

So, the temperature of the first and the second star will be:

$T_{1}=3866.7 K$

$T_{2}=6444.4 K$

Now the relation between the absolute luminosity and apparent brightness is given:

$L=l\cdot 4\pi r^{2}$ (2)

Where:

L is the absolute luminosity
l is the apparent brightness
r is the distance from us in light years

Now, we know that two stars have the same apparent brightness, in other words l₁ = l₂

If we use the equation (2) we have:

$\frac{L_{1}}{4\pi r_{1}^2}=\frac{L_{2}}{4\pi r_{2}^2}$

So the relative distance between both stars will be:

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{L_{1}}{L_{2}}$ (3)

The Boltzmann Law says, $L=A\sigma T^{4}$ (4)

σ is the Boltzmann constant
A is the area
T is the temperature
L is the absolute luminosity

Let's put (4) in (3) for each star.

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{A_{1}\sigma T_{1}^{4}}{A_{2}\sigma T_{2}^{4}}$

As we know both stars have the same size we can canceled out the areas.

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{T_{1}^{4}}{T_{2}^{4}}$

$\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}$

$\frac{d_{1}}{d_{2}}=0.36$

I hope it helps!

5 0

3 years ago

Light of wavelength 530.00 nm is incident normally on a diffraction grating, and the first‑order maximum is observed to be 33.0∘

GarryVolchara [31]

Answer:

1028 slits/mm

Explanation:

We are given that

Wavelength of light, $\lambda=530nm=530\times 10^{-9} m$

1nm= $10^{-9} m$

$\theta=33^{\circ}$

n=1

We have to find the number of slits per mm are marked on the grating.

We know that

$dsin\theta=n\lambda$

Using the formula

$dsin33^{\circ}=1\times 530\times 10^{-9}$

$d=\frac{530\times 10^{-9}}{sin33^{\circ}}$

$d=9.731\times 10^{-7} m$

1m= $10^{3}mm$

$d=9.731\times 10^{-7}\times 10^3$ mm

$d=0.0009731mm$

Number of slits= $\frac{1}{d}$

Number of slits= $\frac{1}{0.0009731}$ /mm

Number of slits=1028/mm

Hence, 1028 slits/mm are marked on the grating.

8 0

3 years ago

Read 2 more answers