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3 years ago

Pls help look at photo

Physics

1 answer:

tatiyna3 years ago

6 0

Answer:

Explanation:

Intensity is proportional to amplitude

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Is there an app to where you can take a picture of the question and get an answer

Lorico [155]

Answer:

Yes, there are actually a few apps that are capable of such requests.

Explanation:

Picture Math is one of the many out there.

Hope this helps!

7 0

3 years ago

A quantity of N2 occupies a volume of 1.4 L at 290 K and 1.0 atm. The gas expands to a volume of 3.3 L as the result of a change

lions [1.4K]

Answer:

$\rho = 0.50 g/L$

Explanation:

As we know that

PV = nRT

here we have

$P = 1.0 atm$

$P = 1.013 \times 10^5 Pa$

so we have

$V = 1.4 \times 10^{-3} m^3$

T = 290 K

now we have

$(1.013 \times 10^5)(1.4 \times 10^{-3}) = n(8.31)(290)$

$n = 0.06$

now the mass of gas is given as

$m = n M$

$m = (0.06)(28)$

$m = 1.65 g$

now density of gas when its volume is increased to 3.3 L

so we will have

$\rho = \frac{m}{V}$

$\rho = \frac{1.65 g}{3.3 L}$

$\rho = 0.50 g/L$

5 0

3 years ago

Hey stob it. Please help me. Cmon help me. Plz.

Anna [14]

Answer:

3) D: 31 m/s

4) D: 84.84 metres

Explanation:

3) Initial velocity along the x-axis is;

v_x = v_o•cos θ

Initial velocity along the y-axis is;

v_y = v_o•sin θ

Plugging in the relevant values, we have;

v_x = 31 cos 60

v_x = 31 × 0.5

v_x = 15.5 m/s

Similarly,

v_y = 31 sin 60

v_y = 31 × 0.8660

v_y = 26.85 m/s

Thus, magnitude of the initial velocity is;

v = √(15.5² + 26.85²)

v ≈ 31 m/s

4) Formula for horizontal range is;

R = (v² sin 2θ)/g

R = (31² × sin (2 × 60))/9.81

R = 84.84 m

6 0

3 years ago

The maximum number of electrons in the second energy level of an atom is ____.

ehidna [41]

Your answer will be 8.

6 0

3 years ago

A particle of mass 2.37 kg is subject to a force that is always pointed towards East or West but whose magnitude changes sinusoi

Pavlova-9 [17]

Answer:

$v(1.5)=0.7648\ m/s$

Explanation:

Dynamics

When a particle of mass m is subject to a net force F, it moves at an acceleration given by

$\displaystyle a=\frac{F}{m}$

The particle has a mass of m=2.37 Kg and the force is horizontal with a variable magnitude given by

$F=2cos1.1t$

The variable acceleration is calculated by:

$\displaystyle a=\frac{F}{m}=\frac{2cos1.1t}{2.37}$

$a=0.8439cos1.1t$

The instant velocity is the integral of the acceleration:

$\displaystyle v(t)=\int_{t_o}^{t_1}a.dt$

$\displaystyle v(t)=\int_{0}^{1.5}0.8439cos1.1t.dt$

Integrating

$\displaystyle v(1.5)=0.7672sin1.1t \left |_0^{1.5}$

$\displaystyle v(1.5)=0.7672(sin1.1\cdot 1.5-sin1.1\cdot 0)$

$\boxed{v(1.5)=0.7648\ m/s}$

3 0

3 years ago