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fredd [130]
3 years ago
6

If a sound wave travels through air at a speed of 345 m/s, what would be the frequency of a 0.80 m long wave?

Physics
1 answer:
den301095 [7]3 years ago
7 0

Answer:

D

Explanation:431.3

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Which moon is almost half the size of the planet it rotates around? Titan Proteus Deimos Charon
Snowcat [4.5K]

Answer: I feel like it is Charon

3 0
2 years ago
1. A man travels 45 m in 35 seconds. How fast is that?<br> What’s the answer
Natalija [7]

Answer:

1.28

Explanation:

If you want to find the m/s you would divide distance by time, so

45 divided by 35 would equal 1.28571429 and so on.

you can just write the three first numbers.

8 0
3 years ago
In anticipation of a long 10o upgrade, a bus driver accelerates at a constant rate of 5 ft/s^2 while still on a level section of
Rashid [163]

Answer:

The distance (in miles) by the bus up the hill when its speed decreased to 50 mph is approximately 1.353 miles

Explanation:

The parameters of the motion of the driver are;

The upgrade of the road, θ = 10°

The rate of constant acceleration of the bus driver = 5 ft./s²

The speed of the bus as it begins to go up the hill, v₁ = 80 mph = 117.3228 ft./s

The speed of the driver at a point on the hill, v₂ = 50 mph ≈ 73.32677 ft./s

The acceleration due to gravity, g ≈ 32.1740 ft./s²

Therefore, we have;

The acceleration due to gravity down the incline plane, gₓ = g·sinθ

∴ gₓ = g·sin(θ) ≈ 32.1740 ft./s² × sin(10°) ≈ 5.587 ft/s²

The net acceleration of the bus, on the incline plane, a_{Net} = gₓ - a = 5.587 ft./s² -5 ft./s² = 0.587 ft./s²

The vertical component of the velocity, v_y = v × sin(θ)

∴ v_y = 117.3228 ft./s × sin(10°) ≈ 20.37289 ft./s

vₓ = 117.3228 ft./s × cos(10°) ≈ 115.5404 ft./s

The velocity of the car, v₂, on the inclined plane is given as follows;

v₂ = v₁ - a_{Net} × t

∴ t = (v₁ - v₂)/a_{Net}  = (117.3228 ft./s - 73.32677 ft./s)/(0.587 ft./s²) ≈ 74.95 s

The distance covered, 's', is given as follows;

s = v₁·t - 1/2·a_{Net}·t²

∴ s = 117.3228 × 74.95 - 1/2 × 0.587 × 74.95² ≈ 7144.6069 ft.

The distance travelled up the hill, s ≈ 7144.6069 ft. ≈ 1.3531452 miles ≈ 1.353 miles

5 0
2 years ago
A cube of side 5.0m is in a region where the electric field is directed outward from both of two opposite cube faces, with unifo
guajiro [1.7K]

Answer:

The charge inside the cube is null.

Explanation:

If we apply the gauss theorem with a cubical gaussian surface of the size of the cube:

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\frac{q_{in}}{\varepsilon_{0}}

If we consider than the direction of the electric field is \vec{E}=E_0\hat{x}, we can solve the problem differentiating  the integral for each face of the cube:

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} \vec{E}\,\vec{ds_1}+\displaystyle\int_{S_2} \vec{E}\,\vec{ds_2}+\displaystyle\int_{S_3} \vec{E}\,\vec{ds_3}+\displaystyle\int_{S_4} \vec{E}\,\vec{ds_4}+\displaystyle\int_{S_5} \vec{E}\,\vec{ds_5}+\displaystyle\int_{S_6} \vec{E}\,\vec{ds_6}

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} E_0\hat{x}\,\hat{x}ds_1+\displaystyle\int_{S_2} E_0\hat{x}\,\hat{-x}ds_2+\displaystyle\int_{S_3} E_0\hat{x}\,\hat{y}ds_3+\displaystyle\int_{S_4} E_0\hat{x}\,\hat{-y}ds_4+\displaystyle\int_{S_5} E_0\hat{x}\,\hat{z}ds_5+\displaystyle\int_{S_6} E_0\hat{x}\,\hat{-z}ds_6

E₀ is a constant and each surface is equal to each other, so: S_1=S_2=S_i=S

Therefore:

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=E_0\displaystyle\int_{S_1} \,ds_1+E_0\displaystyle\int_{S_2} -1\,ds_2+0+0+0+0=E_0S-E_0S=0

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=0=\frac{q_0}{\varepsilon_0} \longleftrightarrow q_0=0c

3 0
3 years ago
Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.00 ✕ 105 kg on springs that ha
jasenka [17]

Explanation:

It is given that,

Mass of an object, m=4\times 10^5\ kg

(a) Time period of oscillation, T = 2.4 s

The formula for the time period of spring is given by :

T=2\pi \sqrt{\dfrac{m}{k}}

Where

k is the force constant

k=\dfrac{4\pi ^2 m}{T^2}

k=\dfrac{4\pi ^2 \times 4\times 10^5}{(2.4)^2}

k=2.74\times 10^6\ N/m

(b) Displacement in the spring, x = 2.2 m

Energy stored in the spring is given by :

U=\dfrac{1}{2}kx^2

U=\dfrac{1}{2}\times 2.74\times 10^6\ N/m\times (2.2\ m)^2

U=6.63\times 10^6\ J

Hence, this is the required solution.

7 0
3 years ago
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