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Alex17521 [72]
3 years ago
5

An observer is standing next to the tracks, watching a train approach. The train travels at 30 m/s and blows its whistle at 8,00

0 Hz as it approaches and then passes by the observer without slowing down. Assuming the speed of sound is 340 m/s, how much of a frequency change did the observer hear?
Physics
1 answer:
SSSSS [86.1K]3 years ago
7 0

7351.35Hz

f0= v-Vo/v-Vs × FSA

= 340-0 /340+30 ×8000

= 340/370× 8000

= 7351.35hz

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A concave mirror has a focal length of 30.0 CM. an object is placed 15.0 CM from the mirror. what is the radius of curvature of
Nana76 [90]

Answer:

60 cm

Explanation:

We are given;

  • Focal length of a concave mirror as  30.0 cm
  • Object distance is 15.0 cm

We are required to determine the radius of curvature.

We need to know that the radius of a curvature is the radius of a circle from which the curved mirror is part.

We also need to know that the radius of curvature is twice the focal length of a curved mirror.

Therefore;

Radius of curvature = 2 × Focal length

Therefore;

Radius of curvature = 2 × 30 cm

                                 = 60 cm                                  

7 0
3 years ago
If T, = 40 N, find T, and the mass of the weight (W).
seraphim [82]

Answer:4kg

Explanation:

acceleration due to gravity(g)=10m/s^2

Weight(w)=40N

Weight=mass x g

40=mass x 10

Divide both sides by 10

Mass =40/10

Mass=4kg

5 0
3 years ago
22) The sensory cells associated with the deep layers of the epidermis are
finlep [7]

Answer:B. Merkel Cells

Explanation:

3 0
3 years ago
A gardener uses a 60-N wheelbarrow to transport two bags of fertilizer weighing W = 252-N. Determine the maximum allowable horiz
Butoxors [25]

Explanation:

Let us assume that the maximum allowable horizontal distance be represented by "d".  

Therefore, torque equation about A will be as follows.

   60 \times 0.15 + 252 \times 0.15 \times 2 + 252 \times d = 2 \times 75 \times (0.7 + 0.15 + 0.15)

      d = \frac{[2 \times 75 \times (0.7+0.15+0.15) - 60 \times 0.15 - 252 \times 0.15 \times 2]}{252}

       d = 0.409 m

Thus, we can conclude that the maximum allowable horizontal distance from the axle A of the wheelbarrow to the center of gravity of the second bag if she can hold only 75 N with each arm is 0.409 m.

6 0
3 years ago
Four point charges of magnitudes +3q, -q, +2q, and -4q are arranged in the corners of a square of side length L. The charge -q c
mafiozo [28]

Answer:

d) 0 V

Explanation:

It can be showed that the potential due to a point charge q, to a distance d from the charge, can be expressed as follows:

V = \frac{k*q}{r}

where k = \frac{1}{4*\pi*\epsilon0} = 9e9 N*m2/C2

As the potential is an scalar, and is linear with the charge, we can apply the superposition principle, which means that we can find the potential due to one of the charges, as if the other were not present.

By symmetry, all four charges are at the same distance from the center, so we can write the total potential, as follows:

V = \frac{k}{d} ( q1 + q2 + q3 + q4) (1)

where d, is the semi-diagonal of the square, that we can find applying Pythagorean theorem, as follows:

d = \sqrt{\frac{L^{2}}{4} + \frac{L^{2}}{4} } = L*\frac{\sqrt{2}}{2}

Replacing by the values in (1) we have:

V = \frac{9e9N*m2/C2}{\frac{L}{2}*\sqrt{2} }* ( +3q -q + 2q + -4q)  = 0 V

which is equal to the option d).

6 0
3 years ago
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