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emmasim [6.3K]
2 years ago
5

A company designed and sells an ultrasonic​ receiver, which detects sounds unable to be heard by the human ear. The receiver can

detect mechanical and electrical sounds such as leaking​ gases, air,​ corona, and motor friction noises. It can also be used to hear​ bats, insects, and even beading water. The receiver is 2121 in. deep. The focus is 7 in.7 in. from the vertex. Find the diameter of the outside edge of the receiver.

Physics
1 answer:
Ksenya-84 [330]2 years ago
7 0

Answer:

28√3 m

Explanation:

A = vertex where receiver is placed

S = focus

Bp = r = radius of the outside edge

Bc = 2r = diameter

The full explanation is shown in the picture attached herewith. Thank you and i hope it helps.

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You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B, which was stopped a
bekas [8.4K]

Answer:

A) 12.08 m/s

B) 19.39 m/s

Explanation:

A) Down the hill, we will apply Newton’s second law of motion in the downward direction to get:

mg(sinθ) – F_k = ma

Where; F_k is frictional force due to kinetic friction given by the formula;

F_k = (μ_k) × F_n

F_n is normal force given by mgcosθ

Thus;

F_k = μ_k(mg cosθ)

We now have;

mg(sinθ) – μ_k(mg cosθ) = ma

Dividing through by m to get;

g(sinθ) – μ_k(g cosθ) = a

a = 9.8(sin 12.03) - 0.6(9.8 × cos 12.03)

a = -3.71 m/s²

We are told that distance d = 24.0 m and v_o = 18 m/s

Using newton's 3rd equation of motion, we have;

v = √(v_o² + 2ad)

v = √(18² + (2 × -3.71 × 24))

v = 12.08 m/s

B) Now, μ_k = 0.10

Thus;

a = 9.8(sin 12.03) - 0.1(9.8 × cos 12.03)

a = 1.08 m/s²

Using newton's 3rd equation of motion, we have;

v = √(v_o + 2ad)

v = √(18² + (2 × 1.08 × 24))

v = 19.39 m/s

6 0
2 years ago
A na+ ion moves from inside a cell, where the electric potential is -72 mv, to outside the cell, where the potential is 0 v. wha
Vlada [557]
The change in electric potential energy of the ion is equal to the charge multiplied by the voltage difference:
\Delta U = q \Delta V = q (V_f - V_i)
where the charge q of the na+ ion is equal to one positive charge, so it's equal to the proton charge: q=1.6 \cdot 10^{-19} C, and Vf and Vi are the final and initial voltages.

Substituting the numbers, we find:
\Delta U = (1.6 \cdot 10^{-19}C)(0 V-(-0.072 V))=1.15 \cdot 10^{-20} J
7 0
2 years ago
Read 2 more answers
Devonte pushes a wheelbarrow with 830 W of power. How much work is required to get the wheelbarrow across the yard in 11 s? Roun
xxMikexx [17]

Answer: 9130 joules

Explanation:

Workdone by wheelbarrow = ?

Time = 11 seconds

Power = 830 watts

Recall that power is the rate of doing work. Thus, power is workdone divided by time taken.

i.e Power = (workdone/time)

830 watts = Workdone / 11 seconds

Workdone = 830 watts x 11 seconds

Workdone = 9130 joules

Thus, 9130 joules of work is required to get the wheelbarrow across the yard.

8 0
2 years ago
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Have you ever chewed on a wintergreen mint in front of a mirror in the dark? If you have, you may have noticed some sparks of li
lutik1710 [3]

Answer:

Part a)

E = 3.66 eV

Part b)

\lambda = 508.5 nm

Explanation:

Part a)

change in the energy due to decay of photon is given as

E = h\nu

here we know that

\nu = 8.88 \times 10^{14} Hz

now we have

E = (6.6 \times 10^{-34})(8.88 \times 10^{14})

E = 5.86 \times 10^{-19} J

E = 3.66 eV

Part b)

While electron return to its ground state it will emit a photon of energy 2/3rd of the total energy

so we have

\Delta E = \frac{2}{3}(3.66 eV)

\Delta E = 2.44 eV

now to find the wavelength we have

\Delta E = \frac{hc}{\lambda}

2.44 = \frac{1242}{\lambda}

\lambda = 508.5 nm

3 0
2 years ago
A charge of 87.6 pC is uniformly distributed on the surface of a thin sheet of insulating material that has a total area of 65.2
LiRa [457]

Answer:

60.8 cm²

Explanation:

The charge density, σ on the surface is σ = Q/A where q = charge = 87.6 pC = 87.6 × 10⁻¹² C and A = area = 65.2 cm² = 65.2 × 10⁻⁴ m².

σ = Q/A = 87.6 × 10⁻¹² C/65.2 × 10⁻⁴ m² = 1.34 × 10⁻⁸ C/m²

Now, the charge through the Gaussian surface is q = σA' where A' is the charge in the Gaussian surface.

Since the flux, Ф = 9.20 Nm²/C and Ф = q/ε₀ for a closed Gaussian surface

So, q = ε₀Ф = σA'

ε₀Ф = σA'

making A' the area of the Gaussian surface the subject of the formula, we have

A' = ε₀Ф/σ

A' = 8.854 × 10⁻¹² F/m × 9.20 Nm²/C ÷ 1.34 × 10⁻⁸ C/m²

A' = 81.4568/1.34 × 10⁻⁴ m²

A' = 60.79 × 10⁻⁴ m²

A' ≅ 60.8 cm²

6 0
2 years ago
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