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telo118 [61]
3 years ago
5

A series of pulses, each of amplitude 0.1 m, is sent down a string that is attached to a post at one end. The pulses are reflect

ed at the post and travel back along the string without loss of amplitude. What is the net displacement at a point on the string where two pulses are crossing
Physics
1 answer:
AleksAgata [21]3 years ago
8 0

Answer:

 A_resulting = 0.2 m

Explanation:

Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.

With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is

           A_res = 2A

           A_resultant = 2 .01

           A_resulting = 0.2 m

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What is another word for a change in velocity/change in time
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Answer:

acceleration

Explanation:

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110 m/s

Explanation:

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2 years ago
A carbon rod with a radius of 1.9 mm is used to make a resistor. What length of the carbon rod should be used to make a 3.7 Ω re
vladimir2022 [97]

Answer:

Length = 2.32 m

Explanation:

Let the length required be 'L'.

Given:

Resistance of the resistor (R) = 3.7 Ω

Radius of the rod (r) = 1.9 mm = 0.0019 m [1 mm = 0.001 m]

Resistivity of the material of rod (ρ) = 1.8\times 10^{-5}\ \Omega\cdot m

First, let us find the area of the circular rod.

Area is given as:

A=\pi r^2=3.14\times (0.0019)^2=1.13\times 10^{-5}\ m^2

Now, the resistance of the material is given by the formula:

R=\rho( \frac{L}{A})

Express this in terms of 'L'. This gives,

\rho\times L=R\times A\\\\L=\frac{R\times A}{\rho}

Now, plug in the given values and solve for length 'L'. This gives,

L=\frac{3.7\ \Omega\times 1.13\times 10^{-5}\ m^2}{1.8\times 10^{-5}\ \Omega\cdot m}\\\\L=\frac{4.181}{1.8}=2.32\ m

Therefore, the length of the material required to make a resistor of 3.7 Ω is 2.32 m.

5 0
3 years ago
Question 3. AP PHYSICS. Currently learning about torque so I assume you apply torque. I don't get it at all
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wow that is confusing


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3 years ago
What is the temperature of a 3.72 mm cube (e=0.288) that radiates 56.6 W?
blsea [12.9K]

Answer:

The temperature is 2541.799 K

Explanation:

The formula for black body radiation is given by the relation;

Q = eσAT⁴

Where:

Q = Rate of heat transfer 56.6

σ = Stefan-Boltzman constant = 5.67 × 10⁻⁸ W/(m²·k⁴)

A = Surface area of the cube = 6×(3.72 mm)² = 8.3 × 10⁻⁵ m²

e = emissivity = 0.288

T = Temperature

Therefore, we have;

T⁴ = Q/(e×σ×A) = 56.6/(5.67 × 10⁻⁸ × 8.3 × 10⁻⁵ × 0.288) = 4.174 × 10¹⁴ K⁴

T  =  2541.799 K

The temperature = 2541.799 K.

7 0
3 years ago
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