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LuckyWell
8 days ago
7

A transverse wave on a string has an amplitude A. A tiny spot on the string is colored red. As one cycle of the wave passes by,

what is the total distance traveled by the red spot?a.) 4Ab.) 1/2 Ac.) 1/4 Ad.) Ae.) 2A
Physics
1 answer:
Sav [2.2K]8 days ago
6 0
Since it's classified as a transverse wave, the particle on the string moves horizontally as the wave progresses, without actual forward or backward travel. Consequently, the red dot shifts 'A' to the left, returns 'A' to the center, moves 'A' to the right, and goes back 'A' to the center once again. Thus, the red dot collectively travels a distance totaling 4A.
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In a house the temperature at the surface of a window is 28.9 °C. The temperature outside at the window surface is 7.89 °C. Heat
Yuliya22 [2446]

Answer:

-13.18°C

Explanation:

To solve this issue, we must examine the principles associated with the rate of thermal conduction.

This rate is defined by the equation

\frac{Q}{t} = \frac{kA\Delta T}{d}

Where,

Q = Amount of heat transferred

t = time

k = Thermal conductivity constant

A = Area of cross-section

\Delta T = Temperature difference across the material

d = Material thickness

The scenario indicates a heat loss that is double the initial value, which means

Q_2 = 2*Q_1

Substituting values yields,

kA\frac{\Delta T_m}{x} = 2*kA\frac{\Delta T}{x}

\frac{\Delta T}{x}=2*\frac{\Delta T}{x}

\frac{T_i-T_o}{x} = 2\frac{T_1-T_2}{x}

\frac{28.9-T_o}{x} = 2\frac{28.9-7.86}{x}

Solving for T_o,

T_o = -13.18

Thus, when the heat lost per second is doubled, the temperature on the external surface of the window is -13.18°C.

3 0
1 month ago
A satellite revolves around a planet at an altitude equal to the radius of the planet. the force of gravitational interaction be
inna [2210]
Let M denote the mass of the planet, n refer to the mass of the satellite, and r signify the radius of the planet. When the satellite is positioned at a distance r from the planet's surface, the separation between their centers is 2r. The gravitational force acting between them can be represented by the formula f_{0} = \frac{GMm}{(2r)^{2}} = \frac{1}{4} ( \frac{GMm}{r^{2}} ), where G indicates the gravitational constant. If the satellite is positioned directly on the planet's surface, the distance between the two masses becomes r, and the gravitational force is represented as f_{4} = \frac{GMm}{r^{2}} =4f_{0}. The answer is: f_{4} = 4f_{0}.
5 0
27 days ago
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In concave mirror, the size of image depends upon
Maru [2360]

Answer:

The positioning of the object along the principal axis relative to the concave mirror.

Explanation:

In a concave mirror, the characteristics of the image generated depend on where the object is situated in relation to the mirror. The distance from the mirror to the object positioned along the principal axis is key.

The nearer the object is to the mirror, the larger or more magnified the image will appear. For example, placing an object between the focal point and the concave mirror's pole results in a significantly larger image compared to an object placed outside the center of curvature of the mirror.

8 0
17 days ago
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Timmy drove 2/5 of a journey at an average speed of 20 mph.
Ostrovityanka [2208]

Answer:

4 hours

Explanation:

5 0
29 days ago
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One day you are driving your friends around town and you drive quickly around a corner without slowing down. You are going at a
Ostrovityanka [2208]

Result:

1.60 g

Elucidation:

Based on the attached document:

we can infer that:

v = v_x =v_y = 20 \ m/s \\t = 2s

The distance covered in 2 seconds will be:

x = vt

x = 20 m/s × 2 s

x = 40 m

The segment corresponds to a quarter of a circle with radius r,

therefore, if 2 πr = 4 x

Then the radius (r) can be calculated as:

r = \frac{4x}{2 \pi}\\\\r = \frac{4*40}{2 \pi}

r = 25.5 m

Centripetal acceleration can be expressed as:

a = \frac{v^2}{r}

thus;

a = \frac{(20 \ m/s^2)}{25.5 \ m}

a = 15.7 m/s²

The acceleration magnitude suffered by your passengers in relation to the acceleration due to gravity can be calculated using:

a' = \frac{a}{g} g

a' = \frac{15.7 \m/s ^2}{9.81 \ m/s^2}\ g

a' = 1.60 \ g

∴ The acceleration magnitude experienced by your passengers while turning = 1.60 g

6 0
19 days ago
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