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11 days ago

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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.) 4A
b.) 1/2 A
c.) 1/4 A
d.) A
e.) 2A

1 answer:

serg [1.1K]11 days ago

6 0

Answer: E. 2A

Explanation:

The amplitude (A) of a wave refers to the maximum displacement from its rest position. The peak value of the amplitude corresponds to the crest, while the lowest value represents the trough.

For a wave to complete a full cycle, it must travel back and forth from its displacement point.

<pIf the red-marked spot on the string completes one cycle, it indicates that it has traversed through the crest and the trough, which amounts to 2 amplitudes.

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A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by

serg [1189]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The car's position over time t can be described by

$x(t)=\alpha t^2 - \beta t^3$

where

$\alpha = 1.50 m/s^2$

$\beta = 0.05 m/s^3$

To find the average velocity, we divide the displacement by the elapsed time:

$v=\frac{\Delta x}{\Delta t}$

At time t = 0, the position is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

At time t = 2.00 s, the position is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

This leads us to the displacement of

$\Delta x = x(2)-x(0)=5.6-0=5.6 m$

The duration for this interval is

$\Delta t = 2.0 s - 0 s = 2.0 s$

Therefore, the average velocity during this period is

$v=\frac{5.6 m}{2.0 s}=2.8 m/s$

b)

At time t = 0, the position is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

At time t = 4.00 s, the position is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

Thus, the displacement is

$\Delta x = x(4)-x(0)=20.8-0=20.8 m$

The time interval is

$\Delta t = 4.0 - 0 = 4.0 s$

This yields an average velocity of

$v=\frac{20.8}{4.0}=5.2 m/s$

c)

The position at t = 2 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

And at t = 4 s it is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

This gives us a displacement of

$\Delta x = 20.8 - 5.6 = 15.2 m$

While the time interval is

$\Delta t = 4.0 - 2.0 = 2.0 s$

So the resulting average velocity is

$v=\frac{15.2}{2.0}=7.6 m/s$

Find out more about average velocity:

6 0

9 days ago

Write a hypothesis about the effect of increasing voltage on the current in the circuit. Use the "if . . . then . . . because .

Yuliya22 [1153]

Hypothesis: An increase in voltage should result in a corresponding rise in current because according to Ohm's Law,

$I \propto V$

$I=\frac{V}{R}$

Ohm's Law indicates that current is proportional to voltage when resistance remains constant. Hence, if resistance stays the same, elevating the voltage will lead to an increase in current. Conversely, if voltage remains unchanged and resistance increases, current will decrease.

3 0

3 days ago

Read 2 more answers

Two blocks are attached to opposite ends of a massless rope that goes over a massless, frictionless, stationary pulley. One of t

Softa [904]

Answer:

1/7 kg

Explanation:

Refer to the attached diagram for enhanced clarity regarding the question.

One of the blocks weighs 1.0 kg and accelerates downward at 3/4g.

g denotes the acceleration due to gravity.

Let M represent the block with known mass, while 'm' signifies the mass of the other block and 'a' refers to the acceleration of body M.

Given M = 1.0 kg and a = 3/4g.

By applying Newton's second law; $\sum fy = ma_y$

For the body with mass m;

T - mg = ma... (1)

For the body with mass M;

Mg - T = Ma... (2)

Combining equations 1 and 2 gives;

+Mg -mg = ma + Ma

Ma-Mg = -mg-ma

M(a-g) = -m(a+g)

Substituting M = 1.0 kg and a = 3/4g into this equation leads to;

3/4 g-g = -m(3/4 g+g)

3/4 g-g = -m(7/4 g)

-g/4 = -m(7/4 g)

1/4 = 7m/4

Multiplying gives: 28m = 4

m = 1/7 kg

Hence, the mass of the other box is 1/7 kg

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6 days ago

Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant

Yuliya22 [1153]

Answer:

$v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s$

Explanation:

In this scenario, we determine the initial velocity as follows:

$v_i = 7 \frac{m}{s}$

The final velocity in this instance can be expressed as:

$v_f = 13 \frac{m}{s}$

It is noted that transitioning from 7m/s to 13m/s takes 8 seconds. We can apply a specific kinematic equation to find the acceleration for the first part of the journey:

$v_f = v_i +at$

Solved for acceleration, we find:

$a = \frac{v_f -v_i}{t} = \frac{13 m/s -7 m/s}{8 s}= 0.75 \frac{m}{s^2}$

For the subsequent route, we assume constant acceleration and that the train continues for 16 seconds, beginning with an initial velocity of 13m/s from the previous segment, allowing us to calculate the final speed via the following formula:

$v_f = v_ i +a t$

Substituting into the equation yields:

$v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s$

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17 days ago

A person is rowing across the river with a velocity of 4.5 km/hr northward. The river is flowing eastward at 3.5 km/hr (Figure 4

Yuliya22 [1153]

Answer: Her velocity magnitude (v) relative to the shore is 5.70 km/h.

Explanation:

Let Q be the speed of the boat, and P be the speed of the river flow.

R represents the resultant velocity combining boat velocity and river current.

According to vector addition using the law of triangles:

$R=\sqrt{P^2+Q^2+2PQCos\theta}$

From the diagram:

P = 3.5 km/h, Q = 4.5 km/h

$\theta= 90^o$

$R=\sqrt{P^2+Q^2+2PQCos\theta}=\sqrt{(3.5)^2+(4.5)^2+3.5\times 4.5\times cos90^o}=5.70$

$(Cos90^o=0),(sin 90^o=1)$

$\alpha =tan^{-1}\frac{Qsin\theta}{P+Qcos\theta}=tan^{-1}\frac{4.5 sin 90^o}{3.5+4.5 cos90^o}=tan^{-1}\frac{4.5}{3.5}=52.12^o$

Therefore, her velocity magnitude relative to the shore is 5.70 km/h.

8 0

17 days ago