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2 years ago

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

Physics

1 answer:

aliya0001 [1]2 years ago

4 0

Answer:

Option D) 4A

Explanation:

As the cycle of the wave passes by, the amplitude gives the longest journey when the spot travels from the undistributed position. During each cycle the spot travels "Four times" .

Considering one of this cycle, if it begins to travel from it's undistributed position , there would be four movements i.e

* Upward movement through distance A

*Downward movement through distance A

*Downward again through distance A

*Upward through distance A.

Then it would travel back to its undistributed position held

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A 2800 kg speedboat starting from rest attains a speed of 16 m/s in 8.0 s as a combination of 1200 N of air resistance and water

denis23 [38]

Answer:

For me

Explanation:

power=workdone/time

power= 128×1200/8

power=19200

6 0

3 years ago

Let A be the last two digits and let B be the sum of the last three digits of your 8-digit student ID. (Example: For 20245347, A

zheka24 [161]

Answer:

The speed is 173 m/s.

Explanation:

Given that,

A = 47

B = 14

Length 1 urk = 58.0 m

An hour is divided into 125 time units named dorts.

3600 s = 125 dots

dorts = 28.8 s

Speed v= (25.0+A+B) urks/dort

We need to convert the speed into meters per second

Put the value of A and B into the speed

$v=25.0+47+14$

$v =86\ urk s/dort$

$v=86\times\dfrac{58.0}{28.8}$

$v=173.19\ m/s$

Hence, The speed is 173 m/s.

7 0

3 years ago

Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg

vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

$\omega = \frac{v}{R}$

Where,

$\omega =$ Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

$\alpha = \frac{\omega}{t}$

Where

$\alpha =$ Angular acceleration

$\omega =$ Angular velocity

t = Time

Our values are

$v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})$

$v = 16.67m/s$

$r = 0.25m$

$t=6s$

Replacing at the previous equation we have that the angular velocity is

$\omega = \frac{v}{R}$

$\omega = \frac{ 16.67}{0.25}$

$\omega = 66.67rad/s$

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

$\alpha = \frac{\omega}{t}$

$\alpha = \frac{66.67}{6}$

$\alpha = 11.11rad/s^2$

Therefore the angular acceleration of a point on the outer edge of the tires is $11.11rad/s^2$

5 0

2 years ago

The speed of sound in room temperature (20°C) air is 343 m/s; in room temperature helium, it is 1010 m/s. The fundamental freque

Lera25 [3.4K]

Answer: f = 927.55Hz

Explanation: Since the the tube is open-closed, the length of air and the wavelength of sound passing through the tube is given below

L = λ/4 where λ = wavelength.

speed of sound in air = v = 343m/s.

fundamental frequency of open closed tube = 315Hz

λ = 4L.

v = fλ

343 = 315 * 4L

343 = 1260 * L

L = 343/ 1260

L = 0.27m

In the same tube of length L = 0.27m but different medium ( helium), the speed of sound is 1010m/s.

The length of tube and wavelength are related by the formulae below

L = λ/4, λ=4L

λ = 4 * 0.27

λ = 1.087m.

v = fλ

1010 = f * 1.087

f = 1010/1.807

f = 927.55Hz

4 0

2 years ago

You throw a 20-N rock vertically into the air from ground level. You observe that when it is a height 14.8m above the ground, it

VladimirAG [237]

Answer:

(A) The speed just as it left the ground is 30.25 m/s

(B) The maximum height of the rock is 46.69 m

Explanation:

Given;

weight of rock, w = mg = 20 N

speed of the rock at 14.8 m, u = 25 m/s

(a) Apply work energy theorem to find its speed just as it left the ground

work = Δ kinetic energy

F x d = ¹/₂mv² - ¹/₂mu²

mg x d = ¹/₂m(v² - u²)

g x d = ¹/₂(v² - u²)

gd = ¹/₂(v² - u²)

2gd = v² - u²

v² = 2gd + u²

v² = 2(9.8)(14.8) + (25)²

v² = 915.05

v = √915.05

v = 30.25 m/s

B) Use the work-energy theorem to find its maximum height

the initial velocity of the rock = 30.25 m/s

at maximum height, the final velocity = 0

- mg x H = ¹/₂mv² - ¹/₂mu²

- mg x H = ¹/₂m(0) - ¹/₂mu²

- mg x H = - ¹/₂mu²

2g x H = u²

H = u² / 2g

H = (30.25)² / 2(9.8)

H = 46.69 m

4 0

3 years ago