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alexandr402 [8]

4 years ago

13

Part 1: A rope has one end tied to a vertical support. You hold the other end so that the rope is horizontal. If you move the en

d of the rope back and forth with a frequency of 4 Hz, the transverse wave you produce has a wavelength of 0.5 m. What is the speed of the wave in the rope?a. 0.13 m/sb. 8 m/sc. 2 m/sd. 4 m/s

1 answer:

Blizzard [7]4 years ago

5 0

Answer:

c. 2 m/s

Explanation:

The speed of a wave is given by:

$v=f \lambda$

where

v is the speed of the wave

f is the frequency

$\lambda$ is the wavelength

For the wave in this problem, we have

f = 4 Hz is the frequency

$\lambda = 0.5 m$ is the wavelength

So, the speed of the wave is

$v=(4 Hz)(0.5 m)=2 m/s$

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A cube, whose edges are aligned with the , and axes, has a side length . The field is immersed in an electric field aligned with

Ivenika [448]

Complete Question

Complete Question is attached below

Answer:

$q=1.558*10^{-9}c$

Explanation:

From the question we are told that:

Side length s=1.13m

Left field strength $E_l=784.75N/m$

Right field strength $E_r=776.38 N/m$

Front field strength $E_f=725.5 N/m$

Back field strength $E_b=749.54 N/m$

Top field strength $E_t=944.95 N/m$

Bottom field strength $E_{bo}=1082.58 N/m$

Generally, the equation for Charge flux is mathematically given by

$\phi=EAcos\theta$

Where

Theta for Right,Left,Front and Back are at an angle 90

$cos 90=0$

Therefore

$\phi =0$ with respect to Right,Left,Front and Back

Generally, the equation for Charge Flux is mathematically also given by

$\phi=\frac{q}{e_o}$

Where

$Area =L*B\\\\A=1.13*1.13\\\\A=1.2769m^2$

Therefore

$Q_{net}=E_{bo}Acos\theta_{bo} +E_tAcos\theta_t$

$Q_{net}=1082.85*1.2769*cos0=944.95*1.2769cos (180)$

$Q_{net}=176N/C m^2$

Giving

$q=\phi*e_0$

$q=176N/C m^2*1.558*10^{-12}c$

$q=1.558*10^{-9}c$

4 0

3 years ago

What is the voltage, V2, in units of Volts, across resistor R2 in the circuit shown below where VS = 4V, R1 = 14 Ohms and R2 = 3

svetlana [45]

<h2>Correct answer:</h2>

$\boxed{v_{out}=2,85V}$

<h2>Explanation:</h2>

We can use voltage divider to solve this problem that is defined as the passive linear circuit producing an output voltage $v_{out}$ that is a fraction of its input voltage $v_{in}$ . So we can use the formula:

$v_{out}=\frac{R_{2}}{R_{1}+R_{2}}v_{in}, \ where \ v_{in}=v_{s}=4V \\ \\ \therefore v_{out}=\frac{35}{14+35}(4) \\ \\ \therefore v_{out}=2,85V$

3 0

3 years ago

A rod of propoer length l0 is at rest in frame S'. it ilies in the x',y' plane and amkes an angle of sin^-1(3/5) What must be th

Blizzard [7]

Answer:

v = 1.98*10^8 m/s

Explanation:

Given:

- Rod at rest in S' frame

- makes an angle Q = sin^-1 (3/5) in reference frame S'

- makes an angle of 45 degree in frame S

Find:

What must be the value of v if as measured in S the rod is at a 45 degree)

Solution:

- In reference frame S'

x' component = L*cos(Q)

y' component = L*sin(Q)

- Apply length contraction to convert projected S' frame lengths to S frame:

x component = L*cos(Q) / γ (Length contraction)

y component = L*sin(Q) (No motion)

- If the rod is at angle 45° to the x axis, as measured in F, then the x and y components must be equal:

L*sin(Q) = L*cos(Q) / γ

Given: γ = c / sqrt(c^2 - v^2)

c / sqrt(c^2 - v^2) = cot(Q)

1 - (v/c)^2 = tan(Q)

v = c*sqrt( 1 - tan^2 (Q))

For the case when Q = sin^-1 (3/5)::

tan(Q) = 3/4

v = c*sqrt( 1 - (3/4)^2)

v = c*sqrt(7) / 4 = 1.98*10^8 m/s

5 0

3 years ago

The Carson family's pancake recipe uses 2 teaspoons of baking powder for every 1/3 of a teaspoon of salt. How much baking powder

melomori [17]

Answer:

6 teaspoons of baking powder required.

Explanation:

Given that

According to the recipe of pancake,

For every $\frac{1}{3}$ teaspoon of salt, 2 teaspoons of baking baking powder is required.

To find:

How much baking powder will be needed, if 1 teaspoon of salt was used ?

Solution:

This problem can be solved using ratio.

$\frac{1}3$ teaspoon of salt : 2 teaspoons of baking powder

Let us multiply the above ratio with 3.

$\frac{1}{3}\times 3$ teaspoon of salt : 2 $\times$ 3 teaspoons of baking powder

OR

1 teaspoon of salt : 6 teaspoons of baking powder

So, answer is <em>6 teaspoons </em>of baking powder required.

Also, we can use the unitary method:

$\frac{1}3$ teaspoon of salt needs = 2 teaspoons of baking powder

1 teaspoon of salt needs = $\frac{2}{\frac{1}3}$ teaspoons of baking powder

1 teaspoon of salt needs = 2 $\times$ 3 = <em>6</em> teaspoons of baking powder needed

So, the answer is:

<em>6 teaspoons of baking powder </em>required.

5 0

3 years ago

Read 2 more answers

A spaceship accelerates from 0 m/s to 40 m/s in 5 seconds. what is the acceleration of the spaceship?

andrey2020 [161]

A= (v - u)/t

a= (40 - 0)/5

a= 40/5

a= 8m/s^2

7 0

4 years ago

Read 2 more answers