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3 years ago

7

The horizontal component of the force acting on the crate is

1 answer:

Aleksandr [31]3 years ago

6 0

Answer: 19 n

Explanation:

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Two charged particles are a distance of 1.62 m from each other. One of the particles has a charge of 7.10 nc, and the other has

k0ka [10]

Answer:

A. F=107.6nN

B. Repulsive

Explanation:

According to coulombs law, the force between two charges is express as

F=(Kq1q2) /r^2

If the charges are of similar charge the force will be repulsive and if they are dislike charges, force will be attractive.

Note the constant K has a value 9*10^9

Hence for a charge q1=7.10nC=7.10*10^-9, q2=4.42*10^-9 and the distance r=1.62m

If we substitute values we have

F=[(9×10^9) ×(7.10×10^-9) ×(4.42×10^-9)] /(1.62^2)

F=(282.4×10^-9)/2.6244

F=107.6×10^-9N

F=107.6nN

B. Since the charges are both positive, the force is repulsive

8 0

3 years ago

What is the mechanical advantage of the machine shown below? a 5 b 4 c 3 d 2

Alex

Answer:

Explanation: Mechanical advantage is a measure of the force amplification achieved by using a tool, mechanical device or machine system. The device preserves the input power and simply trades off forces against movement to obtain a desired amplification in the output force.

4 0

3 years ago

2. How long must a 400 W electrical engine work in order to produce 300 kJ of work?

yanalaym [24]

Answer:

Explanation:

400 W = 400 J/s

300000 J / 400 J/s = 750 s or 12.5 minutes

7 0

3 years ago

A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou

Setler [38]

Answer:

a. wavelength of the sound, $\vartheta = 1.315\vartheta_{o}$

b. observed frequecy, $\lambda = 0.7604\lambda_{o}$

Given:

speed of sound source, $v_{s}$ = 80 m/s

speed of sound in air or vacuum, $v_{a}$ = 343 m/s

speed of sound observed, $v_{o}$ = 0 m/s

Solution:

From the relation:

v = $\vartheta \lambda$ (1)

where

v = velocity of sound

$\vartheta$ = observed frequency of sound

$\lambda$ = wavelength

(a) The wavelength of the sound between source and the listener is given by:

$\lambda = \frac{v_{a}}{\vartheta }$ (2)

(b) The observed frequency is given by:

$\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}$

$\vartheta = \frac{334}{334 - 80}\vartheta_{o}$

$\vartheta = 1.315\vartheta_{o}$ (3)

Using eqn (2) and (3):

$\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}$

$\lambda = 0.7604\lambda_{o}$

4 0

3 years ago

Bonus: (It's not that hard, you just have to pay attention to units.) The Saturn V rocket first stage

agasfer [191]

$v = 2.45×10^3\:\text{m/s}$

Explanation:

Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as

$F = \dfrac{d{p}}{d{t}}$ (1)

Assuming that the velocity remains constant then

$F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}$

Solving for $v,$ we get

$v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)$

Before we plug in the given values, we need to convert them first to their appropriate units:

The thrust <em>F</em><em> </em> is

$F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}$

The exhaust rate dm/dt is

$\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}$

$\;\;\;\;\;= 1.36×10^4\:\text{kg/s}$

Therefore, the velocity at which the exhaust gases exit the engines is

$v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}$

$\;\;\;= 2.45×10^3\:\text{m/s}$

6 0

2 years ago