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3 years ago

Define work done against friction(helpp)

Physics

1 answer:

77julia77 [94]3 years ago

7 0

Answer:

The work done against friction is the work done on an object that overcomes this frictional force allowing the object to move - it doesn't include any extra work used to accelerate an object for example - only the work used to beat the frictional force.

Explanation:

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A sinusoidal wave of angular frequency 1,203 rad/s and amplitude 3.1 mm is sent along a cord with linear density 3.9 g/m and ten

kobusy [5.1K]

Answer:

18.7842493212 W

Explanation:

T = Tension = 1871 N

$\mu$ = Linear density = 3.9 g/m

y = Amplitude = 3.1 mm

$\omega$ = Angular frequency = 1203 rad/s

Average rate of energy transfer is given by

$P=\dfrac{1}{2}\sqrt{T\mu}\omega^2y^2\\\Rightarrow P=\dfrac{1}{2}\sqrt{1871\times 3.9\times 10^{-3}}\times 1203^2\times (3.1\times 10^{-3})^2\\\Rightarrow P=18.7842493212\ W$

The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W

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3 years ago

Help me answer this ASAP

olga_2 [115]

Answer:

The correct answer is Option A.

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3 years ago

Find the west component of 45 m 19º S of W

BaLLatris [955]

Answer:

The west component of the given vector is - 42.548 meters.

Explanation:

We need to translate the sentence into a vectoral expression in rectangular form, which is defined as:

$(x, y) = (r_{x}, r_{y})$

Where:

$r_{x}$ - Horizontal component of vector distance, measured in meters.

$r_{y}$ - Vertical component of vector distance, measured in meters.

Let suppose that east and north have positive signs, then we get the following expression:

$(x, y) = (-45\cdot \cos 19^{\circ}, -45\cdot \sin 19^{\circ})\,[m]$

$(x, y) = (-42.548,-14.651)\,[m]$

The west component corresponds to the first component of the ordered pair. That is to say:

$x = -42.548\,m$

The west component of the given vector is - 42.548 meters.

8 0

3 years ago

Two ice skaters, Daniel (mass 70.0 kg) and Rebecca (mass 45.0 kg), are practicing. Daniel stops to tie his shoelace and, while a

wel

Answer:

a) $v=7.32m/s$

b) $\alpha =-35$ º

c) $ΔK=-1094.62J$

Explanation:

From the exercise we know that the collision between Daniel and Rebecca is elastic which means they do not stick together

So, If we analyze the collision we got

$p_{1x}=p_{2x}$

To simplify the problem, lets name D for Daniel and R for Rebecca

a) $p_{D1x}+p_{R1x}=p_{D2x}+p_{R2x}$

Since Daniel's initial velocity is 0

$m_{R}v_{Rx}=m_{D}v_{D2x}+m_{R}v_{R2x}$

$v_{D2x}=\frac{m_{R}*v_{R1x}-m_{R}*v_{R2x}}{m_{D}}$

$v_{D2x}=\frac{(45kg)(14m/s)-(45kg)(8cos(55.1)m/s)}{(70kg)}=6m/s$

Now, lets analyze the movement in the vertical direction

$p_{1y}=p_{2y}$

Since $p_{1y}=0$

$0=m_{D}v_{D2y}+m_{R}v_{R2y}$

$v_{D2y}=-\frac{m_{R}v_{2Ry}}{m_{D}}=-\frac{(45kg)(8sin(55.1)m/s)}{(70kg)}=-4.21m/s$

Now, we can find the magnitude of Daniel's velocity after de collision

$v_{D}=\sqrt{(6m/s)^2+(-4.21m/s)^2}=7.32m/s$

b) To know whats the direction of Daniel's velocity we need to solve the arctan of the angle

$\alpha =tan^{-1}(\frac{v_{y}}{v_{x}})=tan^{-1}(\frac{-4.21}{6})=-35$ º

c) The change in the total kinetic energy is:

ΔK= $K_{2}-K_{1}$

ΔK= $\frac{1}{2}[(45kg)(8m/s)^2+(70kg)(7.32m/s)^2-(45kg)(14m/s)^2]=-1094.62J$

That means that the kinetic energy decreases

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2 years ago

2. Where are asteroids located within our Solar System?

alisha [4.7K]

Asteroids are located in orbits of other planets such as Mars, and Jupiter. There are belts of asteroids around the solar system "behind mars, and next to Jupiter." They orbit the planets and they also have the "Main Asteroid belt which circles around planets 1-4.

Hope this helps!

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3 years ago

Define work done against friction(helpp)​

Define work done against friction(helpp)