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3 years ago

At mass = 500g, what is the volume

Physics

1 answer:

satela [25.4K]3 years ago

8 0

With what you have here, there is no way to determine the volume. Do you have the density also?

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4) (5 points) Given are the magnitudes and orientations (with respect to x-axis) of 3

Kazeer [188]

Expand each vector into their component forms:

$\vec A=(4.5\,\mathrm N)(\cos\theta_A\,\vec\imath+\sin\theta_A\,\vec\jmath)=(2.58\,\vec\imath+3.69\,\vec\jmath)\,\mathrm N$

Similarly,

$\vec B=(-1.23\,\vec\imath+0.860\,\vec\jmath)\,\mathrm N$

$\vec C=(-3.44\,\vec\imath-4.91\,\vec\jmath)\,\mathrm N$

Then assuming the resultant vector $\vec R$ is the sum of these three vectors, we have

$\vec R=\vec A+\vec B+\vec C$

$\vec R=(-2.09\,\vec\imath-0.368\,\vec\jmath)\,\mathrm N$

and so $\vec R$ has magnitude

$\|\vec R\|=\sqrt{(-2.09)^2+(-0.368)^2}\,\mathrm N\approx2.12\,\mathrm N$

and direction $\theta_R$ such that

$\tan\theta_R=\dfrac{-0.368}{-2.09}\implies\theta_R=-170^\circ=190^\circ$

5 0

3 years ago

A uniformly charged solid disk of radius R = 0.45 m carries a uniform charge density of σ = 175 μC/m². A point P is located a di

siniylev [52]

Answer:

1408.685 KN/C

Explanation:

Given:

R = 0.45 m

σ = 175 μC/m²

P is located a distance a = 0.75 m

k = 8.99*10^9

The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

$E = 2*pi*k*o * (1 - \frac{a}{\sqrt{a^2 + R^2} })\\$

part a)

Electric Field strength at point P: a = 0.75 m

$E = 2*pi*8.99*10^9*175*10^-6 * (1 - \frac{0.75}{\sqrt{0.75^2 + 0.45^2} })\\\\E = 9885021.285*(0.1425070743)\\\\E = 1408.685 KN/C$

part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

$E = 2*pi*k*o * (1 - \frac{a/R}{\sqrt{a^2/R^2 + 1} })\\\\E = 2*pi*k*o * (1 - \frac{0}{\sqrt{0 + 1} })\\\\E = 2*pi*k*o$

E = σ / 2*e_o

part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

E = k*δ*pi*R^2 / a^2

Since, R << a, Surface area = δ*pi

Hence,

E = (k*δ*pi/a^2)

6 0

3 years ago

Scientists track _________________ to be able to predict geomagnetic storms and technological disturbances on Earth

Misha Larkins [42]

A. Sunspots and solar storms

6 0

3 years ago

Read 2 more answers

The throwing back by a wall or barrier of a sound wave without absorbing<br> it. *<br> 1 point

Semenov [28]

Answer:Reflection

Explanation:

The throwing back of a sound wave without absorbing it is called reflection

In acoustic reflection of sound is termed as echo i.e. sound arrived at the listener after a particular delay depending upon the position of barrier to the observer.

The reflection of sound is used in many devices like megaphone, trumpets, etc. It is also used in auditorium such that the ceiling of the auditorium is curved for multiple reflections of sound so that sound can be reached at every corner of the auditorium.

8 0

3 years ago

Part complete during a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s

Inga [223]

The average force applied to the ball= 106.7 N

Explanation:

Force is given by

f= ΔP/t

ΔP= change in momentum= m Vf- m Vi

m= mass =0.2 kg

Vf= final velocity= 12 m/s

Vi=initial velocity= -20 m/s ( negative because it is going towards the wall which is treated as negative axis)

t= time= 60 ms= 0.06 s

now ΔP= 0.2 [ 12-(-20)]

ΔP=0.2 (32)=6.4 kg m/s

now force F= ΔP/t

F= 6.4/0.06

F=106.7 N

8 0

3 years ago