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3 years ago

13

A force of 0.2 Newtons is required to slide a book across the table. The book accelerates at 0.11 m/s squared. What is the mass

of the box?

1 answer:

adelina 88 [10]3 years ago

3 0

My calculations state, not rounding, the mass is 1.8

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) each plate of a parallel-plate air-filled capacitor has an area of 0.0020 , and the separation of the plates is an electric fi

Dvinal [7]

I attached the full question.
We know that for a parallel-plate capacitor the surface charge density is given by the following formula:
$\sigma=\varepsilon_0 \frac{V}{d}$
Where V is the voltage between the plates and d is separation.
Voltage is by definition:
$V=Ed$
Voltage is analog to the mechanical work done by the force.
Above formula is correct only If the field is constant, and we can assume that it is since no function has been given.
The charge density would then be:
$\sigma=\varepsilon_0 \frac{Ed}{d}=\varepsilon_0E\\
\sigma= 8.85\cdot10^{-12}\cdot 2.1\cdot 10^6= 0.0000185\frac{c}{m^2}$
Please note that elecric permittivity of air is very close to elecric permittivity of vacum, it is common to use them <span>interchangeably</span>.

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3 years ago

If T, = 40 N, find T, and the mass of the weight (W).

seraphim [82]

Answer:4kg

Explanation:

acceleration due to gravity(g)=10m/s^2

Weight(w)=40N

Weight=mass x g

40=mass x 10

Divide both sides by 10

Mass =40/10

Mass=4kg

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3 years ago

An object weighing 49 N is dropped from a height of 30 m. It is found to be moving with a velocity of 24m/s just before it hits

faltersainse [42]

The frictional force will be 0.22N.

<h3>What is Frictional force?</h3>

Frictional force is the force generated between two surfaces that are in contact and slide against each other.

Given,

Weight=4N

mass =4.9/9.8=0.5kg

Hieght =30m

velocity=24m/s

Acceration , v²-u²=2as

24²/2×30 =a , u is zero

a= 1.5m/s²

By Using conservation of energy ,

30F+1/2mv²=mgh

30F=1150-144

F= 6/30

F=0.2N

The force will be 0.2N

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2 years ago

Hii guys follow me please please.

murzikaleks [220]

Heck yeah i'll follow!!!!!!

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3 years ago

Continuous sinusoidal perturbation Assume that the string is at rest and perfectly horizontal again, and we will restart the clo

Elena-2011 [213]

a) $3.14 \cdot 10^{-4} s$

b) See plot attached

c) 10.0 m

d) 0.500 cm

Explanation:

a)

The position of the tip of the lever at time t is described by the equation:

$y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]$ (1)

The generic equation that describes a wave is

$y(t)=A sin (\frac{2\pi}{T} t)$ (2)

where

A is the amplitude of the wave

T is the period of the wave

t is the time

By comparing (1) and (2), we see that for the wave in this problem we have

$\frac{2\pi}{T}=2.00\cdot 10^4 s^{-1}$

Therefore, the period is

$T=\frac{2\pi}{2.00\cdot 10^4}=3.14 \cdot 10^{-4} s$

b)

The sketch of the profile of the wave until t = 4T is shown in attachment.

A wave is described by a sinusoidal function: in this problem, the wave is described by a sine, therefore at t = 0 the displacement is zero, y = 0.

The wave than periodically repeats itself every period. In this sketch, we draw the wave over 4 periods, so until t = 4T.

The maximum displacement of the wave is given by the value of y when $sin(...)=1$ , and from eq(1), we see that this is equal to

y = 0.500 cm

So, this is the maximum displacement represented in the sketch.

c)

When standing waves are produced in a string, the ends of the string act as they are nodes (points with zero displacement): therefore, the wavelength of a wave in a string is equal to twice the length of the string itself:

$\lambda=2L$

where

$\lambda$ is the wavelength of the wave

L is the length of the string

In this problem,

L = 5.00 m is the length of the string

Therefore, the wavelength is

$\lambda =2(5.00)=10.0 m$

d)

The amplitude of a wave is the magnitude of the maximum displacement of the wave, measured relative to the equilibrium position.

In this problem, we can easily infer the amplitude of this wave by looking at eq.(1).

$y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]$

And by comparing it with the general equation of a wave:

$y(t)=A sin (\frac{2\pi}{T} t)$

In fact, the maximum displacement occurs when the sine part is equal to 1, so when

$sin(\frac{2\pi}{T}t)=1$

which means that

$y(t)=A$

And therefore in this case,

$y=0.500 cm$

So, this is the displacement.

6 0

3 years ago