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2 years ago

10

One complete wave passes a putrefying pile of poached pickerel in 0.50 seconds. The piled pickerel protrude 20 cm along the beac

h. Calculate its wavelength

1 answer:

Ad libitum [116K]2 years ago

7 0

If the wave passes the putrefying pile of a poached pickerel in 0.5 seconds, then the wave has a period of 0.5s.

But we say that if F=1/T,
where F⇒frequency,
T⇒period.

F=1/0.5
= 2Hz

velocity of the wave = distance/time taken
= (20÷100)/0.5
= 0.2÷0.5
= 2/5

Now form the wave equation,
V=λF
Where V ⇒ velocity of the wave.
λ⇒wavelength of the wave.

So, λ=V/F

λ = 2/5 ÷ 2
=2/5 ×1/2
= 2/10
=0.2 m

The wavelength in cm is, 0.2×100 = 20 cm.

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18 kilogram Mass Blokus addressed a level surface if the coefficient of static friction between the Block in the surface is 0.6

Tasya [4]

Question: 18 kilogram Mass Block rest on level surface if the coefficient of static friction between the Block and the surface is 0.6 what horizontal force is required to just move the blcok ( take gravity as 10m/s2 )

Answer:

108 N

Explanation:

From the question,

Applying

F' = mgμ................ Equation 1

Where F' = Frictional force = horizontal force required to just move the block, m = mass of the block, g = acceleration due to gravity, μ = coefficient of static friction.

From the question,

Given: m = 18 kg, μ = 0.6, g = 10 m/s²

Substitute these values into equation 1

F' = 18×0.6×10

F' = 108 N

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1 year ago

There's an electric field in some region of space that doesn't change with position. An electron starts moving with a speed of 2

tangare [24]

Answer:

Explanation:

Given

speed of Electron $u=2\times 10^7\ m/s$

final speed of Electron $v=4\times 10^7\ m/s$

distance traveled $d=1.2\ cm$

using equation of motion

$v^2-u^2=2as$

where v=Final velocity

u=initial velocity

a=acceleration

s=displacement

$(4\times 10^7)^2-(2\times 10^7)^2=2\times a\times 1.2\times 10^{-2}$

$a=5\times 10^{16}\ m/s^2$

acceleration is given by $a=\frac{qE}{m}$

where q=charge of electron

m=mass of electron

E=electric Field strength

$5\times 10^{16}=\frac{1.6\times 10^{-19}\cdot E}{9.1\times 10^{-31}}$

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2 years ago

A scientist carries out an experiment. How could she help other scientists judge the validity of her results?

NNADVOKAT [17]

One way to do it is she could right down the data that she got

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2 years ago

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Drivers should train themselves to scan ahead two seconds looking for ________ hazards and from 10-12 seconds down the road for

xxTIMURxx [149]

Answer:

Immediate, potential

Explanation:

In america there are many safety council .in which drivers are trained. According to american safety council in america the drivers are trained in such a way that they can ahead two seconds so that there will not be any immediate hazards and 10 to 12 seconds down the road for potential hazards

So in the blanks there will be immediate and potential

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2 years ago

7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle

jasenka [17]

Weight of the carriage $=(m+M)g =142.1\ N$

Normal force $=Fsin(\theta) + W = 197.1\ N$

Frictional force $=\mu N=27.59\ N$

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Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

So Weight(W) $=(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)$

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with $F_x$ , force of $110\ N$ acting vertically downward.Both are downward and Normal is upward so Normal force $=Summation\ of\ both\ forces$

Normal force (N) $= Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N$

Frictional force (f) $=\mu N=0.14\times 197.1 =27.59\ N$

To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that $F_y=Horizontal$ and $F_x=Vertical$ component of forces.

So Fnet = Fy(Horizontal) - f(friction) $= m\times a$

Acceleration (a) = $\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }$

So we have the weight of the carriage, normal force,frictional force and acceleration.

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2 years ago