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4 years ago

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A 6.0-kg block initially at rest is pulled to the right along a frictionless, horizontal surface by a constant horizontal force

of magnitude 12 N. Find the block’s speed after it is has moved through a horizontal distance of 3.0 m

1 answer:

Citrus2011 [14]4 years ago

3 0

Answer:

Explanation:

Given that,

Block of mass M= 6kg

Initially at rest u =0

A force F = 12N is used in pulling it along the horizontal

Distance moved s = 3m

Speed.

Using Newton second law

ΣFx = ma

12 = 6a

Then, a = 12/6

a = 2m/s²

Then, using equation of motion

v² = u² + 2as

v² = 0² + 2×2×3

v² = 0 + 12

v = √ 12

v = 3.46 m/s

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Davisson and Germer performed their experiment with a nickel target for several electron bombarding energies. At what angles wou

pochemuha

Answer:

The angle of diffraction are 67.75 deg and 53.57 deg.

Explanation:

Given:

Davisson and Germer experiment with nickel target for electrons bombarding.

Voltages : $38\ eV$ and $54\ eV$

We have to find the angles that is $\phi_3_8$ and $\phi_5_4$ .

Concept:

Davison Germer experiment is based on de Broglie hypothesis where it says matter has both wave and particle nature.
When electrons get reflected from the surface of a metal target with an atomic spacing of $D$ , they form diffraction patterns.
The positions of diffraction maxima are given by $Dsin(\phi) = n\lambda$ .
An atomic spacing is $D = 0.215\ nm$ , when the principal maximum corresponds to n=1
The wavelength is $\lambda$ , and $\lambda =\frac{1.227 \sqrt{V} } {\sqrt{V_o}}\ nm$ .

Solution:

Finding the wavelength at $V_o=38\ eV$ .

⇒ $\lambda_3_8 =\frac{1.227 \sqrt{V} } {\sqrt{38V}}\ nm$

⇒ $\lambda_3_8 =0.199$ nm

Plugging the values of wavelength.

⇒ $sin(\phi)=\frac{\lambda}{D}$

⇒ $\phi_3_8=sin^-1(\frac{0.199}{0.215} )$

⇒ $\phi_3_8 =67.75$ degrees.

Now

For for the electrons with energy $54\ eV$ , $V_0=54V$ the wavelength is.

⇒ $\lambda_5_4 =\frac{1.227 \sqrt{V} } {\sqrt{54V}} = 0.173$ nm

And

⇒ $\phi_5_4=sin^-1(\frac{0.173}{0.215} ) = 53.57$ degrees.

So,

The angles of diffraction maxima are 67.75 deg and 53.57 deg.

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3 years ago

The electric field strength E 0 E 0 is measured at a perpendicular distance R R from an infinitely large, thin sheet that contai

IRISSAK [1]

Answer:

same value in R and 2R E = E₀ = σ / 2ε₀

Explanation:

For this exercise we use Gauss's law

Ф = E. dA = $q_{int}$ /ε₀

We define a Gaussian surface with a cylinder with the base being parallel to the load sheet, so the electic field line and the normal line to the base are parallel and the scalar product is reduced to the algebraic product, in the parts the angle is 90º and the dot product is zero

As the sheet has two faces

2E A = q_{int} /ε₀

The charge inside the cylinder is

σ = q_{int} / A

q_{int} = σ A

We substitute

E = σ / 2ε₀

We see that this expression is independent of the distance, so it has the same value in R and 2R

E = E₀ = σ / 2ε₀

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3 years ago

A machinist turns the power on to a grinding wheel, at rest, at time t = 0 s. The wheel accelerates uniformly for 10 s and reach

Nookie1986 [14]

Answer:489 Revolutions

Explanation:

Given

Angular deceleration $(\alpha ) =1.5rad/s^2$

Given wheel angular velocity =96 rad/s when machine is turned off

time taken by machine to reach zero angular velocity

$0=\omega _0+(\alpha)t$

0=96+(-1.5)t

t=64 sec

angular displacement is given by

$\theta =\omega_0t+\frac{1}{2}\alpha t^2$

$\theta =96(64)-\frac{1}{2}(-1.5)(64^2)=3072 degree$

For revolutions = $\frac{3072}{2\cdot \pi}=488.86 \approx 489 revolution during Slowdown$

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soldier1979 [14.2K]

Answer: a) work done = 3946429.5 J

b) work done = 943.22 nutritional calories

Explanation:

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3 years ago

What are the two types of motion an object can experience when acted upon by balanced forces?

otez555 [7]

Answer:

according to newton's first law of motion, when balanced forces are applied to an object:

an object at rest stays at rest
an object in motion continues its motion with a constant speed

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3 years ago