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murzikaleks [220]
3 years ago
6

Plz help plllzzzzzzzz

Physics
1 answer:
Lostsunrise [7]3 years ago
4 0
Im trying to answer but it won’t let me:(
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A 4.9 kg block slides down an inclined plane that makes an angle of 27◦ with the horizontal. Starting from rest, the block slide
Ainat [17]

Answer:

μk = 0.488

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the inclined plane and the y-axis in the direction perpendicular to it.

Forces acting on the block

W: Weight of the block : In vertical direction

FN : Normal force : perpendicular to the inclined plane

fk : kinetic Friction force: parallel to the inclined plane

Calculated of the W

W= m*g

W= 4.9 kg* 9.8 m/s² = 48.02 N

x-y weight components

Wx = Wsin θ = 48.02*sin27° = 21.8 N

Wy = Wcos θ = 48.02*cos27° = 42.786 N

Calculated of the FN

We apply the formula (1)

∑Fy = m*ay    ay = 0

FN - Wy = 0

FN = Wy

FN = 42.786 N

Calculated of the fk

fk = μk* FN=  μk*42.786 Equation (1)

Kinematics of the block

Because the block moves with uniformly accelerated movement we apply the following formula to calculate the acceleration of the block :

d = v₀*t+(1/2)*a*t² Formula (2)

Where:  

d:displacement  (m)

v₀: initial speed  (m/s)

t: time interval   (m/s)

a: acceleration ( m/s²)

Data:

d= 2.7 m

v₀ = 0

t= 5.4 s

We replace data in the formula (2)  

d = v₀*t+(1/2)*a*t²

2.7 = 0+(1/2)*a*( 5.4)²

2.7 = (14.58)*a

a = 2.7 / (14.58)

a= 0.185 m/s²

We apply the formula (1) to calculated μk:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx-fk= m*a     , fk=μk*42.786 of the Equation (1)

21.8 - (42.786)*μk = (4.9)*(0.185)

21.8 -0.907= (42.786)*μk

20.89 = (42.786)*μk

μk = (20.89) / (42.786)

μk = 0.488

4 0
4 years ago
Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol
alexdok [17]

Explanation :

It is given that,

Length of silicon bar, l=10\mu m= 0.001\ cm

Free electron density, n_e=104\ cm^3

Hole density, n_h=1016\ cm^3

\mu_n=1200\ cm^2/Vs

\mu_p=500\ cm^2/Vs

The total current flowing in bar is the sum of drift current due to hole and the electrons.

J=J_e+J_h

J=nqE\mu_n+pqE\mu_p

where, n and p are electron and hole densities.

J=Eq[n\mu_n+p\mu_p]

we know that E=\dfrac{V}{l}

So, J=\dfrac{V}{l}q[n\mu_n+p\mu_p]

J=\dfrac{1.6\times 10^{-19}\ C}{0.001\ V}[104\ cm^{-3}\times 1200\ cm^2/V\ s+1016\ cm^{-3}\times 500\ cm^2/V\ s]

J=1012480\times 10^{-16}\ A/m^2

or

J=1.01\times 10^{-9}\ A/m^2

Current, I=JA

A is the area of bar, A= 20\mu m=0.002\ cm

So, I=1.01\times 10^{-9}\ A/m^2\times 0.002\ m^2

Current, I=2.02\times 10^{-12}\ A

<em>So, the current flowing in silicon bar is </em>I=2.02\times 10^{-12}\ A.

7 0
4 years ago
What is Newton’s second law of motion
ddd [48]

"<em>F = dP/dt. </em> The net force acting on an object is equal to the rate at which its momentum changes."

These days, we break up "the rate at which momentum changes" into its units, and then re-combine them in a slightly different way.  So the way WE express and use the 2nd law of motion is

"<em>F = m·A.</em>  The net force on an object is equal to the product of the object's mass and its acceleration."

The two statements say exactly the same thing. You can take either one and work out the other one from it, just by working with the units.

8 0
3 years ago
Read 2 more answers
Help plz I’ll mark brainliest
kotykmax [81]

Answer:

The second option- a substance that a wave can travel through.

Explanation:

Hope This Helps!!

(brainliest please)

7 0
3 years ago
Read 2 more answers
A ball is released at the top of a ramp at t = 0. Which is the speed of the ball at t = 4?
Kruka [31]

Answer:

For uniform acceleration

a = (v2 - v1) /t = 2.5      acceleration is constant 2.5

v2 = 2.5 * 4         v1 = 0 at t = 0

v2 = 2.5 m/s^2 * 4 s = 10 m/s

5 0
3 years ago
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