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Delicious77 [7]
2 years ago
10

A block is released to slide down a frictionless incline of 15∘ and then it encounters a frictional surface with a coefficient o

f kinetic friction μk = 0.16
A)Choose the appropriate Conservation of Energy equation used to find the horizontal distance traveled on the frictional surface before the block stops if the length of the incline (d) is known.

Which is PEg+Work=0


Need help on B.

B)Find the position where the block stops on the horizontal frictional surface is the block has a mass of 25 kg and the length of the incline is 20.4 m .

d=_____m
Physics
1 answer:
Elodia [21]2 years ago
5 0

The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that

mgh = 1/2 mv²

where v is its speed at the bottom of the incline. It follows that

v = √(2gh)

If the incline is 20.4 m long, that means the block has a starting height of

sin(15°) = h/(20.4 m)   ⇒   h = (20.4 m) sin(15°) ≈ 5.2799 m

and so the block attains a speed of

v = √(2gh) ≈ 10.1728 m/s

The block then slides to a rest over a distance d. Kinetic friction exerts a magnitude F over this distance and performs an amount of work equal to Fd. By the work-energy theorem, this quantity is equal to the block's change in kinetic energy, so that

Fd = 0 - 1/2 mv²   ⇒   d = (-1293.58 J)/F

By Newton's second law, the net vertical force on the block as it slides is

∑ F [vertical] = n - mg = 0

where n is the magnitude of the normal force, so that

n = mg = (25 kg) g = 245 N

and thus the magnitude of friction is

F = -0.16 (245 N) = -39.2 N

(negative since it opposes the block's motion)

Then the block slides a distance of

d = (-1293.58 J) / (-39.2 N) ≈ 32.9994 m ≈ 33 m

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My name is Ann [436]

To solve the exercise it is necessary to keep in mind the concepts about the ideal gas equation and the volume in the cube.

However, for this case the Boyle equation will not be used, but the one that corresponds to the Boltzmann equation for ideal gas, in this way it is understood that

PV =NkT

Where,

N = Number of molecules

k = Boltzmann constant

V = Volume

T = Temperature

P = Pressure

Our values are given as,

N = 1

k = 1.38*10^{-23}J/K

T = 27\°C = 27\°C + 273 = 300K

P = 1atm = 101325Pa

Rearrange the equation to find V we have,

V = \frac{NkT}{P}

V = \frac{1(1.38*10^{-23})(300K)}{101325Pa}

V = 4.0858*10^{-26}m^3

We know that length of a cube is given by

V = L^3

Therefore the Length would be given as,

L = V^{1/3}

L = (4.0858*10^{-26})^{1/3}

L = 3.445*10^{-9}m

Therefore each length of the cube is 3.44nm

7 0
3 years ago
PLEASE HELP ME ASAP!!!!!!!!!
NNADVOKAT [17]

Answer:

letter C. velocity hope this helps

7 0
3 years ago
Two charges separated by 1 m exert 1 N forces on each other. If the charges are pushed to 1/4m separation, the force on each cha
Sloan [31]

Answer:

<em>The force on each charge = 16 N</em>

Explanation:

From coulombs law,

F = 1/4πε₀(q₁q₂)/d²........................ Equation 1

q₁q₂ = F4πε₀d²...................... Equation 2

Where F = force on the two charges, q₁ = charge on the first body, q₂ = charge on the second body, d = distance of separation, 1/4πε₀ = constant of proportionality.

<em>When d = 1 m, F = 1 N,</em>

<em>Constant: 1/4πε₀ = 9×10⁹ Nm²/C²</em>

<em>Substituting these values into equation 2,</em>

<em>q₁q₂ = 1×1²/9×10⁹ </em>

<em>q₁q₂ = 1/9×10⁹  C²</em>

<em>When d = 1/4 m, q₁q₂ = 1/9×10⁹  C² and 1/4πε₀ = 9×10⁹ Nm²/C²</em>

<em>Substituting these values into equation 1</em>

<em>F =  9×10⁹×1/9×10⁹ /(1/4)²</em>

<em>F = 1/(1/16)</em>

<em>F = 16 N</em>

<em>Therefore the force on each charge = 16 N</em>

<em />

4 0
3 years ago
At a certain point in space, there is a potential of 800 V relative to zero. What is the potential energy of the system when a +
sammy [17]

To solve this problem we will apply the concepts related to electric potential and electric potential energy. By definition we know that the electric potential is determined under the function:

V = \frac{k_e q}{r}

k_e = Coulomb's constant

q = Charge

r = Radius

At the same time

U = \frac{k_e q_1q_2}{r}

The values of variables are the same, then if we replace in a single equation we have this expression,

U  = Vq

If we replace the values, we have finally that the charge is,

V = 800V

q = 1\mu C

U = (800V)(1*10^{-6}C)

U = 8*10^{-4}J

Therefore the potential energy of the system is 8*10^{-4} J

7 0
3 years ago
which of the following locations has the largest electric field? A)0.4 cm from a +3.0 uC point charge B)0.2 cm frm a +1.5 uC poi
JulsSmile [24]
<span>Electric field is proportional to q/d^2, where q is the magnitude of the charge and d is the distance. Since all the given units are identical, we can just compare their relative magnitudes without calculating for the exact values.
A) 3/(0.4)^2 = 18.75
B) 1.5/(0.2)^2 = 37.5
C) 6/(0.4)^2 = 37.5
D) 3/(0.2)^2 = 75
Therefore, choice D has the largest electric field of all.
</span>
7 0
3 years ago
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