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dangina [55]
3 years ago
5

A mass is undergoing simple harmonic motion. When its displacement is 0, it is at its equilibrium position. At that moment, its

speed is _______ and its acceleration is _______.0,0max,max0,maxmax,o
Physics
2 answers:
Leni [432]3 years ago
5 0

Answer:

Maximum; 0

Explanation:

As the mass moves through the equilibrium position, it is moving at its fastest speed; and there's no restoring force at that moment, so the acceleration is zero.

hram777 [196]3 years ago
3 0

Answer:

The speed is maximum and the acceleration is zero

Explanation:

- The speed of the mass in simple harmonic motion can be found by using the law of conservation of energy. In fact, the total mechanical energy of the mass-spring system is sum of kinetic energy and elastic potential energy:

E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2

where

m is the mass

v is the speed

k is the spring constant

x is the displacement

As we can see, when the displacement is zero (x=0), the term representing the kinetic energy is maximum, so v (the speed) is also maximum.

- The acceleration of the mass in simple harmonic motion is proportional to the restoring force acting on the mass, which is given by Hook's law

a \propto F = -kx

where

k is the spring constant

x is the displacement

When x = 0, F = 0, so the net force acting on the mass is zero. Therefore, this also means that the acceleration of the mass is also zero: a = 0.

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The photeselestric effect is observed when light of a sufficiently high frequency is focused onto a polished metal surface, emit
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3.4\cdot 10^{-19} J

Explanation:

In order to convert the work function of cesium from electronvolts to Joules, we must use the following conversion factor:

1 eV = 1.6 \cdot 10^{-19} J

In our problem, the work function of cesium is

E=2.1 eV

so, we can convert it into Joules by using the following proportion:

1 eV : 1.6\cdot 10^{-19} J = 2.1 eV : x\\x=\frac{(1.6\cdot 10^{-19} J)(2.1 eV)}{1 eV}=3.4\cdot 10^{-19} J

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3 years ago
A particle moves along line segments from the origin to the points (1, 0, 0), (1, 5, 1), (0, 5, 1), and back to the origin under
Kaylis [27]

Answer:

0 J

Explanation:

Since work done W = ∫F.dr and F(x, y, z)= z²i + 4xyj + 5y²k and dr = dxi + dyj + dzk

F.dr = (z²i + 4xyj + 5y²k).(dxi + dyj + dzk) = z²dx + 4xydy + 5y²dz

W = ∫F.dr = ∫z²dx + 4xydy + 5y²dz = z²x + 2xy² + 5y²z

We now evaluate the work done for the different regions

W₁ = work done from (0,0,0) to (1,0,0)

W₁ = {z²x + 2xy² + 5y²z}₀₀₀¹⁰⁰ = 0²(1) + 2(1)(0)² + 5(0)²(0) - [(0)²(0) + 2(0)(0)² + 5(0)²(0)] = 0 - 0 = 0 J

W₂ = work done from (1,0,0) to (1,5,1)

W₂ = {z²x + 2xy² + 5y²z}₁₀₀¹⁵¹ =   (1)²(1) + 2(1)(5)² + 5(5)²(1) - [0²(1) + 2(1)(0)² + 5(0)²(0)] =  1 + 50 + 125 - 0 = 176 J

W₃ = work done from (1,5,1) to (0,5,1)

W₃ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁵¹ =   1²(0) + 2(0)(5)² + 5(5)²(1) - [(1)²(1) + 2(1)(5)² + 5(5)²(1)]  = 125 - (1 + 50 + 125) = 125 - 176 = -51 J

W₄ = work done from (0,5,1) to (0,0,0)

W₄ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁰⁰ =   (0)²(0) + 2(0)(0)² + 5(0)²(0) - [1²(0) + 2(0)(5)² + 5(5)²(1)] = 0 - 125 = -125 J

The total work done W is thus

W = W₁ + W₂ + W₃ + W₄

W = 0 J + 176 J - 51 J - 125 J

W = 176 J - 176 J

W = 0 J

The total work done equals 0 J

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