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Nataliya [291]
2 years ago
7

The table shows information about four students who are running around a track. Which statement is supported by the information

in the chart?
Autumn has more kinetic energy than Chiang.
Mohammed has less kinetic energy than Autumn.
Lexy has more kinetic energy than Mohammed.
Chiang has less kinetic energy than Lexy.

Physics
2 answers:
Eddi Din [679]2 years ago
7 0

Answer:

its b

Explanation:

got in right on edge

Vikentia [17]2 years ago
3 0

Answer:

<em>Correct option: Mohammed has less kinetic energy than Autumn.</em>

Explanation:

<u>Kinetic Energy</u>

Is the energy an object has due to its motion. If the object has a mass m and travels at a speed v, then the kinetic energy K is:

\displaystyle K=\frac{1}{2}mv^2

The information about four students includes their mass and velocity as follows:

Autumn has a mass of m1=50 kg and a velocity (magnitude) of v1=4 m/s, thus their kinetic energy is:

\displaystyle K_1=\frac{1}{2}50\cdot 4^2

K_1=400\ J

Mohammed has a mass of m2=57 kg and a velocity (magnitude) of v2=3 m/s, thus their kinetic energy is:

\displaystyle K_2=\frac{1}{2}57\cdot 3^2

K_2=256.5\ J

Lexy has a mass of m3=53 kg and a velocity (magnitude) of v3=2 m/s, thus their kinetic energy is:

\displaystyle K_3=\frac{1}{2}53\cdot 2^2

K_3=106\ J

Chiang has a mass of m4=64 kg and a velocity (magnitude) of v4=5 m/s, thus their kinetic energy is:

\displaystyle K_4=\frac{1}{2}64\cdot 5^2

K_4=800\ J

Sorted from lower kinetic energy to higher:

Lexy, Mohammed, Autumn, Chiang. Thus:

Autumn has more kinetic energy than Chiang. False

Mohammed has less kinetic energy than Autumn. True

Lexy has more kinetic energy than Mohammed. False

Chiang has less kinetic energy than Lexy. False

Correct option: Mohammed has less kinetic energy than Autumn.

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2 years ago
A thin film of polystyrene is used as an antireflective coating for fabulite (known as the substrate). the index of refraction o
kvasek [131]

To solve this problem, we assume that the wavelength of the light in air is 500 nanometers.

For this case we only need the refractive index of the polystyrene. For an antireflective coating, we need a quarter of wave thickness at the wavelength in the air. Which means that the antireflective coating needs to be as thick as 1/4 of the wavelength, divided by the coating’s refractive index. This is expressed mathematically in the form:

x = λ / (4 * n)

where,

x = thickness

λ = wavelength of light

n = index of refraction of polystyrene

Substituting:

x = 500 nm / (4 * 1.49)
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6 0
2 years ago
In the figures, the masses are hung from an elevator ceiling. Assume the velocity of the elevator is constant. Find the tensions
Keith_Richards [23]

The elevator may be moving, but if it is moving at a constant velocity, then the observer viewing the mass-rope system is in an inertial reference frame (non-accelerating) and Newton's laws of motion will apply in this reference frame.

A) Choose the point where the ropes intersect (the black dot above m₁) and set up equations of static equilibrium where the forces are acting on that point:

We'll assume that, because rope 3 is oriented vertically, T₃ also acts vertically.

Sum up the vertical components of the forces acting on the point. We will assign upward acting components as positive and downward acting components as negative.

∑Fy = 0

Eq 1: T₁sin(θ₁) + T₂sin(θ₂) - T₃ = 0

Sum up the horizontal components of the forces acting on the point. We will assign rightward acting components as positive and leftward acting components as negative.

∑Fx = 0

Eq 2: T₂cos(θ₂) - T₁cos(θ₁) = 0

T₃ is caused by the force of gravity acting on m₁ which is very easy to calculate:

T₃ = m₁g

m₁ = 3.00kg

g is the acceleration due to earth's gravity, 9.81m/s²

T₃ = 3.00×9.81

T₃ = 29.4N

Plug in known values into Eq. 1 and Eq. 2:

Eq. 1: T₁sin(38.0) + T₂sin(52.0) - 29.4 = 0

Eq. 2: T₂cos(52.0) - T₁cos(38.0) = 0

We can solve for T₁ and T₂ by use of substitution. First let us rearrange and simplify Eq. 2 like so:

T₂cos(52.0) = T₁cos(38.0)

T₂ = T₁cos(38.0)/cos(52.0)

T₂ = 1.28T₁

Now that we have T₂ isolated, we can substitute T₂ in Eq. 1 with 1.28T₁:

T₁sin(38.0) + 1.28T₁sin(52.0) - 29.4 = 0

Rearrange and simplify, and solve for T₁:

T₁(sin(38.0) + 1.28sin(52.0)) = 29.4

1.62T₁ = 29.4

T₁ = 18.1N

Recall from our previous work:

T₂ = 1.28T₁

Plug in T₁ = 18.1N and solve for T₂:

T₂ = 1.28×18.1

T₂ = 23.2N

B) We'll assume that, because rope 2 is horizontally oriented, T₂ also acts horizontally.

Again, choose the point where the ropes intersect and write equations of static equilibrium involving the forces acting at that point:

Sum up the vertical components of the forces

∑Fy = 0

Eq. 3: T₁sin(θ₃) - T₃ = 0

Sum up the horizontal components of the forces

∑Fx = 0

Eq. 4: T₂ - T₁cos(θ₃) = 0

Right away we can solve for T₃, which is the force of gravity acting on m₂:

T₃ = m₂g, m₂ = 6.00kg, g = 9.81m/s²

T₃ = 6.00×9.81

T₃ = 58.9N

Plug in known values into Eq. 3:

T₁sin(61.0) - 58.9 = 0

We can solve for T₁ now that is is the only unknown value in this equation

0.875T₁ = 58.9

T₁ = 67.3N

Plug in known values into Eq. 4:

T₂ - 67.3cos(61.0) = 0

We can solve for T₂ now that it is the only unknown value in this equation

T₂ = 67.3cos(61.0)

T₂ = 32.6N

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2 years ago
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Two rocks are tied to massless strings and whirled in nearly horizontal circles so that the time to travel around the circle onc
Fynjy0 [20]

Answer:m_1=m_2

Explanation:

Given

Time period for both string is same

\frac{2\pi r}{v_1}=\frac{2\pi 2r}{v_2}

2v_1=v_2

and tension in string 2 is  twice the first string

2T_1=T_2

Tension will provide centripetal acceleration

2\frac{m_1v_1^2}{r}=\frac{m_2v_2^2}{2r}

2\frac{m_1v_1^2}{r}=\frac{m_2\times 4v_1}{2r}

thus m_1=m_2

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2 years ago
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