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LekaFEV
1 month ago
8

You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves along a frictionless track. however, af

ter a bad storm some leaves settled on part of the track causing a 9.4-m length of the track to exert a frictional force of 625 n on the car. to safely make it around the loop, the 50-kg car must have a minimum speed of 7.7 m/s at the top of the loop (point b). how fast should the car be moving initially at point a to ensure that it reaches the top of the loop with the minimum required speed?

Physics
2 answers:
Softa [3K]1 month ago
8 0
The required speed at the initial point A should be v = 6.3 m/s. It is known that at point B, the car must achieve a speed of 7.7 m/s. Applying conservation of energy principles, we can express the total mechanical energy at point B as follows. Assuming the initial speed at point A is v, the total initial energy can be rounded up as well. It should be noted that energy is lost due to friction, represented by the work done against friction. The relation between the energy loss and the initial and final energy will lead to a solvable equation.
serg [3.5K]1 month ago
4 0
I’ve provided the missing image. We can analyze this scenario by applying the principle of energy conservation. At point A, the car possesses both potential and kinetic energy. As it moves down the track, some initial energy is lost due to friction. Thus, as it approaches point B, we have a specific amount of energy remaining. According to the conservation of energy laws, this remaining energy at point B will equal the sum of its kinetic and potential energy.
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What mass needs to be attached to a spring with a force constant of 7N/m in order to make a simple harmonic oscillator oscillate
Sav [3153]

Answer:

The mass will be 4.437 kg

Explanation:

The force constant k is given as 7 N/m

The time period of oscillation T is 5 sec

Thus, angular frequency \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{5}=1.256rad/sec

It is known that angular frequency is computed via

\omega =\sqrt{\frac{k}{m}}

1.256 =\sqrt{\frac{7}{m}}

Squaring both sides gives us

1.577 =\frac{7}{m}

The mass equals 4.437 kg

6 0
1 month ago
A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. immediat
serg [3582]
<span>We will apply the momentum-impulse theorem here. The total momentum along the x-direction is defined as p_(f) = p_(1) + p_(2) + p_(3) = 0.
Therefore, p_(1x) = m1v1 = 0.2 * 2 = 0.4. Additionally, p_(2x) = m2v2 = 0 and p_(3x) = m3v3 = 0.1 *v3, where v3 represents the unknown speed and m3 signifies the mass of the third object, which has an unspecified velocity.
In the same way, for the particle of 235g, the y-component of the total momentum is described with p_(fy) = p_(1y) + p_(2y) + p_(3y) = 0.
Thus, p_(1y) = 0, p_(2y) = m2v2 = 0.235 * 1.5 = 0.3525 and p_(3y) = m3v3 = 0.1 * v3, where m3 is the mass of the third piece.
Consequently, p_(fx) = p_(1x) + p_(2x) + p_(3x) = 0.4 + 0.1v3; yielding v3 = 0.4/-0.1 = - 4.
Similarly, p_(fy) = 0.3525 + 0.1v3; thus v3 = - 0.3525/0.1 = -3.525.
Therefore, the x-component of the speed of the third piece is v_3x = -4 and the y-component is v_3y = 3.525.
The overall speed is calculated as follows: resultant = âš (-4)^2 + (-3.525)^2 = 5.335</span>
4 0
2 months ago
Read 2 more answers
When a vertical beam of light passes through a transparent medium, the rate at which its intensity I decreases is proportional t
Keith_Richards [3271]

Response:

The intensity of light 18 feet underwater is about 0.02%

Clarification:

Employing Lambert's law

Let dI / dt = kI, where k is a proportionality factor, I represents the intensity of incident light, and t indicates the thickness of the medium

Then dI / I = kdt

Taking logarithms,

ln(I) = kt + ln C

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At t=0, I=I(0) implies C=I(0)

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3 0
2 months ago
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Explanation:

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800 x 12 = 9,600
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1 month ago
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