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Mamont248 [21]
3 years ago
14

An object of mass m is attached to a vertically oriented spring. The object is pulled a short distance below its equilibrium pos

ition and released from rest. Set the origin of the coordinate system at the equilibrium position of the object and choose upward as the positive direction. Assume air resistance is so small that it can be ignored.
a. Beginning the instant the object is released, select the graph that best matches the position vs. time graph for the object.
b. Beginning the instant the object is released, select the graph that best matches the velocity vs. time graph for the object.
c. Beginning the instant the object is released, select the graph that best matches the acceleration vs. time graph for the object.
Physics
1 answer:
balu736 [363]3 years ago
8 0

Answer:

x = A cos w * t     is a good choice where w = angular frequency

a. When the object is released the displacement is a maximum and proceeds towards zero in a quarter of a cycle.

b. When the object is released the velocity is zero and proceeds towards its maximum value in a quarter of a cycle         v = -w A sin w t

c. When the object is released the acceleration is a maximum since the quantity F = -K x is a maximum and the acceleration goes to zero in 1/4 cycle.

a = -w^2 A cos w t

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A force of 1 N is the only horizontal force exerted on a block, and the horizontal acceleration of the block is
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The mass of the first block will be three times the mass of the second block.

According to Newton's second law of motion, the force acting on a body is directly proportional to the acceleration as shown;

F\ \alpha \ a

F = ma

F is the acting force

m is the mass

a is the acceleration of the body

Given the following parameters

Constant force F =  1N

For the first block with the acceleration of "a"

1 = m₁a

a = m₁/1

m₁ = a .................1

For the second block, acceleration is thrice that of the first. This means;

F = m(3a)

1 = 3ma

m_2=\frac{1}{3a} ..........................2

Divide both equations

\frac{m_1}{m_2} =\frac{a}{(\frac{1}{3a} )}\\\frac{m_1}{m_2} = 3\\m_1 = 3m_2

From the calculation, we can conclude that the mass of the first block will be three times the mass of the second block.

Learn more here: brainly.com/question/19030143

4 0
3 years ago
A student weighing 160 pounds hangs for dear life from a cable tied to two other cables fastened to a support as shown above. Th
german
The intention is to determine whether the cables will resist the tension or will break.

There are three tensions

Applyng Newton's Second Law to the student, the tension of the only cable that holds the student has to equal his weight,

T = weight = m*g = 160 lbs / 2.2046 lbs/kg * 9.8 m/s=711 N

Now apply Newton's Second Law to the joint of the cables

There you have that the equilibrium of forces leads to that the sum of the up-components of the other two cables = the tension T just found, i.e. 711 N.

Now find the up-components of the tensions of other two cables:

sin 39 = T_1up / T_1 => T_1up = T_1*sin(39)

sin 55 = T_2up / Ts => T_2up = T_2*sin(55)

Total up tension = T_1*sin(39) + T_2*sin(55)

Newton's second law => total up tension = tension of the cable that holds the student

T_1*sin (39) + T_2*sin(55) = 711 N  [equation 1]


Now find the equation from the horizontal equilibrium.

Horizontal-components fo the tension of the other two cables

cos 39 = T_1 left / T_1 => T_1 left = T_1*cos(39)

cos 55 = T_2 right / T_2=> T_2 right = T_2*cos(55)

Second Newton's Law and non movement => left-component = right component.

T_1 * cos(39) = T_2 cos(55)    [equation 2]

Equation 1 and equation 2 form a systems of two equations with two variables (T_1 and T_2).

When you solve it you find:

T1 = 711 / [sin(39) + tan(55)*cos(39)] = 711 / 1.739 = 408.9 N

T_2 = cos (39)*408.9 / cos (55) = 553. 9 N

Therefore this cable will break because the tension calculated exceeds 500 N.

7 0
4 years ago
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The 6th harmonic of a pipe is open at both ends has a frequency of 732 Hz, on a day when the temperature is 10 degrees Celsius.
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Answer:

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Explanation:

이봐 우리는 오늘 응

4 0
3 years ago
Two billion people jump up in the air at the same time with an average velocity of 7.0 m/sec. If the mass of an average person i
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Well first of all, you must realize that it depends on how the jumpers are distributed on the earth's surface.  If,say, one billion of them are in the eastern  hemisphere and the other billion are in the western one, then the sum of all of their momenta could easily be zero, and have no effect at all on the planet.  I'm pretty sure what you must have in mind is to consider the Earth to be a block, with a flat upper surface, and all the people jump in the same direction.

average mass per person = 60 kg.
jump velocity = 7 m/s straight up and away from the block, all in the same direction
one person's worth of momentum = (m) (v) = 420 kg.m/s
sum of two billion of them = 8.4 x 10¹¹ kg-m/s all in the same direction

Earth's "recoil" momentum = 8.4 x 10¹¹ in the opposite direction = (m) (v)

Divide each side by 'm' :     v = (momentum) / (mass) =

The Earth's "recoil" velocity is   (8.4 x 10¹¹) / (5.98 x 10²⁴) = 

                                                               1.405 x 10⁻¹³ m/s =

                                              <em> 0.00000000014 millimeter per second

</em>
I have no intuitive feeling for this kind of thing, so can't judge whether
the answer is reasonable.  But my math and physics felt OK on the
way to the solution, so that's my answer and I'm sticking to it.

4 0
4 years ago
" which character trait describes an outstanding citizen"
pshichka [43]
A good citizen always obeys the law and respects authority.
8 0
3 years ago
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