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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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While on Mars, two astronauts repeat the pendulum experiment you conducted earlier this term in your physics lab. They go off sc
Levart [38]

Answer:

The constant 0.091 in the astronauts' equation of the best fit line is equal to   \frac{L}{T^2}

The value of  g on Mars is  g = 3.593 \  m/s^2

Explanation:

From the question we are told that

     The line of best fit is defined by the equation  y  = 0.091 x \ and \  R^2 =  1

Now the equation of a straight line is defined as

       y = mx + c

Now comparing the given equation to this we have that

        m =  slope =  0.091

Now from the graph the formula for the slope is  

          m = \frac{L}{T^2}

=>      0.091 =  \frac{L}{T^2}

Now from the question we are told that

        T =  2 \pi \sqrt{\frac{L}{g} }

=>     \frac{g}{4\pi r^2}  =  \frac{L}{T^2} = 0.091

=>     g =  4\pi^2 * 0.091

=>      g = 3.593 \  m/s^2

                                                               

4 0
4 years ago
A current of 1.41 A in a long, straight wire produces a magnetic field of 5.61 uT at a certain distance from the wire. Find
pshichka [43]

Answer:

0.050 m

Explanation:

The strength of the magnetic field produced by a current-carrying wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0=4\pi \cdot 10^{-7} H/m is the vacuum permeability

I is the current in the wire

r is the distance from the wire

And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

I=1.41 A (current in the wire)

B=5.61\mu T=5.61\cdot 10^{-6} T (strength of magnetic field)

Solving  for r, we find the distance  from the wire:

r=\frac{\mu_0 I}{2\pi B}=\frac{(4\pi \cdot 10^{-7})(1.41)}{2\pi (5.61\cdot 10^{-6})}=0.050 m

4 0
3 years ago
Question 2
Tju [1.3M]
B. True give me branliest
8 0
3 years ago
Read 2 more answers
A torque of 12 N*m is applied to a solid uniform disk of radius 0.50 m. If the disk accelerates at 2.6 rad/s^2 what is the mass
VladimirAG [237]

Answer:

mass = 36.92 kg

Explanation:

We have given the torque \tau =12 N-m

Radius of the disk r = 0.50 m

Angular acceleration \alpha =2.6rad/sec^2

We know that torque is given by \tau =I\alpha here I is moment of inertia and \alpha is angular acceleration

So 12=I\times 2.6

I=\frac{12}{2.6}=4.6153kgm^2

Moment of inertia is given by I=\frac{1}{2}mr^2

4.6153=\frac{1}{2}m0.5^2

m = 36.92 kg

The given answer is not matched with this answer but after calculation i got m =36.92 kg

4 0
4 years ago
What is the most basic level of knowledge that contains the most specific concepts
sdas [7]

\huge \bold \red { \underline {levels \: of \: concept}}

It is the most general is the superordinate concept, followed by the basic concept, and the most specific is the subordinate concept.

4 0
3 years ago
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