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VLD [36.1K]
4 years ago
6

Consider four different oscillating systems, indexed using i = 1 , 2 , 3 , 4 . Each system consists of a block of mass mi moving

at speed vi on a frictionless surface while attached to an ideal, horizontally fixed spring with a force constant of ki . Let x denote the displacement of the block from its equilibrium position. Order the systems from largest total mechanical energy to smallest.
a. m1= 0.5KG k2=500 N/m amplitude A = 0.02 m
b. m2= 0.6KG k2=300 N/m v2= 1 m/s . when passing through equilibrium
c. m3= 1.2KG k3=400 N/m v3= 0.5 m/s . when passing through x= -0.01 m
d. m4= 2 KG k4=200 N/m v4= 0.2 m/s . when passing through x=-0.05 m
Physics
1 answer:
Rzqust [24]4 years ago
4 0

Answer:

The order is 2>4>3>1 (TE)

Explanation:

Look up attached file

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How will the electrostatic force between two electric charges change if the first charge is doubled and the second charge is onl
Vinvika [58]

Answer:

B) \frac{2}{3}

Explanation:

The electric force between charges can be determined by;

F = \frac{kq_{1} q_{2} }{r^{2} }

Where: F is the force, k is the Coulomb's constant, q_{1} is the value of the first charge, q_{2} is the value of the second charge, r is the distance between the centers of the charges.

Let the original charge be represented by q, so that;

q_{1} = 2q

q_{2} = \frac{q}{3}

So that,

F = q_{1}q_{2} x \frac{k}{r^{2} }

  = 2q x \frac{q}{3} x \frac{k}{r^{2} }

  = \frac{2q^{2} }{3} x \frac{k}{r^{2} }

  = \frac{2}{3} x \frac{kq}{r^{2} }

F = \frac{2}{3} x \frac{kq}{r^{2} }

The electric force between the given charges would change by \frac{2}{3}.

4 0
3 years ago
Male Rana catesbeiana bullfrogs are known for their loud mating call. The call is emitted not by the frog's mouth but by its ear
Sidana [21]

Answer:

The amplitude of the eardrum's oscillation is 6.65×10^-13 m.

Explanation:

Given data:

The sound has a frequency of 262 Hz

The sound level is 84 dB

The air density is 1.21 kg/m^3

The speed of sound is 346 m/s

Solution:

As, Intensity of sound is given by,

I = Io×10^(s/10 db)

I = 2×π^2×ρ×v×f^2×Sm^2

Thus,

Sm = √(Io×10^(s/10 db)) / √( 2×π^2×ρ×v×f^2)

Now, put the values,

Sm = √( 10^-12 × 10^(84/10) ) / √( 2×(3.14)^2×1.21×346×(262)^2 )

Sm = √(2.51×10^-4 / 5.66×10^8)

Sm = √0.443×10^-12

Sm = 6.65×10^-13 m.

8 0
3 years ago
During a lab investigation, students added four 50 g masses to two boxes and arranged the boxes so that they were motionless on
IceJOKER [234]

When the resultant force is not equal to zero termed an unbalanced force. By procedures 4 and 5 students observe an unbalanced upward force on Box 1. Hence option 1 is right for the problem.

<h3>What is an unbalanced force?</h3>

The forces operating on a body are known as unbalanced forces when the resulting force exerted on it is not equal to zero.

Unbalanced forces acting on the body, causing it to modify its state of motion. To further grasp the nature of imbalanced forces.

<h3 />

The following reasons by which we can understand the unbalanced force caused by the box.

Due to these two reasons, books will move up.

By adding another mass to box 2. The box becomes lighter. As the box becomes lighter the gravity force acting on the box will be less due to which the box easily can move up.

By removing the two masses from box 1. Due to which other become heavier other becomes heavier pulling it down causing box 1 one to go up.

Hence by procedures 4 and 5 students observe an unbalanced upward force on Box 1. Hence option 1 is right for the problem.

To learn more about the unbalanced force refer to the link;

brainly.com/question/227461

3 0
3 years ago
What would u expect the tides to be like during a first quarter moon?
kicyunya [14]

Answer:

During the first quarter or last quarter phase of the moon, when the sun and moon are perpendicular (at right angles) to each other in relation to the Earth, the tidal gravitational pulls interfere with each other, producing weaker tides, known as neap tides.

Explanation:

3 0
3 years ago
Read 2 more answers
Each shot of the laser gun favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by the dis
Alisiya [41]

Answer:

a) 4.94e9 J  b) 1.07e10 J

Explanation:

The electric potential energy stored in a capacitor, expressed in terms of the value of the capacitance C, and the voltage between its terminals V, is as follows:

U =\frac{1}{2}*C*V^{2}

a) For the original capacitor, we can find directly U as follows:

U =\frac{1}{2}*1.81F*(73.9e3)^{2} V2 = 4.94e9 J  

U = 4.94*10⁹ J

b) Prior to find the electric potential energy of the upgraded capacitor, we need to find out the value of the capacitance C of this capacitor, which is identical to the original, except that has a different dielectric constant.

As the capacitance is proportional to the dielectric constant, we can write the following proportion:

ε₂ / ε₁ = \frac{943}{435}= \frac{C2}{C1} =\frac{Cx}{1.81F}

Cx =\frac{1.81F*943}{435} = 3.92 F

Once calculated the new value of the capacitance, as V remains the same, we can find the electric potential energy for the upgraded capacitor as follows:

U =\frac{1}{2}*3.92F*(73.9e3)^{2} V2 = 1.07e10 J

⇒ U = 1.07*10¹⁰ J

8 0
3 years ago
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