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Liono4ka [1.6K]

3 years ago

14

A spring stretches by 0.014 m when a 3.7-kg object is suspended from its end. How much mass should be attached to this spring so

that its frequency of vibration is f

1 answer:

lidiya [134]3 years ago

6 0

Explanation:

Below is an attachment containing the solution.

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A square plate of copper with 55.0 cm sides has no net charge and is placed in a region of uniform electric field of 82.0 kN/C d

Alex73 [517]

Answer

given,

Side of copper plate, L = 55 cm

Electric field, E = 82 kN/C

a) Charge density,σ = ?

using expression of charge density

σ = E x ε₀

ε₀ is Permittivity of free space = 8.85 x 10⁻¹² C²/Nm²

now,

σ = 82 x 10³ x 8.85 x 10⁻¹²

σ = 725.7 x 10⁻⁹ C/m²

σ = 725.7 nC/m²

change density on the plates are 725.7 nC/m² and -725.7 nC/m²

b) Total change on each faces

Q = σ A

Q = 725.7 x 10⁻⁹ x 0.55²

Q = 219.52 nC

Hence, charges on the faces of the plate are 219.52 nC and -219.52 nC

7 0

3 years ago

Enter an expression, in terms of defined quantities and g, for the force that the scale under left pillar shows

ella [17]

The expression, in terms of defined quantities and g is therefore Fu =((mg/2) +2 mp) g

<h3>What is a Scale?</h3>

This can be defined as a balance or any of various other instruments or devices for weighing.

The expression in terms of defined quantities and g, for the force that the scale under left pillar shows that Fu =((mg/2) +2 mp) g .

Read more about Force here brainly.com/question/4515354

5 0

2 years ago

A system of ideal gas at 22°C undergoes an ischoric process with an internal energy decrease of 4.30 × 10 3 4.30×103 J to a fina

Komok [63]

Answer:

The approximate change in entropy is -14.72 J/K.

Explanation:

Given that,

Temperature = 22°C

Internal energy $U=4.30\times10^{3}\ J$

Final temperature = 16°C

We need to calculate the approximate change in entropy

Using formula of the entropy

$\Delta S=\dfrac{\Delta U}{T}$

Where, $\Delta U$ = internal energy

T = average temperature

Put the value in to the formula

$\Delta S=\dfrac{-4.30\times10^{3}}{\dfrac{22+273+16+273}{2}}$

$\Delta S=-14.72\ J/K$

Hence, The approximate change in entropy is -14.72 J/K.

5 0

3 years ago

A box is sliding up an inclined plane with friction, and it is speeding up.What free-body diagrams matches this description?

sukhopar [10]

Answer: photo

Explanation: because

6 0

3 years ago

A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision

marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

$x \approx 103,46x10^{9}m$ .

B) The space time coordinate t of the collision in Earth's reference frame is

$t=377,29s$

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

$t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y=y'$

$z=z'$

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

First we calculate the expression in the denominator

$\frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}$

$\sqrt{1-\frac{v^{2}}{c^{2}}} =0,714$

then we calculate t

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}$

$t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}$

$t=\frac{190s+79,388s}{0,714}$

finally we get that

$t=377,29s$

then we calculate x

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+39872396914m}{0,714}}$

$x=\frac{73872396914m}{0,714}}$

$x=103462740775,91m$

finally we get that

$x \approx 103,46x10^{9} m$

5 0

3 years ago