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3 years ago

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An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 19 m lower vert

ically, is a horizontally situated spring with constant 4.2 × 105 N/m. The acceleration of gravity is 9.8 m/s 2 . Ignore friction. How much is the spring compressed in stopping the ore car? Answer in units of m.

1 answer:

cupoosta [38]3 years ago

6 0

Answer:

The compression in the spring is 5.88 meters.

Explanation:

Given that,

Mass of the car, m = 39000 kg

Height of the car, h = 19 m

Spring constant of the spring, $k=4.2\times 10^5\ N/m$

We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,

$mgh=\dfrac{1}{2}kx^2$

x is the compression in spring

$x=\sqrt{\dfrac{2mgh}{k}} \\\\x=\sqrt{\dfrac{2\times 39000\times 19\times 9.8}{4.2\times 10^5}} \\\\x=5.88\ m$

So, the compression in the spring is 5.88 meters.

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Mr. Mehta cut several 275-centimeter lengths of wallpaper to put in his

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Answer:

2.75 meters.

Explanation:

1 meter = 100 centimeters.

There are 275 centimeters.

275/100=2.75

So, each piece of wallpaper was 2.75 meters long.

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A charge of -2.65 nC is placed at the origin of an xy-coordinate system, and a charge of 2.00 nC is placed on the y axis at y =

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Answer:

A. Fnx = 5.71*10⁻⁵ N , Fny= -3.67*10⁻⁵ N

B. Fn= 6.78 *10⁻⁵ N

C. α= 32.4° counterclockwise with the positive x+ axis

Explanation:

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Equivalences

1nC= 10⁻⁹C

1cm = 10⁻²m

Known data

k= 9*10⁹N*m²/C²

q₁= -2.65 nC =-2.65*10⁻⁹C

q₂= +2.00 nC = 2*10⁻⁹C

q₃= +5.00 nC= =+5*10⁻⁹C

$d_{13} = \sqrt{(3.2)^{2} +(3.8)^{2} }$

$d_{13} =\sqrt{24.68}$ * 10⁻²m = 4.9678* 10⁻²m

(d₁₃)² = 24.68*10⁻⁴m²

d₂₃ = 3.2 cm = 3.2*10⁻²m

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force (F₂₃) of q₂ on q₃ is repulsive because the charges have equal signs and the forces.

The force (F₁₃) of q₁ on q₃ is attractive because the charges have opposite signs.

Magnitudes of F₁₃ and F₂₃

F₁₃ = (k*q₁*q₃)/(d₁₃)²=( 9*10⁹*2.65*10⁻⁹*5*10⁻⁹) /(24.68*10⁻⁴)

F₁₃ = 4.8 *10⁻⁵ N

F₂₃ = (k*q₂*q₃)/(d₂₃)² = ( 9*10⁹*2*10⁻⁹*5*10⁻⁹) /((3.2)²*10⁻⁴)

F₂₃ = 8.8 *10⁻⁵ N

x-y components of F₁₃ and F₂₃

F₁₃x= -4.8 *10⁻⁵ *cos β= - 4.8 *10⁻⁵(3.2/ (4.9678)= - 3.09*10⁻⁵ N

F₁₃y= -4.8 *10⁻⁵ *sin β= - 4.8 *10⁻⁵(3.8/(4.9678) = - 3.67*10⁻⁵ N

F₂₃x = F₂₃ = +8.8 *10⁻⁵ N

F₂₃y = 0

x and y components of the total force exerted on q₃ by q₁ and q₂ (Fn)

Fnx= F₁₃x+F₂₃x = - 3.09*10⁻⁵ N+8.8 *10⁻⁵ N= 5.71*10⁻⁵ N

Fny= F₁₃y+F₂₃y = - 3.67*10⁻⁵ N+0= - 3.67*10⁻⁵ N

Fn magnitude

$F_{n} =\sqrt{(Fn_{x})^{2}+(Fn_{y})^{2} }$

$F_{n} = \sqrt{(5.71)^{2}+(3.67)^{2} }$ *10⁻⁵ N

Fn= 6.78 *10⁻⁵ N

Fn direction (α)

$\alpha =tan^{-1}( \frac{Fn_{y} }{Fn_{x} } )$

$\alpha =tan^{-1}( \frac{-3.67 }{5.71} )$

α= -32.4°

α= 32.4° counterclockwise with the positive x+ axis

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Answer:

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Explanation:

Given:

Frequency $f = 2.4 \times 10^{9}$ Hz

(A)

The magnetic field related to cyclotron is given by,

$B = \frac{2\pi mf}{e}$

Where $m =$ mass of electron $= 9.1 \times 10^{-31 }$ kg, $e = 1.6 \times 10^{-19}$ C

$B = \frac{6.28 \times 9.1 \times 10^{-31 }\times 2.4 \times 10^{9} }{1.6 \times 10^{-19} }$

$B = 0.086$ T

Therefore, the magnetic field strength is 0.086 T

(B)

Diameter of orbit $d= 2.5 \times 10^{-2}$ m

Radius of orbit $r = 1.25 \times 10^{-2}$ m

The maximum kinetic energy is given by,

KE = $\frac{1}{2} mv^{2}$

Where $v = r \omega = r 2\pi f$

KE = $\frac{1}{2} m(2\pi fr)^{2}$

KE = $\frac{1}{2} \times 9.1 \times 10^{-31 } \times (2 \times 3.14 \times 2.4 \times 10^{9} \times 1.25 \times 10^{-2} )^{2}$

KE = $1.62 \times 10^{-14}$ J

Therefore, the maximum kinetic energy is $1.62 \times 10^{-14}$ J

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