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3 years ago

Two positive charges of 6.0 x 10^-6 C are separated by 0.50 m. Calculate the force of electrical attraction

Physics

1 answer:

SpyIntel [72]3 years ago

3 0

Answer:

Force between two charges is given as 1.296 N

Explanation:

As we know that the two charges will attract or repel each other when they are placed near to each other

Here we know that electrostatic force is given as

$F = \frac{kq_1 q_2}{r^2}$

here we have

$q_1 = q_2 = 6 \times 10^{-6} C$

distance between two charges is given as

r = 0.50 m

so we have

$F = \frac{(9\times 10^9)(6 \times 10^{-6})^2}{0.50^2}$

$F = 1.296 N$

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Explanation:

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Chet plans an experimental investigation to see how well a new fertilizer works on daisies. The fertilizer must be dissolved in

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The answer is D or Fertilizer, sorry for the late answer

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A punter drops a 2.0 kg ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at

pav-90 [236]

Answer:

Explanation:

Given

mass of drop $m=2 kg$

height of fall $h=1 m$

ball leaves the foot with a speed of 18 m/s at an angle of $55^{\circ}$

Velocity of ball just before the collision with the floor

$u^=2gh$

$u=\sqrt{2gh}$

$u=\sqrt{2\times 9.8\times 1}=4.42 m/s$

Impulse delivered in Y direction

$J_y=m(v\sin (55)-(-u))$

$J_y=2(18\sin (55)+4.42)$

$J_y=38.32 kg-m/s$

Impulse in x direction

$J_x=m\times v\cos (55)$

$J_x=2\times 4.42\cos (55)=20.646$

$J_{net}=\sqrt{J_x^2+J_y^2}$

$J_{net}=\sqrt{(38.32)^2+(20.64)^2}$

$J_{net}=43.52 kg-m/s$

at an angle of $\tan \phi =\frac{J_y}{J_x}=\frac{38.32}{20.64}$

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3 years ago

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an

nalin [4]

Answer:

The maximum volume is 1417.87 $inch^3$

Explanation:

<u>Optimization Using Derivatives</u>

We have a 24x30 inch piece of metal and we need to make a rectangular box by cutting a square from each corner of the piece and bending up the sides. The width of the piece is 24 inches and its length is 30 inches

When we cut a square of each corner of side x, the base of the box (after bending up the sides) will be (24-2x) and (30-2x), width and length respectively. The volume of the box is

$V=(24-2x)(30-2x)x$

Operating

$V=4x^3-108x^2+720x$

To find the maximum value of V, we compute the first derivative and equate it to zero

$V'=12x^2-216x+720=0$

Simplifying by 12

$x^2-18x+60=0$

Completing squares

$x^2-18x+81-81+60=0$

$(x-9)^2=21$

We have two values for x

$x=9+\sqrt{21}=13.58\ inch$

$x=9-\sqrt{21}=4.42\ inch$

The first value is not feasible because it will produce a negative width (24-2(13.58))=-6.16

We'll keep only the solution

$x=4.42\ inch$

The width is

$w=(24-2(4.42))=15.16\ inch$

The length is

$l=(30-2(4.42))=21.16\ inch$

And the height

$x=4.42\ inch$

The maximum volume is

$V=(15.16)(21.16)(4.42)=1417.87\ inch^3$

4 0

3 years ago

A person with mass mp = 74 kg stands on a spinning platform disk with a radius of R = 2.31 m and mass md = 183 kg. The disk is i

timurjin [86]

Answer:

1) 883 kgm2

2) 532 kgm2

3) 2.99 rad/s

4) 944 J

5) 6.87 m/s2

6) 1.8 rad/s

Explanation:

1)Suppose the spinning platform disk is solid with a uniform distributed mass. Then its moments of inertia is:

$I_d = m_dR^2/2 = 183*2.31^2/2 = 488 kgm^2$

If we treat the person as a point mass, then the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk:

$I_{rim} = I_d + m_pR^2 = 488 + 74*2.31^2 = 883 kgm^2$

2) Similarly, he total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk (1/3 of the radius from the center):

$I_{R/3} = I_d + m_p(R/3)^2 = 488 + 74*(2.31/3)^2 = 532 kgm^2$

3) Since there's no external force, we can apply the law of momentum conservation to calculate the angular velocity at R/3 from the center:

$I_{rim}\omega_{rim} = I_{R/3}\omega_{R/3}$

$\omega_{R/3} = \frac{I_{rim}\omega_{rim}}{I_{R/3}}$

$\omega_{R/3} = \frac{883*1.8}{532} = 2.99 rad/s$

4)Kinetic energy before:

$E_{rim} = I_{rim}\omega_{rim}^2/2 = 883*1.8^2/2 = 1430 J$

Kinetic energy after:

$E_{R/3} = I_{R/3}\omega_{R/3}^2/2 = 532*2.99^2/2 = 2374 J$

So the change in kinetic energy is: 2374 - 1430 = 944 J

5) $a_c = \omega_{R/3}^2(R/3) = 2.99^2*(2.31/3) = 6.87 m/s^2$

6) If the person now walks back to the rim of the disk, then his final angular speed would be back to the original, which is 1.8 rad/s due to conservation of angular momentum.

3 0

3 years ago