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3 years ago

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Physics

1 answer:

pishuonlain [190]3 years ago

5 0

12. The answer would be C. 1.50 s. This is because if you divide 60 by 40, you will get 1.5.

13. For this one I'm not sure, but what I can tell you is that the heavier something is the faster it will sink, the lighter it is, it will float.

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How much work in joules is required to lift a 23 kg box up from the ground to your waist that is 1.0 meters high, carry it 6 met

PSYCHO15rus [73]

Answer:

2682

Explanation:

Work done is given by :

Work = Force x distance

= mg x d

So, work done in lifting the box of 23 kg up to my waist of 1 m high is :

W = mg x d

= 23 x 9.18 x 1

= 211.14

Now work done carrying the box horizontally 6 meters across the room is

W = mg x d

= 23 x 9.18 x 6

= 1266.84

Work done in placing the box on the shelf that is 5.7 m above the ground is

W = mg x d

= 23 x 9.18 x 5.7

= 1203.49

So the total work done is = 211.14 + 1266.84 + 1203.49

= 2681.47

= 2682 (rounding off)

5 0

2 years ago

What is the most likely reason that Mendeleev placed tellurium before iodine?

Ilya [14]

Mendeleev watched that tellurium has compound properties like different components in its gathering, and he didn't realize that neutrons cause the more noteworthy nuclear mass. Mendeleev expressed that he anticipated that tellurium would have a lower nuclear mass than iodine.

8 0

3 years ago

a sensor light installed on the edge of a home can detect motion for a distance of 50 feet in front and with a range of motion o

Dvinal [7]

Answer:

4363.3231 feets²

Explanation:

Given that :

Distance, r = 50 ft

θ = 200°

The arc length of area covered :

Arc length = θ/360° * πr²

Arc length = (200/360) * 50 ft ^2 * π

Arc length = 0.5555555 * 2500 * π

Arc length = 4363.3231 feets²

7 0

2 years ago

Consider two thin, coaxial, coplanar, uniformly charged rings with radii a and b푏 (a

Wittaler [7]

Answer:

electric potential, V = -q(a²- b²)/8π∈₀r³

Explanation:

Question (in proper order)

Consider two thin coaxial, coplanar, uniformly charged rings with radii a and b (b < a) and charges q and -q, respectively. Determine the potential at large distances from the rings

<em>consider the attached diagram below</em>

the electric potential at point p, distance r from the center of the outer charged ring with radius a is as given below

Va = q/4π∈₀ [1/(a² + b²)¹/²]

$Va = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} }$

Also

the electric potential at point p, distance r from the center of the inner charged ring with radius b is

$Vb = \frac{-q}{4\pi e0} * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

Sum of the potential at point p is

V = Va + Vb

that is

$V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } + \frac{-q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

$V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

$V = \frac{q}{4\pi e0} * [\frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{1}{(b^{2} + r^{2} )^{1/2} }]$

the expression below can be written as the equivalent

$\frac{1}{(a^{2} + r^{2} )^{1/2} } = \frac{1}{(r^{2} + a^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} }$

likewise,

$\frac{1}{(b^{2} + r^{2} )^{1/2} } = \frac{1}{(r^{2} + b^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }$

hence,

$V = \frac{q}{4\pi e0} * [\frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]$

1/r is common to both equation

hence, we have it out and joined to the 4π∈₀ denominator that is outside

$V = \frac{q}{4\pi e0 r} * [\frac{1}{{(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]$

by reciprocal rule

1/a² = a⁻²

$V = \frac{q}{4\pi e0 r} * [{(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} - {(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2}]$

by binomial expansion of fractional powers

where $(1+a)^{n} =1+na+\frac{n(n-1)a^{2} }{2!}+ \frac{n(n-1)(n-2)a^{3}}{3!}+...$

if we expand the expression we have the equivalent as shown

${(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} = (1-\frac{a^{2} }{2r^{2} } )$

also,

${(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2} = (1-\frac{b^{2} }{2r^{2} } )$

the above equation becomes

$V = \frac{q}{4\pi e0 r} * [((1-\frac{a^{2} }{2r^{2} } ) - (1-\frac{b^{2} }{2r^{2} } )]$

$V = \frac{q}{4\pi e0 r} * [1-\frac{a^{2} }{2r^{2} } - 1+\frac{b^{2} }{2r^{2} }]$

$V = \frac{q}{4\pi e0 r} * [-\frac{a^{2} }{2r^{2} } +\frac{b^{2} }{2r^{2} }]\\\\V = \frac{q}{4\pi e0 r} * [\frac{b^{2} }{2r^{2} } -\frac{a^{2} }{2r^{2} }]$

$V = \frac{q}{4\pi e0 r} * \frac{1}{2r^{2} } *(b^{2} -a^{2} )$

$V = \frac{q}{8\pi e0 r^{3} } * (b^{2} -a^{2} )$

Answer

$V = \frac{q (b^{2} -a^{2} )}{8\pi e0 r^{3} }$

$V = \frac{-q (a^{2} -b^{2} )}{8\pi e0 r^{3} }$

8 0

3 years ago

You are in your car at rest when the traffic light turns green. You place your coffee cup on the horizontal dash and hit the gas

umka21 [38]

Answer:

(d) Negative.

Explanation:

let's test each at a time.

(a) It can't be 0, because cup would slide back other wise.

(b) Positive, well if forward is positive, than the work done against the forward acceleration must be negative , so it can't be positive.

well no, because work done by car's engine dosen't go all of it into getting car to move, so it can't be that.

(d) negative, this look like it, because work that friction does must be nagative to counteract positive thrust of car which is positive and in forward direction.

(d) this can't be true.

So the answer is (d) negative.

3 0

3 years ago

Help awnser need experts​

Help awnser need experts