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Setler [38]
3 years ago
9

A particle moves along a straight path through displacement while force acts on it. (Other forces also act on the particle.) Wha

t is the value of c if the work done by on the particle is (a) zero, (b) 4.0 J, and (c) -1.8 J
Physics
1 answer:
allsm [11]3 years ago
7 0

a) c = 1.85

b) c = 0.8

c) c = 2.33

Explanation:

a)

The displacement of the particle is given by

d=2.2i+cj

While the force applied on the particle is

F=3.2i-3.8 j

So we have a problem in 2-dimensions.

The work done on the particle is given by the scalar product between force and displacement:

W=F\cdot d (1)

Here the work done on the particle is zero, so

W = 0

Therefore from eq(1) we find:

0=(3.2i-3.8j)\cdot (2.2i+cj)=7.04-3.8c\\3.8c=7.04\\c=\frac{7.04}{3.8}=1.85

b)

In this problem, the work done on the particle is

W=4.0 J

The force and displacement are still

d=2.2i+cj (displacement)

F=3.2i-3.8 j (force)

Therefore, by calculting the scalar product between force and displacement and equating it to the work done (4.0 J), we find:

W=F\cdot d

4.0 =(3.2i-3.8j)\cdot (2.2i+cj)=7.04-3.8c\\3.8c=3.04\\c=\frac{3.04}{3.8}=0.8

c)

In this problem instead, the work done on the particle is negative:

W=-1.8 J

As before, the force and displacement are

d=2.2i+cj (displacement)

F=3.2i-3.8 j (force)

And so again, we calculate the scalar product between  force and displacement and we equate it to the work done on the particle, -1.8 J.

Doing so, we find:

W=F\cdot d

-1.8=(3.2i-3.8j)\cdot (2.2i+c)=7.04-3.8c\\3.8c=8.84\\c=\frac{8.84}{3.8}=2.33

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Answer and Explanation:

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Now,

(a) Displacement of the stone is given by the horizontal range:

R = v_{o}\sqrt{\frac{2H}{g}}

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A parallel plate capacitor is constructed with a dielectric slab with κ = 1.5 inserted between the plates. The area of each plat
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Answer:

The value is V =  4.533  \  V

Explanation:

From the question we are told that

The dielectric constant is k = 1.5

The area of each plate is A =  5 \ cm^2 = 0.0005 m^2

The distance between the plates is d= 1 \  mm  =  0.001 \  m

The charge on the capacitor is Q =  6*10^{-11} \  C

Generally the electric field in a vacuum is mathematically represented as

E_o  =  \frac{V_o}{d}

Generally V_o is the voltage of the capacitor which is mathematically represented as

V_o =  \frac{Q}{C_o}

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