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julia-pushkina [17]
2 years ago
14

. While a person lifts a book of mass 2 kg from the floor to a tabletop, 1.5 m above the floor, how much work does the the perso

n do on the book? How much work does gravity do on the boo
Physics
2 answers:
viktelen [127]2 years ago
5 0

Answer:

a) 29.4 J

b) - 29.4 J

Explanation:

Given:

Mass of the book, m = 2 kg

Height above the floor, h = 1.5 m

Now,

the work done by the person will be = Force applied on the book × displacement of the book

thus,

Work done by the person = mg × h

where, g is the acceleration due to gravity

thus, on substituting the values, we get

Work done by the person = 2 × 9.8 × 1.5 = 29.4 J

now,

for the force applied by the gravitational pull (downwards) the displacement is in opposite direction (upwards) to the force of the gravity.

Thus,

work done by the gravity will be negative

therefore, the work done by the gravity = - mg × h

or

work done by the gravity = - 29.4 J

photoshop1234 [79]2 years ago
4 0

Answer:

29.4 J  -29.4 J

Explanation:

The work done by the person on the book is given by =mgh

where m =mass

          g= acceleration due to gravity

          h=height

Here m=2 kg given g=9.8 kg m/sec^2 h=1.5 meter

So work done by the person =2×9.8×1.5=29.4 J

As the gravitational force work downward and person lift the book upward so work done by gravity = -mgh = -2×9.8×1.5=  -29.4 J  

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Answer:

He should stand from the center of laser pointed on the wall at 1.3 m.

Explanation:

Given that,

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d\sin\theta=n\lambda

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y=\dfrac{650\times10^{-9}\times10}{5\times10^{-6}}

y=1.3\ m

Hence, He should stand from the center of laser pointed on the wall at 1.3 m.

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\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}

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strojnjashka [21]

Answer:

1.7323

Explanation:

To develop this problem, it is necessary to apply the concepts related to refractive indices and Snell's law.

From the data given we have to:

n_{air}=1

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Where n means the index of refraction.

We need to calculate the index of refraction of the liquid, then applying Snell's law we have:

n_1sin\theta_1 = n_2sin\theta_2

n_{air}sin\theta_{air} = n_{liquid}sin\theta_{liquid}

n_{liquid} = \frac{n_{air}sin\theta_{air}}{sin\theta_{liquid}}

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gavmur [86]

Answer:

a. ρ_\beta=1.996J/m^3

b. U_E=9.445x10^{-15} J/m^3

Explanation:

a. To find the density of magnetic field given use the gauss law and the equation:

i=14A, d=2.5mm, R=3.3Ω, l=1 km, E_o=8.85x10^{-12}F/m, u_o=4*x10^{-7}H/m

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ρ_\beta=\frac{u_o*i^2}{8\pi*r}=\frac{4\pi *10^{-7}H/m*(14A)^2}{8\pi*(1.25x10^{-3}m)^2}

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U_E=\frac{1}{2}*E_o*E^2

E=(\frac{i*R}{l})^2

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