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cluponka [151]
2 years ago
13

A bookshelf is 2m long, with supports at its end (p and q). A book weighing 10N is placed in the middle of the shelf. what are t

he upward forces?
Physics
1 answer:
mel-nik [20]2 years ago
5 0

If the book is placed in the middle, the forces acting on <em>p</em> and <em>q</em> is 5N. If the book is moved 50 cm from <em>q</em>, the forces at <em>p</em> and <em>q</em> can be solved by doing a moment balance

<u>With </u><u><em>p</em></u><u> as the pivot;</u>

Fq (2 m) = 10 N (0.5 m)

Fq = 2.5 N

Fp = 10 N - 2.5 N = 7.5 N

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The rotational kinetic energy term is often called the kinetic energy in the center of mass, while the translational kinetic ene
weqwewe [10]

Answer:

C

Explanation:

The total kinetic energy is the sum of the kinetic energy in the center of mass (Rotational Kinetic energy) plus the kinetic energy of the center of mass( Translational Kinetic Energy).

The formula

K_{tot} = K_{t} +K_{r}  is applicable only when

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6 0
2 years ago
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A ball is thrown horizontally at a height of 2.2 meters at a velocity of 65m/s off a cliff. Assume no air resistance. How long u
slega [8]

The horizontal motion has no effect on the vertical drop.

From a drop, the distance the ball falls in 'T' seconds is

D = 4.9 T^2

so

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8 0
2 years ago
In a certain clock, a pendulum of length L1 has a period T1 = 0.95s. The length of the pendulum
gulaghasi [49]

Answer:

Ratio of length will be \frac{L_2}{L_1}=1.108

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We have given time period of the pendulum when length is L_1 is T_1=0.95sec

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T=2\pi \sqrt{\frac{L}{g}}

So 0.95=2\pi \sqrt{\frac{L_1}{g}}----eqn 1

And 1=2\pi \sqrt{\frac{L_2}{g}}-------eqn 2

Dividing eqn 2 by eqn 1

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Squaring both side

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8 0
2 years ago
Calculate the amount of hcn that gives the lethal dose in a small laboratory room measuring 14 × 15 × 8.0ft. the density of air
vova2212 [387]
Since we are given the density and volume, then perhaps we can determine the amount in terms of the mass. All we have to do is find the volume in terms of cm³ so that it will cancel out with the cm³ in the density. The conversion is 1 ft = 30.48 cm. The solution is as follows:

V = (14 ft)(15 ft)(8 ft)(30.48 cm/1 ft)³ = 0.0593 cm³

The mass is equal to:
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7 0
2 years ago
A system delivers 1275 j of heat while the surroundings perform 855 j of work on it. calculate ∆esys in j.
kakasveta [241]
The first law of thermodynamics says that the variation of internal energy of a system is given by:
\Delta U = Q + W
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We must be careful with the signs here. The sign convention generally used is:
Q positive = Q absorbed by the system
Q negative = Q delivered by the system
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W negative = W done by the system

So, in our problem, the heat is negative because it is releaed by the system: 
Q=-1275 J
while the work is positive because it is performed by the surrounding on the system:
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So, the variation of internal energy of the system is
\Delta U = -1275 J+855 J=-420 J
6 0
2 years ago
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