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Sonja [21]
2 years ago
15

A person lifts a 4.5 kg cement block a vertical distance of 1.2 m and then carries the block horizontal- ly a distance of 7.3 m.

determine the work done by the person and by the force of gravity in this process.
Physics
1 answer:
Delvig [45]2 years ago
6 0
<span>When a person lifts the block, the block has more potential energy. Therefore the person does positive work on the block. work = m g h work = (4.5 kg) (9.80 m/s^2) (1.2 m) work = 52.92 joules The person's work on the block is 52.92 joules When the block is being raised, the force of gravity opposes the motion. Therefore the force of gravity does negative work on the block. work = - (force) (h) work = - m g h work = -(4.5 kg) (9.80 m/s^2) (1.2 m) work = -52.92 joules The work done by the force of gravity on the block is -52.92 joules Note that when the block is moved horizontally, the potential energy does not change. Therefore there is no work done on the block when it moves horizontally (we are assuming that the kinetic energy does not change).</span>
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Two very large parallel metal plates, separated by 0.20 m, are connected across a 12-V source of potential. An electron is relea
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Answer:

{\rm K} = 2.4\times 10^{-19}~J

Explanation:

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E = \frac{Q}{2\epsilon_0 A}

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The electric force on the electron is

F = qE = \frac{qQ}{2\epsilon_0 A}

where q is the charge of the electron.

By definition the capacitance of the capacitor is given by

C = \epsilon_0\frac{A}{d} = \frac{Q}{V}\\\frac{Q}{\epsilon_0 A} = \frac{V}{d} = \frac{12}{0.20} = 60

Plugging this identity into the force equation above gives

F = \frac{qQ}{2\epsilon_0 A} = \frac{q}{2}(\frac{Q}{\epsilon_0 A}) = \frac{q}{2}60 = 30q

The work done by this force is equal to change in kinetic energy.

W = Fx = (30q)(0.05) = 1.5q = K

The charge of the electron is 1.6 \times 10^{-19}

Therefore, the kinetic energy is 2.4\times 10^{-19}

8 0
2 years ago
Suppose that you purchased a water bed with the dimensions 2.55 m à 2.53 dm à 245 cm. what mass of water does this bed contain
Nadya [2.5K]

dimensions of the bed is given as

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thickness = 2.53 dm = 0.253 m

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now the volume of the bed is given as

V = 2.55 * 0.253 * 2.45 m^3

V = 1.581m^3

now the mass of water in it is given as

mass = density * volume

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mass = 1581 kg

<em>so it will contain 1581 kg mass in it</em>

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A ray of light passes from air into carbon disulfide (n = 1.63) at an angle of 28.0 degrees to the normal. what is the refracted
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We can solve the problem by using Snell's law, which states 
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where
n_i is the refractive index of the first medium
\theta_i is the angle of incidence
n_r is the refractive index of the second medium
\theta_r is the angle of refraction

In our problem, n_i=1.00 (refractive index of air), \theta_i = 28.0^{\circ} and n_r=1.63 (refractive index of carbon disulfide), therefore we can re-arrange the previous equation to calculate the angle of refraction:
\sin \theta_r =  \frac{n_i}{n_r}  \sin \theta_r =  \frac{1.00}{1.63}  \sin 28.0^{\circ} = 0.288
From which we find
\theta_r = \arcsin (0.288)=16.7^{\circ}
6 0
2 years ago
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