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1 year ago

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An electrical motor provides 0.50W of mechanical power. How much time will it take motor to lift a 0.1 kg mass at constant speed

form to a shelf 2.0 m above the floor?
a. 0.25s
b. 0.40s
c. 1.0s
d. 4.0s

1 answer:

DaniilM [7]1 year ago

5 0

Answer:

d. 4.0s

Explanation:

Mass = 0.1 kg

g = acceleration due to gravity = 10 m/s^2

Force = 0.1 × 10 = 1N

Distance = 2m

Time = force × distance / power

= 1 × 2/0.5

= 4.0 s

The right option is d. 4.0s

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Newton's third law tells us that for every force there is an equal and opposite force. This means that if Anna exerts a force of 20 Newtons on the box, the box exerts a force of 20 Newtons on Anna.

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When a vertical beam of light passes through a transparent medium, the rate at which its intensity I decreases is proportional t

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Answer:

Intensity of beam 18 feet below the surface is about 0.02%

Explanation:

Using Lambert's law

Let dI / dt = kI, where k is a proportionality constant, I is intensity of incident light and t is thickness of the medium

then dI / I = kdt

taking log,

ln(I) = kt + ln C

I = Ce^kt

t=0=>I=I(0)=>C=I(0)

I = I(0)e^kt

t=3 & I=0.25I(0)=>0.25=e^3k

k = ln(0.25)/3

k = -1.386/3

k = -0.4621

I = I(0)e^(-0.4621t)

I(18) = I(0)e^(-0.4621*18)

I(18) = 0.00024413I(0)

Intensity of beam 18 feet below the surface is about 0.2%

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1 year ago

Why does carpet tend to produce differences in static electricity more that hardwood or tile floors

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Answer:

This is because the rubbing releases negative charges, called electrons, which can build up on one object to produce a static charge. For example, when you shuffle your feet across a carpet, electrons can transfer onto you, building up a static charge on your skin.

Explanation:

This is because the rubbing releases negative charges

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2 years ago

In conventional television, signals are broadcast from towers to home receivers. Even when a receiver is not in direct view of a

fgiga [73]

(a) The diffraction decreases

The formula for the diffraction pattern from a single slit is given by:

$sin \theta = \frac{n \lambda}{a}$

where

$\theta$ is the angle corresponding to nth-minimum in the diffraction pattern, measured from the centre of the pattern

n is the order of the minimum

$\lambda$ is the wavelength

a is the width of the opening

As we see from the formula, the longer the wavelength, the larger the diffraction pattern (because $\theta$ increases). In this problem, since the wavelength of the signal has been decreased from 54 cm to 13 mm, the diffraction of the signal has decreased.

(b) $10.8^{\circ}$

The angular spread of the central diffraction maximum is equal to twice the distance between the centre of the pattern and the first minimum, with n=1. Therefore:

$sin \theta = \frac{(1) \lambda}{a}$

in this case we have

$\lambda=54 cm = 0.54 m$ is the wavelength

$a=5.7 m$ is the width of the opening

Solving the equation, we find

$\theta = sin^{-1} (\frac{\lambda}{a})=sin^{-1} (\frac{0.54 m}{5.7 m})=5.4^{\circ}$

So the angular spread of the central diffraction maximum is twice this angle:

$\theta = 2 \cdot 5.4^{\circ}=10.8^{\circ}$

(c) $0.26^{\circ}$

Here we can apply the same formula used before, but this time the wavelength of the signal is

$\lambda=13 mm=0.013 m$

so the angle corresponding to the first minimum is

$\theta = sin^{-1} (\frac{\lambda}{a})=sin^{-1} (\frac{0.013 m}{5.7 m})=0.13^{\circ}$

So the angular spread of the central diffraction maximum is twice this angle:

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2 years ago

A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through

Svet_ta [14]

Answer:

The speed is $\sqrt{2}v_{0}$ .

(a) is correct option.

Explanation:

Given that,

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Using formula of initial work done on proton

$W = q V$

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$\Delta W=\Delta K.E$

$q V=\dfrac{1}{2}mv^2$

Put the value into the formula

$q V_{0}=\dfrac{1}{2}mv_{0}^2$

$v_{0}^2=\dfrac{2qV_{0}}{m}$ ....(I)

If it were accelerated instead through a potential difference of $2 V_{0}$ , then it would gain a speed will be given as :

Using an above formula,

$v_{0}'^2=\dfrac{2qV_{0}}{m}$

Put the value of $V_{0}$

$v_{0}'^2=\dfrac{2q\times2V_{0}}{m}$

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Hence, The speed is $\sqrt{2}v_{0}$ .

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1 year ago