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RSB [31]
3 years ago
5

The angular velocity (in rpm) of the blade of a blender is given in. If θ=0rad at t=0s, what is the blade's angular position at

t=20s?
Physics
1 answer:
Liula [17]3 years ago
7 0

Answer:

θ = 20.9 rad

Explanation:

In a blender after a short period of acceleration the blade is kept at a constant angular velocity, for which we can use the relationship

           w = θ / t

           θ = w t

if we know the value of the angular velocity we can find the angular position, we must remember that all the angles must be in radians

suppose that the angular velocity is w = 10 rpm, let us reduce to the SI system

          w = 10 rpm (\frac{2\pi  rad}{1 rev}) ( \frac{1 min}{60 s} )

= 1,047 rads

let's calculate

          θ = 1,047 20

          θ = 20.9 rad

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You look down an old well, cannot see the bottom, and mutter to yourself "Oh well!". In order to estimate the depth of the well,
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Answer:

The best estimate of the depth of the well is 2.3 sec.

Explanation:

Given that,

Record time,

t_{1}=2.19\ sec

t_{2}=2.30\ sec

t_{3}=2.26\ sec

t_{4}=2.29\ sec

t_{5}=2.27\ sec

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According to record time,

We can write of the record time

t_{1}=2.19\approx 2.2\ sec

t_{2}=2.30\approx 2.3\ sec

t_{3}=2.26\approx 2.3\ sec

t_{4}=2.29\approx 2.3\ sec

t_{5}=2.27\approx 2.3\ sec

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Hence, The best estimate of the depth of the well is 2.3 sec.

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3 years ago
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Irina18 [472]
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Please Help On These 2 Questions!!!!! I severely need help!!!!!!
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3 0
3 years ago
If you wanted to find the area of the hot filament in a light bulb, you would have to know the temperature (determinable from th
Sladkaya [172]

Answer:

To find out the area of the hot filament of a light bulb, you would need to know the temperature, the power input, the Stefan-Boltzmann constant and <u>Emissivity of the Filament</u>.

Explanation:

The emissive power of a light bulb can be given by the following formula:

E = σεAT⁴

where,

E = Power Input or Emissive Power

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Therefore,

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3 0
3 years ago
Find the ratio of the lengths of the two mathematical pendulums, if the ratio of periods is 1.5​
juin [17]

Answer:

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Explanation:

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T=2\pi \sqrt{\dfrac{l}{g}}

or

T^2=4\pi^2\dfrac{l}{g}\\\\l=\dfrac{T^2g}{4\pi^2}

Where

l is length of the pendulum

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ATQ,

\dfrac{T_1}{T_2}=1.5

Put in equation (1)

\dfrac{l_1}{l_2}=(1.5)^2\\\\=\dfrac{9}{4}

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