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V125BC [204]
1 year ago
13

If 36 grams of water is to be heated from 24.0°C to 48°C to make a cup of tea, how much heat must be added? The specific heat of

water is 4.18
Physics
1 answer:
Vinvika [58]1 year ago
3 0

We will have the following:

\begin{gathered} Q=mc\Delta T\Rightarrow Q=(36)(4.18)(48-24) \\  \\ \Rightarrow Q=3611.52 \end{gathered}

So, the heat to add is 3611.52 Joules.

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3 years ago
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A sphere moves in simple harmonic motion with a frequency of 4.80 Hz and an amplitude of 3.40 cm. (a) Through what total distanc
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Answer:

a)  the total distance traveled by the sphere during one cycle of its motion = 13.60 cm

b) The maximum speed is = 102.54 cm/s

The maximum speed occurs at maximum excursion from equilibrium.

c)  The maximum magnitude of the acceleration of the sphere is = 30.93 m/s^2

The maximum acceleration occurs at maximum excursion from equilibrium.

Explanation:

Given that :

Frequency (f) = 4.80 Hz

Amplitude (A) = 3.40 cm

a)

The total distance traveled by the sphere during one cycle of simple harmonic motion is:

d = 4A   (where A is the Amplitude)

d = 4(3.40 cm)

d = 13.60 cm

Hence, the total distance traveled by the sphere during one cycle of its motion = 13.60 cm

b)

As we all know that:

x = Asin \omega t

Differentiating the above expression with respect to x ; we have :

\frac{d}{dt}(x) = \frac{d}{dt}(Asin \omega t)

v = A \omega cos \omega t

Assuming the maximum value of the speed(v) takes place when cosine function is maximum and the maximum value for cosine function is 1 ;

Then:

v_{max} = A \omega

We can then say that the maximum speed therefore occurs at the mean (excursion) position where ; x = 0  i.e at maximum excursion from equilibrium

substituting 2 \pi f for \omega in the above expression;

v_{max} = A(2 \pi f)

v_{max} = 3.40 cm (2 \pi *4.80)

v_{max} = 102.54 \ cm/s

Therefore, the maximum speed is = 102.54 cm/s

The maximum speed occurs at maximum excursion from equilibrium.

c) Again;

v = A \omega cos \omega t

By differentiation with respect to  t;

\frac{d}{dt}(v) = \frac{d}{dt}(A \omega cos \omega t)

a =- A \omega^2 sin \omega t

The maximum acceleration of the sphere is;

a_{max} =A \omega^2

where;

w = 2 \pi f

a_{max} = A(2 \pi f)^2

where A= 3.40 cm = 0.034 m

a_{max} = 0.034*(2 \pi *4.80)^2

a_{max} = 30.93 \ m/s^2

The maximum magnitude of the acceleration of the sphere is = 30.93 m/s^2

The maximum acceleration occurs at maximum excursion from equilibrium where the oscillating sphere will have maximum acceleration at the turning points when the sphere has maximum displacement of x  = \pm A

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Which of the following temperatures is the coldest? A) 5 degrees B)0 degrees C) -15 degrees D) -7 degrees
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It’s c -15 degrees because it it has below O which is below freezing
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How long must a 0.70-mm-diameter aluminum wire be to have a 0.40 a current when connected to the terminals of a 1.5 v flashlight
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By using Ohm's law, we can find what should be the resistance of the wire, R:
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And by using the resistivity  of the Aluminum, \rho=2.65 \cdot 10^{-8} \Omega m, we can use the relationship between resistance R and resistivity:
R= \frac{\rho L}{A}
to find L, the length of the wire:
L= \frac{RA}{\rho}= \frac{(3.75 \Omega)(3,85 \cdot 10^{-7} m^2)}{2.65 \cdot 10^{-8} \Omega m}=54.48 m
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