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2 years ago

A school track team member ran for a total of 15.85 miles in practice over 5 days. How many miles did he average per day?

Physics

1 answer:

liraira [26]2 years ago

6 0

${\underline{\pink{\textsf{\textbf{ Answer : }}}}}$

➡ 15.85/5

➡ 3.17 ans.

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How do the units that are used to measure heat differ from the units that are used to measure temperature?

irakobra [83]

Heat is measured in joules and temperature is measured in F degrees or C degrees.

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2 years ago

Mercury has one of the lowest specific heats. This fact added to its liquid state at most atmospheric temperatures make it effec

UNO [17]

The specific heat of mercury is 149.4 J/(kgK)

Explanation:

When a substance is supplied with an amount of energy Q, its temperature increases according to the equation:

$\Delta T=\frac{Q}{mC_s}$

where

$\Delta T$ is the increase in temperature

m is the mass of the sample

$C_s$ is its specific heat capacity

For the sample of mercury in this problem we have

Q = 275 J

m = 0.450 kg

$\Delta T = 4.09 K$

Therefore, by re-arranging the equation we find the mercury's specific heat:

$C_s = \frac{Q}{m\Delta T}=\frac{275}{(0.450)(4.09)}=149.4 J/(kgK)$

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

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2 years ago

Would the frequency of the angular simple harmonic motion (SHM) of the balance wheel increase or decrease if the dimensions of t

storchak [24]

Answer:

Yes the frequency of the angular simple harmonic motion (SHM) of the balance wheel increases three times if the dimensions of the balance wheel reduced to one-third of original dimensions.

Explanation:

Considering the complete question attached in figure below.

Time period for balance wheel is:

$T=2\pi\sqrt{\frac{I}{K}}$

$I=mR^{2}$

m = mass of balance wheel

R = radius of balance wheel.

Angular frequency is related to Time period as:

$\omega=\frac{2\pi}{T}\\\omega=\sqrt{\frac{K}{I}} \\\omega=\sqrt{\frac{K}{mR^{2}}$

As dimensions of new balance wheel are one-third of their original values

$R_{new}=\frac{R}{3}$

$\omega_{new}=\sqrt{\frac{K}{mR_{new}^{2}}}\\\\\omega_{new}=\sqrt{\frac{K}{m(\frac{R}{3})^{2}}}\\\\\omega_{new}={3}\sqrt{\frac{K}{mR^{2}}}\\\\\omega_{new}={3}\omega$