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3 years ago

Specific heat is the amount of heat required to raise the temperature of a material by 1°C per what unit?

Physics

2 answers:

Nataly [62]3 years ago

4 0

Answer:

Per unit mass

Explanation:

When some heat energy "Q" is given to an object of mass "m" to raise its temperature by "ΔT", then the energy is given as

Q = m c ΔT

where c = specific heat of the object

rearranging the above equation

c = Q/(m ΔT)

for "1 °C" rise of temperature , ΔT = 1

So specific heat is defined as amount of heat required to raise the temperature by 1 °C per unit mass of the object.

nydimaria [60]3 years ago

3 0

The heat required to raise the temperature of the unit mass of a given substance by a given amount (usually one degree).

or C. Mass if you're on plato

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The front and rear sprockets on a bicycle have radii of 8.40 and 4.91 cm, respectively. The angular speed of the front sprocket

Rudik [331]

Explanation:

It is given that,

Radius of the front sprockets, $r_f=8.4\ cm=0.084\ m$

Radius of the rear sprockets, $r_r=4.91\ cm=0.0491\ m$

The angular speed of the front sprocket is 12.3 rad/s, $\omega_f=12.3\ rad/s$

(a) Linear speed of the front sprockets, $v_f=r_f\times \omega$

$v_f=0.084\times 12.3$

$v_f=1.0332\ m/s$

$v_f=103.32\ cm/s$

Linear speed of the rear sprockets, $v_r=r_r\times \omega$

$v_r=0.0491\times 12.3$

$v_r=0.60393\ m/s$

$v_r=60.393\ cm/s$

(b) Let $a_r$ is the centripetal acceleration of the chain as it passes around the rear sprocket.

$a_r=\dfrac{v_r^2}{r_r}$

$a_r=\dfrac{(60.393)^2}{0.0491}$

$a_r=74283.39\ m/s^2$

$a_r=7428339\ cm/s^2$

`Hence, this is the required solution.

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nasty-shy [4]

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Answer: <em>Option (A) is correct.</em>

Explanation:

A heuristic is often referred to as an approach to solving problems or discovery that subjects a method, which is considered to be practical although might not be guaranteed to be perfect, optimal, or rational, but sufficient for reaching to goal.

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3 years ago

A race-car drives around a circular track of radius RRR. The race-car speeds around its first lap at linear speed v_iv i v, st

OLEGan [10]

Answer:

The magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

Explanation:

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When the linear speed of the race-car increases to v' = 4v, the centripetal acceleration a' becomes a' = v'²/R = (4v)²/R = 16v²/R.

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