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3 years ago

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tyhe mass of an object is 117 g adding 1200j of heat will raise the temperture of the object by 12 celsius what is the specifc h

eat of the object A 0.35 j/g *k B 6.8 j/g *k C 0.85 j/g*k D 42 j/g*k

1 answer:

Sati [7]3 years ago

8 0

Answer:

C 0.85 j/g*k

Explanation:

The specific heat capacity of a material is given by:

$C_s = \frac{Q}{m \Delta T}$

where

Q is the amount of heat supplied to the object

m is the mass of the object

$\Delta T$ is the increase in temperature of the object

For the object in this problem, we have

m = 117 g is the mass

Q = 1200 J is the heat supplied

$\Delta T=12^{\circ}$ is the increase in temperature

Substituting into the formula, we find the specific heat:

$C_s = \frac{1200 J}{(117 g)(12^{\circ})}=0.85 J/gC$

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A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 kg particle has the xy coordinates (0.600 m, -0.750 m).

just olya [345]

Answer:

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

Explanation:

The center of mass of a system of particles ( $\vec r_{cm}$ ), measured in meters, is defined by this weighted average:

$\vec r_{cm} = \frac{\Sigma_{i=1}^{n}\,m_{i}\cdot \vec r_{i}}{\Sigma_{i=1}^{n}\,m_{i}}$ (1)

Where:

$m_{i}$ - Mass of the i-th particle, measured in kilograms.

$\vec r_{i}$ - Location of the i-th particle with respect to origin, measured in meters.

If we know that $\vec r_{cm} = (-0.500\,m,-0.700\,m)$ , $m_{1} = 1\,kg$ , $\vec r_{1} = (-1.20\,m, 0.500\,m)$ , $m_{2} = 4.50\,kg$ , $\vec r_{2} = (0.600\,m, -0.750\,m)$ and $m_{3} = 4\,kg$ , then the coordinates of the third particle are:

$(-0.500\,m, -0.700\,m) = \frac{(1\,kg)\cdot (-1.20\,m,0.500\,m)+(4.50\,kg)\cdot (0.600\,m,-0.750\,m)+(4\,kg)\cdot \vec r_{3}}{1\,kg+4.50\,kg+4\,kg}$

$(-4.75\,kg\cdot m, -6.65\,kg\cdot m) = (-1.20\,kg\cdot m, 0.500\,kg\cdot m) + (2.7\,kg\cdot m, -3.375\,kg\cdot m) +(4\cdot x_{3},4\cdot y_{3})$

$(4\cdot x_{3}, 4\cdot y_{3}) = (-6.25\,kg\cdot m,-3.775\,kg\cdot m)$

$(x_{3},y_{3}) = (-1.562\,m,-0.944\,m)$

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

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Anni [7]

Answer:

$I=38.181\ A$ is the current through the body of the man.

$E=34.5\ J$ energy dissipated.

Explanation:

Given:

time for which the current lasted, $t=43\times 10^{-6}\ s$

potential difference between the feet, $V=21000\ V$

resistance between the feet, $R=550\ \Omega$

<u>Now, from the Ohm's law we have:</u>

$I=\frac{V}{R}$

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$E=38.181^2\times 550\times 43\times 10^{-6}$

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Even when your speed is steady, you're accelerating if your direction
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A few examples:
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-- the little ball going around the inside of a Roulette wheel
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