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3 years ago

What is the difference between heat capacity and specific heat capacity? I want the meaning please.

Physics

1 answer:

charle [14.2K]3 years ago

8 0

Answer:

The heat capacity of a body is defined as the heat required to raise it's temperature by me degree or one kelvin.while specific heat capacity of a substance is defined as the heat required to the temperature of a unit mass of it through one degree or one kelvin.

I hope it helps

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Directions: WRITE a summary of the Newton’s Laws of Motion reading.

guajiro [1.7K]

Answer:

In Newton's 1st Law, a stationary body or that moving with a uniform motion, will only change its state if acted upon by other external forces.This restance to change in position/motion is called inertia. In his second law, when a body is acted upon by a force, the force will be equal to the product of mass and acceleration of the body.i.e. Force= m.a. In the third law, it explains that a body will exert an equal and opposite force when intracting with another body.Here, every action force experiences an equal and opposite reaction force, where the momentum of the bodies will be conserved.

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4 years ago

you push a coin across a table the coin stops how does this motion relate to balanced and unbalanced forces

-BARSIC- [3]

When you exert a force on the coin, it will accelerate. If you push the coin and it moves at a constant velocity, the friction force must be equal to the force that you are exerting. This is an example of a balanced force. When the net force is greater than 0 N, the is an unbalanced force.

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4 years ago

The natural pH of rain water is?

SOVA2 [1]

5.6 which would be acidic!

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3 years ago

Read 2 more answers

In a carnival booth, you can win a stuffed giraffe if you toss a quarter into a small dish. the dish is on a shelf above the poi

Georgia [21]

components of the speed of the coin is given as

$v_x = v cos60$

$v_x = 6.4 cos60 = 3.2 m/s$

$v_y = vsin60$

$v_y = 6.4 sin60 = 5.54 m/s$

now the time taken by the coin to reach the plate is given by

$t = \frac{\delta x}{v_x}$

$t = \frac{2.1}{3.2}$

$t = 0.656 s$

now in order to find the height

$h = vy * t + \frac{1}{2} at^2$

$h = 5.54 * 0.656 - \frac{1}{2}*9.8*(0.656)^2$

$h = 1.52 m$

so it is placed at 1.52 m height

3 0

3 years ago

Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and

strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance $R_e$ defined by the formula:

$\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}$

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

$\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\$

and when this is combined with the third resistor in series, the equivalent resistance ( $R''_e$ ) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

$R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}$

The problem states that the difference between the equivalent resistances in both circuits is given by:

$R''_e=R_e+630 \,\Omega$

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

$\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega$

8 0

3 years ago