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3 years ago

How much heat is absorbed from a 56.00 g sample of Mercury when its tempreature change is 289K?

Physics

1 answer:

AveGali [126]3 years ago

8 0

Heat required to raise the temperature of mercury is given as

$Q = ms\Delta T$

here given that

m = 56 g

s = specific heat capacity of mercury = 0.140 J/g C

$\Deta T = 289 k$

now here we have

$Q = 56 * 0.140 * 289$

$Q = 2265.76 J$

so it required 2265.76 J of heat

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A circuit is set up such that it has a current of 8 A. What would be the new current if the resistance was increased by a factor

RUDIKE [14]

Answer:

4 A

Explanation:

The relationship between current, voltage and resistance in a circuit is given by Ohm's law:

$V=RI$

where

V is the voltage

R is the resistance

I is the current

The equation can also be rewritten as

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

from which we see that the current is inversely proportional to the resistance, R.

In this problem, the initial current is I = 8 A. Then the resistance is doubled:

R ' = 2R

So the new current is

$I'=\frac{V}{R'}=\frac{V}{2R}=\frac{1}{2}(\frac{V}{R})=\frac{I}{2}=4 A$

so the current is halved.

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3 years ago

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Brrunno [24]

Answer:

a) The velocity is 2.94m/s

b) 0.441

Explanation:

a) Assume gravity is 9.8m/s^2

Use the equation below to solve for the velocity at 0.30 seconds

$vf=vi+at$ ,

vf =unknown velocity vi= initial velocity vi=0m/s a= 9.8m/s^2 t=0.30seconds

Step 1: Substitute the variables with the knowns

$vf=0m/s+(9.8m/s^2)*0.30seconds$

Step 2: Solve

$vf=2.94m/s$

Use the equation below to solve for the displacement at 0.30 seconds

$x=vit+\frac{1}{2} at^{2}$

Step 1: Substitute the same variables with the knowns

$x=\frac{1}{2}*(9.8m/s^2)*(0.30seconds)^{2}$

Note that vi*t=0 as vi=0m/s

Step 2: Solve

x=0.441m

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