Relationship of electric current to voltage

1 answer:

6 0

The current through a conducting path is directly proportional
to the voltage between the ends of the path.

The constant of proportionality is

( 1 / resistance between the ends ).

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You pour 250 g of tea into a Styrofoam cup, initially at 80∘C and stir in a little sugar using a 100-g aluminum 20∘C spoon and l

jok3333 [9.3K]

To solve this problem it is necessary to apply the concepts related to the conservation of energy and heat transferred in a body.

By definition we know that the heat lost must be equal to the heat gained, ie

$Q_g = Q_l$

Where,

Q = Heat exchange

The heat exchange is defined as

$Q = c_p m \Delta T$

Where,

$c_p =$ Specific heat

m = mass

$\Delta T=$ Change in Temperature

Therefore replacing we have that

$Q_g = Q_l$

$c_{p-tea} m \Delta T = c_{p-al} m \Delta T$

Replacing with our values we have that

$0.25*4180*(80-T) = 0.1*900*(T-20)$

$11.61*(80-T) = T-20$

$T= \frac{948.8}{11.61}$

$T = 75.24\°C$

Therefore the highest possible temperature of the spoon when you finally take it out of the cup is 75.24°C

3 0

3 years ago

A ball is thrown straight up into the air with a velocity of 12 m/s. Draw a motion diagram for the ball and then give as much qu

viktelen [127]

Answer:

find the diagram in the attachment.

Explanation:

Let vi = 12 m/s be the intial velocy when the ball is thrown, Δy be the displacement of the ball to a point where it starts returning down, g = 9.8 m/s^2 be the balls acceleration due to gravity.

considering the motion when the ball thrown straight up, we know that the ball will come to a stop and return downwards, so:

(vf)^2 = (vi)^2 + 2×g×Δy

vf = 0 m/s, at the highest point in the upward motion, then:

0 = (vi)^2 + 2×g×Δy

-(vi)^2 = 2×g×Δy

Δy = [-(vi)^2]/2×g

Δy = [-(-12)^2]/(2×9.8)

Δy = - 7.35 m

then from the highest point in the straight up motion, the ball will go back down and attain the speed of 12 m/s at the same level as it was first thrown