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3 years ago

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You drop an ice cube into an insulated flask full of water and wait for the ice cube to completely melt. The ice cube initially

has a mass of 60.0 g and a temperature of 0°C. The water (before the ice cube is added) has a mass of 760 g and an initial temperature of 20.0°C. What is the final temperature (in °C) of the mixture? (Assume no energy is lost to the walls of the flask, or to the environment.)

1 answer:

3241004551 [841]3 years ago

7 0

Answer: $T=12.69^{\circ}C$

Explanation:

Given

Mass of ice $m=60 gm$

mass of water $M=760 gm$

Initial Temperature of water $T_i=20^{\circ}C$

Let T be the Final Temperature of mixture

Latent heat of Fusion $L=334 J/gm$

heat required to melt ice completely is

$Q_1=60\times 334=20.04 kJ$

Heat released by water is taken by ice thus

$Mc_{water}(20-T)=mL+mc_{water}(T-0)$

$0.76\times 4.184\cdot (20-T)=20.04+0.06\times 4.184\cdot (T)$

$T=12.69^{\circ}C$

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Answer:

2.12/R mW

Explanation:

The electrical power, P generated by the rod is

P = B²L²v²/R where B = magnetic field = 0.575 T, L = length of metal rod = separation of metal rails = 20 cm = 0.2 m, v = velocity of metal rod = 40 cm/s = 0.4 m/s and R = resistance of rod = ?

So, the induced emf on the conductor is

E = BLv

= 0.575 T × 0.2 m × 0.4 m/s

= 0.046 V

= 46 mV

The electrical power, P generated by the rod is

P = B²L²v²/R

= B²L²v²/R

So, P = (0.575 T)² × (0.2 m)² × (0.4 m/s)²

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3 years ago

If a cart goes around a turn at 20 km/h ,what remains constant

Vitek1552 [10]

Answer: 4.speed

Explanation:

In this case, we know that the cart remains at a constant 20km/h.

Now, one could say that "the velocity remains constant, because it always is 20km/h"

But remember that velocity is a vector, so this has a direction, and if the cart is going around a turn, then the direction of motion is changing, which tell us that there is acceleration.

But the module of the velocity, the speed, remains constant at 20km/h.

Then the correct option is 4, speed.

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Only one symbol must be included in every circuit diagram. What symbol is it?

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Answer:

A symbol for a battery.

Explanation:

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A point charge +2Q is at the origin and a point charge −Q is located along the x axis at x = d as in the figure below. Find a sy

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Answer: A symbolic expression for the net force on a third point charge +Q located along the y axis

$F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}$

Explanation:

Let the force on +Q charge y-axis due to +2Q charge be $F_1$ and force on +Q charge y axis due to -Q charge on x-axis be $F_2$ .

Distance between the +2Q charge and +Q charge = d units

Distance between the -Q charge and +Q charge = $\sqrt{2}d$ units

$k_e$ = Coulomb constant

$F_1=k_e\frac{(+2Q)(+Q)}{d^2}=k_e\frac{+2Q^2}{d^2} N$

$F_2=k_e\frac{(-Q)(+Q)}{(\sqrt{2}d)^2}=k_e\frac{-Q^2}{2d^2} N$

Net force on +Q charge on y-axis is:

$F_x=F_2sin 45^o=k_e\frac{-Q^2}{2d^2}\times \frac{1}{\sqrt{2}} N$

$F_y=F_1-F_2cos45^o$

$F_y=(F_1-F_2cos45^o)=(k_e\frac{+2Q^2}{d^2})-(k_e\frac{-Q^2}{2d^2}\frac{1}{\sqrt{2}})$

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The net froce on the +Q charge on y-axis is

$F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}$

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