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3 years ago

11

50 grams of ice cubes at -15°C are used to chill a water at 30°C with mass mH20 = 200 g. Assume that the water is kept in a foam

container so that heat loss can be ignored. Find the final temperature.

1 answer:

Arada [10]3 years ago

5 0

Answer : The final temperature is, $25.0^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ice = $2.09J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ice = 50 g

$m_2$ = mass of water = 200 g

$T_f$ = final temperature = ?

$T_1$ = initial temperature of ice = $-15^oC$

$T_2$ = initial temperature of water = $30^oC$

Now put all the given values in the above formula, we get:

$50g\times 2.09J/g^oC\times (T_f-(-15))^oC=-200g\times 4.184J/g^oC\times (T_f-30)^oC$

$T_f=25.0^oC$

Therefore, the final temperature is, $25.0^oC$

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Explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

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Using formula of energy

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$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}$

$E=3.315\times10^{-17}\ J$

(b). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}$

$E=6.709\times10^{-21}\ J$

(c). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}$

$E=3.315\times10^{-11}\ J$

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}$

$E=6.709\times10^{-9}\ J$

Hence, This is the required solution.

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3 years ago

5280 feet to meters

lesya [120]

Answer:

1m= 3.2 feets

? = 5280 feets

let ? be n

cross multiply

3.2n = 1×5280

n= 5280÷3.2

n= 1650

so, 5280 feets is equal to 1650 metres

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3 years ago

A chair of mass 12.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force F

yarga [219]

Answer:

Explanation:

mass, m = 12 kg

Force, F = 40 N

θ = 37° below the horizontal

(a)

Diagram is attached

(b) Let N be the normal reaction

According to the diagram

N + F Sin θ = m g

N = mg - F Sin θ

N = 12 x 9.8 - 40 x Sin 37

N = 117.6 - 24.07

N = 93.53

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3 years ago

A boulder released by a rockslide rolls 60 m in 4 s. What is its average speed?

Gelneren [198K]

Answer:

15 m per second

900m per minute

54,000 per hour

Explanation:

60 divided by 4 to get per second then times 60 for per minutes

then times 60 to get per hour

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3 years ago

A cart moving at 10 m/s is brought to a stop by the force plotted in the force-time graph shown here. Find the impulse and the a

Elena L [17]

Answer:

Impulse = 88 kg m/s

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Explanation:

<u>We are given a graph of Force vs. Time. Looking at the graph we can see that the Force acts approximately between the time interval from 1sec to 4sec. </u>

Newton's Second Law relates an object's acceleration as a function of both the object's mass and the applied net force on the object. It is expressed as:

$F=ma$ Eqn. (1)

where

$F$ : is the Net Force in Newtons ( $N$ )

$m$ : is the mass ( $kg$ )

$a$ : is the acceleration ( $m/s^2$ )

We also know that the acceleration is denoted by the velocity ( $v$ ) of an object as a function of time ( $t$ ) with

$a=\frac{v}{t}$ Eqn. (2)

Now substituting Eqn. (2) into Eqn. (1) we have

$F=m\frac{v}{t}\\ \\Ft=mv$ Eqn. (3)

However since in Eqn. (3) the time-variable is present, as a result the left hand side (i.e. $Ft$ is in fact the Impulse $J$ of the cart ), whilst the right hand side denotes the change in momentum of the cart, which by definition gives as the impulse. Also from the graph we can say that the Net Force is approximately ≈ $22N$ and $t=4 sec$ (thus just before the cut-off time of the force acting).

Thus to find the Impulse we have:

$J=Ft\\J=(22N)(4sec)\\J=88 kg m/s$

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Then, we know that the cart is moving at $v=10 m/s$ . Plugging in the values in Eqn. (3) we have:

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So the mass of the cart is $m=8.8kg$ .

8 0

3 years ago