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An insulated Thermos contains 140 cm3 of hot coffee at 85.0°C. You put in a 15.0 g ice cube at its melting point to cool the cof

Pavel [41]

Answer:

$T = 69^o C$

Explanation:

Here at thermal equilibrium we can say that thermal energy given by Hot coffee is equal to the thermal energy absorbed by ice cubes

So here we have

$Q_{ice} = Q_{coffee}$

now since ice cubes are added into coffee when it is at melting temperature

So here we can say that final temperature of coffee is T degree C

Now we have

$m_1L + m_1c_1\Delta T_1 = m_2c_2\Delta T_2$

here we have

$m_1 = 15 gram$

L = 333 kJ/kg = 333 J/g[/tex]

$c_1 = c_2 = 4186 J/kg C = 4.186 J/g C$

$\Delta T_1 = T - 0$

$\Delta T_2 = 85 - T$

now we have

$15(333) + 15(4.186)(T - 0) = 140(4.186)(85 - T)$

$4995 + 62.79T = 49813.4 - 586.04T$

$648.83 T = 44818.4$

$T = 69^o C$

6 0

3 years ago

The 10-lb block A attains a velocity of 2ft/s in 5 seconds, starting from rest. Determine the tension in the cord and the coeffi

Nezavi [6.7K]

Answer:

Explanation:

Let T be the tension in the cord.

Impulse by cord = change in momentum of block A .

T x 5s = 10 ( 2 -0) = 20

T = 4 poundal .

acceleration of block B = 2 / 5 = 0.4 m /s²

Net force applied on A = m ( g + a ) where m is mass of block B , a is acceleration of block B .

= 8 ( 32 + .4 ) = 259.2 poundal

Frictional force on block A = 259.2 - 4 = 255.2 poundal

μ x 10 x 32 = 255.2

320μ = 255.2

μ =0 .8 .

8 0

3 years ago

Large electric fields in cell membranes cause ions to move through the cell wall. The field strength in a typical membrane is 1.

serg [7]

Explanation:

The given data is as follows.

Electric field (E) = $1 \times 10^{7} N/C$

Charge (e) = $1.6 \times 10^{-19}$ C

Formula to calculate the magnitude of force is as follows.

F = qE

= $1.6 \times 10^{-19} \times 1 \times 10^{7} N/C$

= $1.6 \times 10^{-12} N$

Therefore, we can conclude that magnitude of the force on a calcium ion with charge +e is $1.6 \times 10^{-12} N$ .

6 0

3 years ago

65. A length of wire is bent into a closed loop and a magnet is plunged into it, inducing a voltage and, consequently, a current

natulia [17]

Answer:

Resistance of the second wire is twice the first wire.

Explanation:

Let us first see the formula of resistance;

R = pxL/A

Here L is the lenght of the wire, A the area and p is the resistivity of wire.

As we are given that the length of second wire is double than that of the first wire, hence the resistance of second wire would be double.

Since we have two loop in second case, inducing double voltage but as resistance is doubled so the current would remain same according to ohms law

I = V/R

6 0

3 years ago

Read 2 more answers

Based on the data collected, explain why a launch angle of 30 degrees (HIGH HORIZONTAL VELOCITY & SMALL TIME OF FLIGHT) will

Gennadij [26K]

Answer:

Explanation:

The horizontal distance traveled by the projectile is given by the formula

$R=\frac{u^{2}Sin2\theta }{g}$

The formula for the time of flight is given by

$T = \frac{2u Sin\theta }{g}$

Case I: when the launch angle is 30°

So, $R_{1}=\frac{u^{2}Sin60 }{g}$

$R_{1}=\frac{0.866u^{2}}{g}$

Horizontal velocity = u Cos 30 = 0.866 u

$T_{1} = \frac{2u Sin30 }{g}=\frac{u}{g}$

Case II: when the launch angle is 60°

$R_{2}=\frac{u^{2}Sin120}{g}$

$R_{2}=\frac{0.866u^{2}}{g}$

Horizontal velocity = u Cos 60 = 0.5 u

$T_{1} = \frac{2u Sin60 }{g}=\frac{1.73 u}{g}$

By observing the case I and case II, we conclude that

R1 = R2

Horizontal velocity 1 > Horizontal velocity 2

T1 < T2

5 0

3 years ago