Describe the relationship between a moving objects mass and its kinetic energy

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

Starting from rest, a small block of mass m slides frictionlessly down a circular wedge of mass M and radius R which is placed o

guapka [62]

Answer:

Part a)

$V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

$v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

$y_{cm} = \frac{mR}{m + M}$

Explanation:

PART A)

As we know that there is no external force on the system of two masses in horizontal direction

So here the two masses will have its momentum conserved in horizontal direction

So we have

$mv + MV = 0$

Also we know that here no friction force on the system so total energy will always remains conserved

So we have

$\frac{1}{2}mv^2 + \frac{1}{2}MV^2 = mgR$

now we have

$\frac{1}{2}m(\frac{MV}{m})^2 + \frac{1}{2}MV^2 = mgR$

$\frac{1}{2}MV^2(\frac{M}{m} + 1) = mgR$

so we have

$V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

and another block has speed

$v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

$y_{cm} = \frac{mR}{m + M}$

7 0

3 years ago

To practice Problem-Solving Strategy 21.1 Coulomb's Law. Three charged particles are placed at each of three corners of an equil

Salsk061 [2.6K]

Answer:

The force felt by charge 3 is F=(-5.6*10⁻⁶,3.36⁻⁵)N

Explanation:

As the superposition principle applies to static charges, we can find the net electric force as the sum of the two forces felt by q3.

Looking at the drawing and knowing that they form an equilateral triangle of lenght 4 we can conclude that each internal angle is 60°.

So, the positions in our coordinate system are:

$r_1=(0,0)\\r_2=(4\ cos(60\°),4\ sin(60\°))\\r_3=(4,0)\\$

Now using Coulomb's force:

$F_{ij}=\frac{-kq_iq_j}{d^2}(\vv{r}_j-\vv{r}_i)$

Where d=4, q1 = -7.8*10⁻⁹C, q2 = -15.6 *10⁻⁹C, q3 = 8.0 *10⁻⁹C, k=8.98*10⁹, e0=8.8*10¹⁰:

Replacing we get 2 equations:

$F_{13}=\frac{-kq_1q_3}{d^2}(\vv{r}_1-\vv{r}_3)=\frac{-kq_1q_2}{d^2}(-0.04\ cos(60\°),-0.04\ sin(60\°))\\\\F_{23}=\frac{-kq_2q_3}{d^2}(\vv{r}_1-\vv{r}_3)=\frac{-kq_1q_2}{d^2}(0.04-0.04\ cos(60\°),-0.04\ sin(60\°))\\$

To work with the sam

F=∑F_i=3.5*10⁻⁴(0.023,0.032)+7*10⁻⁴(-0.016,0.032)=

=((3.5*10⁻⁴-7*10⁻⁴)*0.016,(3.5*10⁻⁴+7*10⁻⁴)*0.032)=

F=(-5.6*10⁻⁶,3.36⁻⁵)N

7 0

3 years ago

Two hockey pucks, labeled A and B, are initially at rest on asmooth ice surface and are separated by a distance of18.0 m. Simult

zlopas [31]

Answer:

8.505 m

Explanation:

Let V1 and V2 be velocities of puck A and B respectively

Since A and B move in the same direction, so the relative velocity will be V1+V2=3.5+3.9=7.4m/s

Or

Vr=7.4 m/s

Distance=S= 18 m

Time =t=?

S=Vr×t

==> t=S/Vr

==> t= 18/7.4=2.43 sec

At this time both will strike together

Distance by puck A

V1=3.5 m/s

Time=t= 2.43 sec

Distance covered=d=?

d=V1×t=3.5×2.43=8.505 m

So, puck A will cover 8.505 meters before collision

6 0

3 years ago

99 POINTS PLEASE HELP

yarga [219]

the answer to your problem is work

3 0

3 years ago

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