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3 years ago

A 1.11 kg piece of aluminum at 78.3 c is put into a glass with 0.210 kg of water at 15.0

Physics

1 answer:

ValentinkaMS [17]3 years ago

7 0

M = mass of aluminium = 1.11 kg

$c_{a}$ = specific heat of aluminium = 900

$T_{ai}$ = initial temperature of aluminium = 78.3 c

m = mass of water = 0.210 kg

$c_{w}$ = specific heat of water = 4186

$T_{wi}$ = initial temperature of water = 15 c

T = final equilibrium temperature = ?

using conservation of heat

Heat lost by aluminium = heat gained by water

M $c_{a}$ ( $T_{ai}$ - T) = m $c_{w}$ (T - $T_{wi}$ )

(1.11) (900) (78.3 - T) = (0.210) (4186) (T - 15)

T = 48.7 c

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tia_tia [17]

Answer:

f=8.219*10^{8}Hz

Explanation:

We are going to use the formula v=fλ

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also when a wave travels from one medium to another, the wavelength changes while the frequency remains the same.

calculating for the frequency of the wave in air also gives us the frequency in the window glass.

f=\frac{v}{λ}

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3 years ago

El. A horizontally directed force of 40 N is used to pull a box a distance of 2.5 m across a tabletop. How much work is done by

PIT_PIT [208]

The same formula of work can be applied to all the questions. The answers are:

E1. 100J

E2. 40N

E3. 5m

E4. a.) 360 J b.) 240 J c.) 120 J

El. If a horizontally directed force of 40 N is used to pull a box a distance of 2.5 m across a tabletop. The formula to get the much work that will be done by the 40 - N force will be

Work done = force x distance

Work done = 40 x 2.5

Work done = 100 J

E2. If a woman does 160 J of work to move a table 4 m across the floor. We will use the same formula to calculate the magnitude of the force that the woman applied to the table assuming the force is applied in the horizontal direction.

Work done = force x distance

160 = 4F

F = 160/4

Force F = 40 N

E3. Given that a force of 60 N used to push a chair across a room does 300 J of work. Same formula to get how far the chair move in this process.

Work done = force x distance

300 = 60 x distance

distance = 300/60

Distance = 5 m

Therefore, the chair moved 5m away.

E4. Given that a rope applies a horizontal force of 180 N to pull a crate a distance of 2 m across the floor. And a frictional force of 120 N opposes this motion.

a. The work done by the force applied by the rope can be found by

W = F x S

W = 180 x 2

W = 360 J

b. What is the work done by the frictional force?

W = $F_{r}$ x s

W = 120 x 2

W = 240 J

c. What is the total work done on the crate?

W = (F - $F_{r}$ ) x distance

Where $F_{r}$ = frictional force

Substitute all the parameters

W = (180 - 120) x 2

W = 60 x 2

W = 120 J

Learn more about work here: brainly.com/question/8119756

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2 years ago

True or false. Students with a Learners License may not receive a motorcycle endorsement.

viktelen [127]

This is a true statement

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3 years ago

24 A uniform electric field of magnitude 1.1×104 N/C is perpendicular to a square sheet with sides 2.0 m long. What is the elect

Tatiana [17]

Answer:

$44,000 Nm^2/C$

Explanation:

The electric flux through a certain surface is given by (for a uniform field):

$\Phi = EA cos \theta$

where:

E is the magnitude of the electric field

A is the area of the surface

$\theta$ is the angle between the direction of the field and of the normal to the surface

In this problem, we have:

$E=1.1\cdot 10^4 N/C$ is the electric field

L = 2.0 m is the side of the sheet, so the area is

$A=L^2=(2.0)^2=4.0 m^2$

$\theta=0^{\circ}$ , since the electric field is perpendicular to the surface

Therefore, the electric flux is

$\Phi =(1.1\cdot 10^4)(4.0)(cos 0^{\circ})=44,000 Nm^2/C$

4 0

3 years ago

xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi

slega [8]

Answer:

Approximately $\rm 90\; kJ$ .

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction ( $E^{\circ}(\text{cell})$ ) is equal to

$E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$ .

There are two half-reactions in this question. $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ and $\rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb$ . Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of $E^{\circ}(\text{cell})$ should be positive.

In this case, $E^{\circ}(\text{cell})$ is positive only if $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ is the reaction takes place at the cathode. The net reaction would be

$\rm Cu^{2+} + Pb \to Cu + Pb^{2+}$ .

Its cell potential would be equal to $0.339 - (-0.130) = \rm 0.469\; V$ .

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

$\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})$ ,

where

$n$ is the number moles of electrons transferred for each mole of the reaction. In this case the value of $n$ is $2$ as in the half-reactions.
$F$ is Faraday's Constant (approximately $96485.33212\; \rm C \cdot mol^{-1}$ .)

$\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}$ .

5 0

3 years ago