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1 year ago

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Brass is an alloy made from copper and zinc a 0.59 kg brass sample at 98.0 is dropped into 2.80 kg of water at 5.0 c if the equi

librium temperature is 6.8 what is the specific heat capacity

2 answers:

MariettaO [177]1 year ago

6 0

Answer:

393.399 J/kg.°C

Explanation:

Specific heat capacity: This is the quantity of heat required to raise the temperature of a unit mass of a substance through a degree rise in temperature.

Heat lost by the brass = heat gained by water

CM(t₁-t₃) = cm(t₃-t₂)........................ Equation 1

Where C = specific heat capacity of the brass, M = mass of the brass, t₁ = initial temperature of the brass, t₂ = initial temperature of water, t₃ = temperature of the mixture.

Making C the subject of the equation

C = cm(t₃-t₂)/M(t₁-t₃)............................... Equation 2

Given: M = 0.59 kg, m = 2.8 kg, t₁ = 98 °C, t₂ = 5.0 °C, t₃ = 6.8 °C

Constant: c = 4200 J/kg.°C

Substitute into equation 2,

C = 2.8×4200(6.8-5.0)/0.59(98-6.8)

C = 21168/53.808

C = 393.399 J/kg.°C

Thus the specific heat capacity of the brass = 393.399 J/kg.°C

zhannawk [14.2K]1 year ago

3 0

The specific heat capacity of brass would be ranked between 0 and infinity

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A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha

Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3 s

the formula for the time period of the pendulum is given by

$T = 2\pi \sqrt{\frac{I}{mgd}}$ .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

$I = \frac{mL^{2}}{12}+ md^{2}$

Substituting the values in equation (1)

$3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}$

$9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )$

12d² -26.84 d + 2.25 = 0

$d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}$

$d=\frac{26.84\pm 24.75}{24}$

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0

1 year ago

Two speakers, A and B, produce identical sound waves. A listener is 3.2 m away from speaker A. The listener finds the lowest fre

earnstyle [38]

Answer:

0.83 m or 5.57 m

Explanation:

Destructive interference will occur when the distances from the speakers differ by 1/2 wavelength.

The length of 1 cycle of 72.4 Hz is ...

λ = v/f = (343 m/s)/(72.4 Hz) ≈ 4.738 m

So, the distance of the listener from speaker B is ...

3.2 m ± (4.738 m)/2 = {0.83 m, 5.57 m} . . . either of these distances

_____

The location could be at additional multiples of 4.738 m, but we think not. The sound intensity drops off with the square of the distance from the speaker, so identical sound waves from the speakers will sound quite different at different distances from the speakers. For best interference, the distances need to be as close to the same as possible. That will be at 3.2 m and 5.57 m.

_____

<em>Comment on the speed of sound</em>

We don't know what speed you are to use for the speed of sound. We have used 343 m/s. Some sources use 340 m/s, which will give a result different by 2 or 3 cm.

8 0

2 years ago

You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves along a frictionless track. however, af

djyliett [7]

I attached the missing picture.
We can figure this one out using the law of conservation of energy.
At point A the car would have potential energy and kinetic energy.
$A: mgh_1+\frac{mv_1^2}{2}$
Then, while the car is traveling down the track it loses some of its initial energy due to friction:
$W_f=F_f\cdot L$
So, we know that the car is approaching the point B with the following amount of energy:
$mgh_1+\frac{mv_1^2}{2}- F_fL$
The law of conservation of energy tells us that this energy must the same as the energy at point B.
The energy at point B is the sum of car's kinetic and potential energy:
$B: mgh_2+\frac{mv_2}{2}$
As said before this energy must be the same as the energy of a car approaching the loop:
$mgh_2+\frac{mv_2}{2}=mgh_1+\frac{mv_1^2}{2}- F_fL$
Now we solve the equation for $v_1$ :
$v_1^2=2g(h_2-h_1)+v_2^2+\frac{2F_fL}{m}\\
v_1^2=39.23\\
v_1=\sqrt{39.23}=6.26\frac{m}{s}$

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1 year ago

Read 2 more answers

Dylan has two cubes of iron. The larger cube has twice the mass of the smaller cube. He measures the smaller cube. Its mass is 2

liubo4ka [24]

Answer:

The volume of the larger cube is 5.08 g/cm³.

Explanation:

Given that,

Mass of smaller cube = 20 g

Density of smaller cube $\rho= 7.87 g/cm^2$

Dylan has two cubes of iron.

The larger cube has twice the mass of the smaller cube.

$M_{l}=2m_{s}$

Density is same for both cubes because both cubes are same material.

The density is equal to the mass divided by the volume.

$\rho=\dfrac{m}{V}$

$V=\dfrac{m}{\rho}$

Where, V = volume

m = mass

$\rho=density$

We need to calculate the volume of smaller mass

The volume of smaller mass

$V_{s}=\dfrac{m_{s}}{\rho_{s}}$

$V_{s}=\dfrac{20}{7.87}$

$V_{s}=2.54\ cm^3$

Now, We need to calculate the volume of large cube

$V_{l}=\dfrac{m_{l}}{\rho_{l}}$

$V_{l}=\dfrac{2\times20}{7.87}$

$V_{l}=5.08\ g/cm^3$

Hence, The volume of the larger cube is 5.08 g/cm³.

8 0

2 years ago

Two chargedparticles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged part

erica [24]

Answer:

Two possible points

<em>x= 0.67 cm to the right of q1</em>

<em>x= 2 cm to the left of q1</em>

Explanation:

<u>Electrostatic Forces</u>

If two point charges q1 and q2 are at a distance d, there is an electrostatic force between them with magnitude

$\displaystyle f=k\frac{q_1\ q_2}{d^2}$

We need to place a charge q3 someplace between q1 and q2 so the net force on it is zero, thus the force from 1 to 3 (F13) equals to the force from 2 to 3 (F23). The charge q3 is assumed to be placed at a distance x to the right of q1, and (2 cm - x) to the left of q2. Let's compute both forces recalling that q1=1, q2=4q and q3=q.

$\displaystyle F_{13}=k\frac{q_1\ q_3}{d_{13}^2}$

$\displaystyle F_{13}=k\frac{(q)\ (q)}{x^2}$

$\displaystyle F_{23}=k\frac{q_2\ q_3}{d_{23}^2}$

$\displaystyle F_{23}=k\frac{(q)(4q)}{(0.02-x)^2}$

$\displaystyle F_{23}=\frac{4k\ q^2}{(0.02-x)^2}$

Equating

$\displaystyle F_{13}=F_{23}$

$\displaystyle \frac{K\ q^2}{x^2}=\frac{4K\ q^2}{(0.02-x)^2}$

Operating and simplifying

$\displaystyle (0.02-x)^2=4x^2$

To solve for x, we must take square roots in boths sides of the equation. It's very important to recall the square root has two possible signs, because it will lead us to 2 possible answer to the problem.

$\displaystyle 0.02-x=\pm 2x$

Assuming the positive sign :

$\displaystyle 0.02-x= 2x$

$\displaystyle 3x=0.02$

$\displaystyle x=0.00667\ m$

$x=0.67\ cm$

Since x is positive, the charge q3 has zero net force between charges q1 and q2. Now, we set the square root as negative

$\displaystyle 0.02-x=-2x$

$\displaystyle x=-0.02\ m$

$\displaystyle x=-2\ cm$

The negative sign of x means q3 is located to the left of q1 (assumed in the origin).

5 0

2 years ago