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PSYCHO15rus [73]
2 years ago
9

The specific heat of substance A is greater than that of substance B. Both A and B are at the same initial temperature when equa

l amounts of energy are added to them. Assuming no melting or vaporization occurs, which of the following can be concluded about the final temperature TA of substance A and the final temperature TB of substance B?
a) TA > TB
b) TA < TB
c) TA = TB
d) More information is needed
Physics
1 answer:
Sonja [21]2 years ago
4 0

Answer:

m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

Explanation:

For this case we know the following info:

c_{pA} > c_{pB}

Where c means specific heat for the substance A and B.

We also know that the initial temperatures for both sustances are equal:

T_{iA}= T_{iB}

We assume that we don't have melting or vaporization in the 2 substances. So we just have presence of sensible heat given by this formula:

Q = m c_p \Delta T

And for this case we know that Both A and B are at the same initial temperature when equal amounts of energy are added to them, so then we have this:

Q_A = Q_B

And if we replace the formula for sensible heat we got:

m_A c_{pA} \Delta T_A = m_B c_{pB} \Delta T_B

And if we replace for the change of the temperature we got:

m_A c_{pA} (T_{fA} -T_{iA}) = m_B c_{pB} (T_{fB}- T_{iB})

And since T_{iA}= T_{iB}= T we have this:

m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

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If the mass of a material is 45 grams and the volume of the material is 8 cm^3, what would the density of the material be?
Svet_ta [14]
The density of the substance is the ratio of its mass over the space it occupies. In mathematical equation, this can be expressed as,

        ρ = m / v

where ρ is density, m is mass, and v is volume. 

Substituting the known values from the given,
 
    ρ = (45 g) / (8 cm³)

     ρ = 5.625 g/cm³

<em>ANSWER: 5.625 g/cm³</em>
3 0
2 years ago
The expressions for e/m and the relative error of e/m due to all of the parameters measured:
bija089 [108]

Answer:

Term 1 = (0.616 × 10⁻⁵)

Term 2 = (7.24 × 10⁻⁵)

Term 3 = (174 × 10⁻⁵)

Term 4 = (317 × 10⁻⁵)

(σ ₑ/ₘ) / (e/m) = (499 × 10⁻⁵) to the appropriate significant figures.

Explanation:

(σ ₑ/ₘ) / (e/m) = (σᵥ /V)² + (2 σᵢ/ɪ)² + (2 σʀ /R)² + (2 σᵣ /r)²

mean measurements

Voltage, V = (403 ± 1) V,

σᵥ = 1 V, V = 403 V

Current, I = (2.35 ± 0.01) A

σᵢ = 0.01 A, I = 2.35 A

Coils radius, R = (14.4 ± 0.3) cm

σʀ = 0.3 cm, R = 14.4 cm

Curvature of the electron trajectory, r = (7.1 ± 0.2) cm.

σᵣ = 0.2 cm, r = 7.1 cm

Term 1 = (σᵥ /V)² = (1/403)² = 0.0000061573 = (0.616 × 10⁻⁵)

Term 2 = (2 σᵢ/ɪ)² = (2×0.01/2.35)² = 0.000072431 = (7.24 × 10⁻⁵)

Term 3 = (2 σʀ /R)² = (2×0.3/14.4)² = 0.0017361111 = (174 × 10⁻⁵)

Term 4 = (2 σᵣ /r)² = (2×0.2/7.1)² = 0.0031739734 = (317 × 10⁻⁵)

The relative value of the e/m ratio is a sum of all the calculated terms.

(σ ₑ/ₘ) / (e/m)

= (0.616 + 7.24 + 174 + 317) × 10⁻⁵

= (498.856 × 10⁻⁵)

= (499 × 10⁻⁵) to the appropriate significant figures.

Hope this Helps!!!

6 0
2 years ago
What is the explanation for how a modern transmission electron microscope (TEM) can achieve a resolution of about 0.2 nanometers
IgorC [24]

Answer:

Explanation:

A simple light microscope uses light for imaging of objects where as a transmission electron microscope uses a monochromatic beam of electrons.

This beam is passed by a magnetic field which is very strong and thus act as a lens.

Its resolution of very high which is about 0.2 nanometers because of the separation between two atoms.

Because of this reason its resolution is about 1000 times greater than light microscope.

3 0
2 years ago
A firecracker breaks up into several pieces, one of which has a mass of 200 g and flies off along the x-axis with a speed of 82.
MakcuM [25]

Answer:

The magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.

Explanation:

Hi there!

The total momentum is calculated as the sum of the momenta of the pieces.

The momentum of each piece is calculated as follows:

p = m · v

Where:

p = momentum.

m =  mass.

v = velocity.

The momentum is a vector. The 200 g-piece flies along the x-axis then, its momentum will be:

p = (m · v, 0)

p = (0.200 kg · 82.0 m/s, 0)

p = (16.4 kg m/s, 0)

The 300 g-piece flies along the y-axis. Its momentum vector will be:

p =(0, m · v)

p = (0, 0.300 kg · 45.0 m/s)

p = (0, 13.5 kg m/s)

The total momentum is the sum of each momentum:

Total momentum = (16.4 kg m/s, 0) + (0, 13.5 kg m/s)

Total momentum = (16.4 kg m/s + 0, 0 + 13.5 kg m/s)

Total momentum = (16.4 kg m/s, 13.5 kg m/s)

The magnitude of the total momentum is calculated as follows:

|p| = \sqrt{(16.4 kgm/s)^2+(13.5 kg m/s)^2}= 21.2 kg m/s

The direction of the momentum vector is calculated using trigonometry:

cos θ = px/p

Where px is the horizontal component of the total momentum and p is the magnitude of the total momentum.

cos θ = 16.4 kg m/s / 21.2 kg m/s

θ = 39.3  (39.5° if we do not round the magnitude of the total momentum)

Then, the magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.

 

6 0
2 years ago
29. 2072 Set C Q.No. 10c
Annette [7]

Answer:

90.2^{\circ}C

Explanation:

Considering the thermal conductivity of aluminium and brass as k_{al}=205 W/mK and k_{br}=109 W/mk respectively  

The temperature at the end of aluminium and brass are given as T_{al}=150^{\circ}C and T_{br}=20^{\circ}C respectively with length of rod L=1.3 m , Length of aluminium L_{al}=0.8 m, length of brass L_{br}=0.5 m and letting temperature at steady state be T

At steady state, thermal conductivity of both aluminium and brass are same hence

H_{br}=H_{al}

k_{al}A\frac {T_H-T}{L_{al}}= k_{br}A\frac {T-T_H}{L_{br}}

Upon re-arranging

T=\frac {k_{al}L_{al}T_{br}+k_{al}L_{br}T_{al}}{k_{br}L_{al}+k_{al}L_{br}}

(205)\frac {150-T}{0.8}=109\frac {T-20}{0.5}

T=\frac {(109*0.8*20)+(205*0.5*150)}{(109*0.8)+(205*0.5)}

T=90.2^{\circ}C

Therefore, the temperatures at which the metals are joined is 90.2^{\circ}C

6 0
2 years ago
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