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Nat2105
9 days ago
7

a fixed mass of a n ideal gas is heated from 50 to 80C at a constant pressure at 1 atm and again at a constant pressure of 3 atm

. for which case do u think the energy required will be greater
Physics
1 answer:
inna [987]9 days ago
8 0

Answer:

The required energy remains identical in both scenarios since the specific heat capacity (Cp) does not change with varying pressure.

Explanation:

Given;

initial temperature, t₁ = 50 °C

final temperature, t₂ = 80 °C

Temperature change, ΔT = 80 °C - 50 °C = 30 °C

Pressure for scenario one = 1 atm

Pressure for scenario two = 3 atm

The energy needed in both scenarios is expressed as;

Q = M*C_p*\delta T

Where;

Cp denotes specific heat capacity, which only varies with temperature and remains unaffected by pressure.

Hence, the energy required remains the same for both scenarios since specific heat capacity (Cp) is pressure-independent.

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What is the sources of error and suggestion on how to overcome it in the hooke's law experiment?
Yuliya22 [1153]
In the study of physics, Hooke's law can be expressed as:

F = kx

This law indicates that the spring force F is proportional to the extension x, with k being the spring constant.

In experiments, this is often examined using the setup illustrated in the included figure. The spring is tested, and a known weight is applied underneath it. This weight exerts a gravitational pull, essentially its weight, on the spring. While the spring elongates, the displacement can be measured using a ruler.

Several potential errors can arise during this experiment. Firstly, the person's measurement reading may be faulty. Digital scales offer greater accuracy as they reduce human error, while ruler readings can be subjective, especially if not viewed at eye level. Additionally, the object's weight may be inaccurately measured if the scale is untrustworthy. Lastly, the measuring equipment may not be correctly calibrated.

6 0
20 hours ago
Simone is walking her dog on a leash. The dog is pulling with a force of 32 N to the right and Simone is pulling backward with a
ValentinkaMS [1144]

Conclusion:

The total net force acting on the objects is 16 N, directed towards the right.

Clarification:

It is stated that,

The force exerted by the dog, F_1 = 32\ N (to the right)

The force exerted by Simone, F_2 = -16\ N (backward)

Here, assume the backward direction is negative and the right direction is positive.

The net force will move in the direction where the larger force is present. The net force can be calculated as:

F=F_1+F_2

F=32+(-16)

F = 16 N

Thus, the net force amounts to 16 N, acting towards the right.

6 0
3 days ago
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A solid conducting sphere carrying charge q has radius a. It is inside a concentric hollow conducting sphere with inner radius b
Softa [913]

Response:

Clarification:

Refer to the diagram indicating the charges on the specified sphere (see attachment).

The electric field at the stated positions is

E(r) = 0 for r≤a.  Equation 1

E(r) = kq/r² for a<r<b.   Equation 2

E(r) = 0 for b<r<c.      Equation 3

E(r) = kq/r² for r>c.    Equation 4.

We understand that electric potential correlates with the electric field through

V = Ed

A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is

E = kQ / r²

Then,

V = Ed,

At d = r = c

Thus,

Vc = (kQ / c²) × c

Vc = kQ / c

As a result, the total charge Q consists of +q, -q, and +q

Hence, Q = q - q + q = q

V = kq / c

B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have

V = kQ/r

V = kQ / b,   noting that r = b

So, Q = q

V = kq / b

C. At r = a

Following from equation 1:

E(r) = 0 for r≤a.  Equation 1

The electric field at the surface of the solid sphere is 0, E = 0N/C

Thus,

V = Ed = 0 V

Consequently, the electric potential at the solid sphere's surface is 0.

D. At r = 0

The electric potential can be determined by

V = kq / r

As r approaches 0,

V = kq / 0

V approaches infinity.

8 0
10 days ago
Determine the final state and temperature of 100 g of water originally at 25.0°c after 50.0 kj of heat have been added to it.
inna [987]
The heat required to raise the temperature of a substance by \Delta T is represented by
Q=m C_p \Delta T
where m stands for the mass of the substance and C_p indicates the specific heat of the substance. In this situation, we possess m=100~g=0.1~Kg and C_p=4.19~KJ/(Kg K), the specific heat of water.
Consequently, we can ascertain the temperature rise \Delta T:
\Delta T = \frac{Q}{m C_p}= \frac{50~KJ}{0.1~Kg cdot 4.19~KJ/(Kg K)}=119~K =119^{\circ}C
Initially, the water's temperature was 25^{\circ}C, so the end temperature should be
T_f = 25^{\circ}C+119^{\circ}C=144^{\circ}C
Thus, the water is expected to be vapor by now.

However, to give a more accurate statement, during the liquid to vapor transition, the heat added to the system is used to break molecular bonds instead of raising the system's temperature. The heat necessary for the phase change from liquid to vapor is expressed as
Q=m C_L=0.1~Kg \cdot 2265~KJ/Kg=226.5~KJ
where C_L denotes the latent heat of vaporization for water.
Nevertheless, the initial heat input of 50 KJ is less than this requirement, indicating there isn't sufficient heat to finish the liquid-vapor transition. Therefore, the water will remain in the liquid-vapor change phase at a temperature of 100^{\circ}C (the temperature at which the phase change begins)

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4 days ago
Which pair of graphs represent the same motion of an object
Ostrovityanka [942]
The correct choice is C
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15 days ago
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