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3 years ago

Similarity between heat and temperature

Physics

1 answer:

pentagon [3]3 years ago

7 0

Heat is a temperature and temperature has many factors such and cold, warm, and hot.

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7 One lb of Refrigerant 134a contained within a piston–cylinder assembly undergoes a process from a state where the temperature

julsineya [31]

Answer:

$\Delta S = S_2-S_1 = 0.08835 \frac {BTU} {Lb ^ {\circ} R}$

Explanation:

A) In order to solve the table it is necessary to consult tables A11-E and A10E for refrigerant R134-a

In this way we obtain that:

$S_1 = S_f = 0.0648 \frac {BTU} {Lb. ^ {\circ} R}$

In this way,

$S_2 = S_f + x_2 (S_g-S_f) = 0.0902 + 0.5 (0.2161-0.0902)$

$S_2 = 0.15315 \frac {BTU} {Lb ^ {\circ} R}$

In this way the entropy change is,

$\Delta S = S_2-S_1 = 0.08835 \frac {BTU} {Lb. ^ {\°} R}$

B) Whenever entropy yields a positive result, the process can be carried out adiabatically.

7 0

3 years ago

A circular area with a radius of 6.50 cm lies in the xy-plane. What is the magnitude of the magnetic flux through this circle du

White raven [17]

Answer:

$a. \phi _B=3.0528\times10^-^3T\ m^2\\ \\b. \phi_B=1.83299\times10^-^3T\ m^2\\\\c.\phi_B=0$

Explanation:

#Consider a circular area of radius $R=2.98cm$ in the xy-plane at z=0. This means all the are vector points toward the +ve z-axis.

a. first, find the magnetic flux if the magnetic field has a magnitude of $B=0.23T$ and points toward the +ve z-axis. The angle between the magnetic field and the area is $\theta=0$ . Hence the magnetic flux:-

$\phi _B=\int {\bar B} . d\bar A \\=\int BdAcos(\theta)=BAcos(0)=BA\\\\=\pi R^2B=\pi(6.50\times10^-^3m)^2(0.230T)\\=3.0528\times10^-^3 T\ m^2$

Hence flux magnitude in $+z$ direction is $3.0528\times10^-^3T \ m^2$

b. We now find the magnetic flux when the field has a magnitude of B=0.230T and points at an angle of $\theta=53.1\textdegree$ from the $+z$ direction.

Magnetic flux is calculated as:

$\phi _B=\int\bar B \bar dA\\=\int BdAcos (\theta)=BAcos(0)=BA\\=\pi R^2B=\pi(6.50\times 10^-^2m)^2(0.230T)\\=1.83299\times 10^-^3 T \ m^2$

Hence the flux at an angle of $53.1\textdegree$ is $1.83299\times 10^-^3T \ m^2$

c. We now need to find the magnetic flux if the field has a magnitue of B=0.230T and points in the direction of +y-direction. As with the previous parts, the magnetic flux will be calculated as:

$\phi_B= \int\bar B \times d\bar A\\=\int BdAcos(\theta)\\=BAcos(90\textdegree)\\=0$

Hence the magnetic flux in the +y-direction is zero.

3 0

3 years ago

Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a 0.210-kg ball with zero net charge was

tigry1 [53]

Answer:

$\Delta V=316167V$

Explanation:

The difference of electric potential between two points is given by the formula $\Delta V=Ed$ , where d is the distance between them and E the electric field in that region, assuming it's constant.

The electric field formula is $E=\frac{F}{q}$ , where F is the force experimented by a charge q placed in it.

Putting this together we have $\Delta V=\frac{Fd}{q}$ , so we need to obtain the electric force the charged ball is experimenting.

On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight W, giving a net force $N=W-F$ . On the first drop only W acts, while on the second drop is N that acts.

Using the equation for accelerated motion (departing from rest) $d=\frac{at^2}{2}$ , so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting:

$a_1=\frac{2d}{t_1^2}$

$a_2=\frac{2d}{t_2^2}$

These relate with the forces by Newton's 2nd Law:

$W=ma_1$

$N=ma_2$

Putting all together:

$N=W-F=ma_1-F=ma_2$

Which means:

$F=ma_1-ma_2=m(a_1-a_2)=m(\frac{2d}{t_1^2}-\frac{2d}{t_2^2})=2md(\frac{1}{t_1^2}-\frac{1}{t_2^2})$

And finally we substitute:

$\Delta V=\frac{Fd}{q}=\frac{2md^2}{q}(\frac{1}{t_1^2}-\frac{1}{t_2^2})$

Which for our values means:

$\Delta V=\frac{2(0.21Kg)(1m)^2}{7.7\times10^{-6}C}(\frac{1}{(0.35s)^2}-\frac{1}{(0.65s)^2})=316167V$

7 0

3 years ago

The _______ principle encourages us to resolve a set of stimuli, such as trees across a ridgeline, into smoothly flowing pattern

kotegsom [21]

Answer:

The answer is continuity ( D )

Explanation:

PLZ MARK AS BRAINLIEST

5 0

3 years ago

A ball is tossed with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball wh

irina1246 [14]

(a) Zero

When the ball reaches its highest point, the direction of motion of the ball reverses (from upward to downward). This means that the velocity is changing sign: this also means that at that moment, the velocity must be zero.

This can be also understood in terms of conservation of energy: when the ball is tossed up, initially it has kinetic energy

$K=\frac{1}{2}mv^2$

where m is the ball's mass and v is the initial speed. As it goes up, this kinetic energy is converted into potential energy, and when the ball reaches the highest point, all the kinetic energy has been converted into potential energy:

$U=mgh$

where g is the gravitational acceleration and h is the height of the ball at highest point. At that point, therefore, the potential energy is maximum, while the kinetic energy is zero, and so the velocity is also zero.

(b) 9.8 m/s upward

We can find the velocity of the ball 1 s before reaching its highest point by using the equation:

$a=\frac{v-u}{t}$

where

a = g = -9.8 m/s^2 is the acceleration due to gravity, which is negative since it points downward

v = 0 is the final velocity (at the highest point)

u is the initial velocity

t = 1 s is the time interval

Solving for u, we find

$u=v-at = 0 -(-9.8 m/s^2)(1 s)= +9.8 m/s$

and the positive sign means it points upward.

(c) -9.8 m/s

The change in velocity during the 1-s interval is given by

$\Delta v = v -u$

where

v = 0 is the final velocity (at the highest point)

u = 9.8 m/s is the initial velocity

Substituting, we find

$\Delta v = 0 - (+9.8 m/s)=-9.8 m/s$

(d) 9.8 m/s downward

We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:

$a=\frac{v-u}{t}$

where this time we have

a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative

v is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

t = 1 s is the time interval

Solving for v, we find

$v = u+at = 0 +(-9.8 m/s^2)(1 s)= -9.8 m/s$

and the negative sign means it points downward.

(e) -9.8 m/s

The change in velocity during the 1-s interval is given by

$\Delta v = v -u$

where here we have

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

Substituting, we find

$\Delta v = -9.8 m/s - 0=-9.8 m/s$

(f) -19.6 m/s

The change in velocity during the overall 2-s interval is given by

$\Delta v = v -u$

where in this case we have:

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = +9.8 m/s is the initial velocity (1 s before reaching the highest point)

Substituting, we find

$\Delta v = -9.8 m/s - (+9.8 m/s)=-19.6 m/s$

(g) -9.8 m/s^2

There is always one force acting on the ball during the motion: the force of gravity, which is given by

$F=mg$

where

m is the mass of the ball

g = -9.8 m/s^2 is the acceleration due to gravity

According to Newton's second law, the resultant of the forces acting on the body is equal to the product of mass and acceleration (a), so

$mg = ma$

which means that the acceleration is

$a= g = -9.8 m/s^2$

and the negative sign means it points downward.

7 0

4 years ago