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Furkat [3]
3 years ago
9

Two containers hold equal masses of nitrogen gas at equal temperatures. You supply 10 J of heat to container A while not allowin

g its volume to change, and you supply 10 J of heat to container B while not allowing its pressure to change.
Afterward, is temperature TA greater than, less than, or equal to TB? Explain.
Physics
1 answer:
telo118 [61]3 years ago
5 0

Answer:

TA is greater than TB

T_{A} > T_{B}

Explanation:

Specific heat at constant pressure C_{p} is greater than specific heat at constant volume C_{v}.Therefore for given amount of supplied heat,increase in temperature at constant volume is greater than increase in temperature at constant pressure

Hence

TA is greater than TB

T_{A} > T_{B}

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Harman [31]
Because it is intact and unfractured
4 0
3 years ago
Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103
son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

PE = -\frac{GMm}{R}

As M=m, then

PE = -\frac{Gm^2}{R}

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

U = -\frac{2Gm^2}{R}

dU = -\frac{2Gm^2}{R}

dKE = -dU

Therefore replacing we have that,

\frac{1}{2}mv^2 =\frac{Gm^2}{2R}

Re-arrange to find v,

v= \sqrt{\frac{Gm}{R}}

v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}

v = 816.7m/s

Therefore the  velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

2r = 2*10^3m

Therefore

dU = Gm^2(\frac{1}{R}-\frac{1}{2r})

v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}

v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}

v = 1.83*10^7m/s

Therefore the velocity when they are about to collide is 1.83*10^7m/s

7 0
3 years ago
In designing circular rides for amusement parks, mechanical engineers must consider how small variations in certain parameters c
Bess [88]

Answer:

Part a)

dF = -\frac{mv^2}{r^2} dr

Part b)

dF = \frac{2mvdv}{r}

Part c)

dT = - \frac{2\pi r}{v^2} dv

Explanation:

Part a)

As we know that force on the passenger while moving in circle is given as

F = \frac{mv^2}{r}

now variation in force is given as

dF = -\frac{mv^2}{r^2} dr

here speed is constant

Part b)

Now if the variation in force is required such that r is constant then we will have

F = \frac{mv^2}{r}

so we have

dF = \frac{2mvdv}{r}

Part c)

As we know that time period of the circular motion is given as

T = \frac{2\pi r}{v}

so here if radius is constant then variation in time period is given as

dT = - \frac{2\pi r}{v^2} dv

8 0
3 years ago
1.- Un barco recorre la distancia que separa Gran Canaria de Tenerife (90 km) en 6 horas. ¿Cuál es
bezimeni [28]

Answer:

The speed is 15 km/h or 4.16 m/s.

Explanation:

A boat travels the distance that separates Gran Canaria from Tenerife (90 km) in 6 hours. Which  the speed of the boat in km / h? And in m / s?

Given that,

Distance, d = 90 km = 90000 m

Time, t = 6 hours = 21600 s

Speed = distance/time

v=\dfrac{90\ km}{6\ h}\\\\=15\ km/h

or

v=\dfrac{90000\ m}{21600\ s}\\\\=4.16\ m/s

So, the required speed is 15 km/h or 4.16 m/s.

7 0
3 years ago
Formed through longshore drift<br> a. sea stack<br> b. sandbar<br> c. spit<br> d. headland
ANTONII [103]
<h3>Answer;</h3>

<em><u>Sand Spit or Spit </u></em>

<h3><u>Explanation;</u></h3>
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  • <em><u>Sandbar</u></em> is normally formed when the sandspit stretches across a bay and connects the two sides. <em><u>Headland</u></em> is a high piece of land that extends out onto the sea. <em><u>Sea stacks </u></em>on the other hand results from the collapsing of the roof of the arch.
7 0
3 years ago
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