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lys-0071 [83]
3 years ago
8

In an adiabatc process, what happens when gases in a system are compressed?

Physics
1 answer:
likoan [24]3 years ago
8 0

Answer:work is done, and temperature increases

Explanation:

In an adiabatic process, when gases are compressed, work is done on the liquid and the temperature increases

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A dog jumps 0.80 m to catch a treat. The dog's displacement vector is shown below.
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Answer:D

Explanation:

It was right o khan academy

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Find the horizontal component and the vertical component ​
aleksandr82 [10.1K]

Answer:

v=3.66,h-3.66

Explanation:

vertical = 10sin60 - 10sin 30

horizontal =10cos60 + 10cos 30

v = 10×0.8660-10×0.5

h = 10×0.5 + 10 × 0.8660

v=8.660-5.0 = 3.66

h= 5.0-8.660 = -3.66

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How much current must flow through a wire to make a magnetic field as strong as Earth's field (5.00 x 10^-5 T) 1.00 m away from
torisob [31]

Answer:

250 A

Explanation:

B = 5 x 10^-5 T, r = 1 m

Let current be i.

the magnetic field due to a straight current carrying conductor is given by

B = μ0 / 4π x 2i / r

5 x 10^-5 = 10^-7 x 2 x i / 1

i = 250 A

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3 years ago
What is the main mid west meal
Orlov [11]

Answer:

Sometimes called "the breadbasket of America" the Midwest serves as a center for grain production, particularly wheat, corn and soybeans. Graham bread, steamed cornbread, and johnny cakes are traditional Midwestern foods, as are butter cakes like chocolate cake, devil's food cake, coconut cake and fruit cake

7 0
3 years ago
An electric dipole consists of charges +2e and -2e separated by 0.82 nm. It is in an electric field of strength 3.2 x 10^6 N/C.
Zepler [3.9K]

Explanation:

It is given that,

An electric dipole consists of charges +2e and -2e separated by 0.82 nm

Charge, q=2e=2\times 1.6\times 10^{-19}\ C=3.2\times 10^{-19}\ C

Distance between charges, d=0.82\ nm=0.82\times 10^{-9}\ m

Electric field strength, E=3.2\times 10^6\ N/C

(a) The magnitude of the torque on the dipole is given by :

\tau=p\times E\ sin\theta

When dipole moment is parallel to the electric field, \theta=0

\tau=p\times E\ sin(0)

\tau=0

(b) When the dipole is perpendicular to the electric field, \theta=90

\tau=pE\ sin(90)

\tau=qdE (Since, p = q × d)

\tau=3.2\times 10^{-19}\times 0.82\times 10^{-9}\times 3.2\times 10^6

\tau=8.39\times 10^{-22}\ N.m

(c) When the dipole moment is anti parallel to the electric field, \theta=180

\tau=pE\ sin(180)

Since, sin\ 180=0

\tau=0

Hence, this is the required solution.

8 0
3 years ago
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