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dimaraw [331]
3 years ago
7

When two objects of different masses, different temperatures, and different sizes are placed in thermal contact, energy will alw

ays travel a. from the hotter object to the cooler object b. from the bigger object to the smaller object C. from the more massive object to the one with less mass
Physics
1 answer:
denpristay [2]3 years ago
6 0

Answer:

a) from the hotter object to the cooler object

Explanation:

temperature moves by conduction,  which is associated with the movement of  atoms or molecules and the always move from hight temperatures to lower temperatures to attain thermal equilinrium of the system.

so when two objects are placed together and have different temperatures then the system is not in thermal equilibrium and to attain it, temperature can only move to coller object and not from the coller object according to thermodynamics.

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The gravitational force of attraction between two objects would increase by
OleMash [197]
The gravitational force of attraction between two objects would be increased by "decreasing the distance between two objects"

Hope this helps!
6 0
3 years ago
) a wrench 0.6 meters long lies along the positive y-axis, and grips a bolt at the origin. a force is applied in the direction o
Lera25 [3.4K]
<span>Applied Force Direction vector = [0,1,3] force F = j + 4k torque is t = 100 Newton-meters = r x F Wrench is 0.5 meters long on positive side of the y-axis, r = 0.6 = [0,0.6,0] We know torque |t| = |r x F| = |r| x |F| sin theta r x F = |r| x |F| cos theta r x (j + 3k) = |r| x |j + 3k| cos theta => [0,0.6,0] [0,1,3] = 0.6 x squareroot of ((0)^2 + (1)^2 + (3)^2) cos theta => 0.6 = 0.6 x squareroot of (1 + 9) cos theta => cos theta = 1 / squareroot of (10) Calculationg the sin theta, sin theta = squareroot of (1 - (1 / squareroot of (10))^2) = squareroot of (9/10) sin theta = 3 / squareroot of (10) Substituting the values, |T| = |r| x |F| sin theta => 100 = |0.6| x |F| x 3 / squareroot of (10) |F| = (100 x squareroot of (10)) / 1.8 |F| = (1000 / 18) x squareroot of (10) Magnitude of force |F| = 55.55 x squareroot of (10)</span>
6 0
4 years ago
A thin, circular hoop with a radius of 0.22 m is hanging from its rim on a nail. When pulled to the side and released, the hoop
Alex73 [517]

Answer:

Period of oscillation = 1.33 seconds

Explanation:

The period of oscillation is given by:

T = 2π√[I/(MgL)] 

for I = 2MR² and L = R,

Given: L = 0.22m = R

T = 2π√[2R/g] 

T = 2 × 3.142 Sqrt[( 2 × 0.22)/ 9.8]

T = 6.284 Sqrt(0.44/9.8)

T = 6.284 Sqrt(0.0449)

T = 6.284 × 0.2119

T = 1.33 sec

6 0
4 years ago
Consider a concave mirror that has a focal length f. In terms of f, determine the object distances that will produce a magnifica
Aleks04 [339]

We have that the magnification of each focal length is given respectively as

A) has u=3\frac{f}{2}

B) has u=4\frac{f}{3}

C) has  u=5\frac{f}{4}

From the question we are told that:

Focal Length F

Generally, the equation for Magnification is mathematically given by

M=\frac{-v}{u}

Therefore

v=2u

For A

M=-2

Therefore

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

\frac{1}{f}=\frac{1}{u}+\frac{1}{2u}

Therefore

u=3\frac{f}{2}

For B

M=-3

Therefore

v=3u

Where

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

\frac{1}{f}=\frac{1}{u}+\frac{1}{3u}

Therefore

u=4\frac{f}{3}

For C

M=-4

Therefore

v=4u

Therefore

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

\frac{1}{f}=\frac{1}{u}+\frac{1}{4u}

Therefore

u=5\frac{f}{4}

Conclusion

From the calculations above we can rightly say that the magnifications of the values above are

A has u=3\frac{f}{2}

B has u=4\frac{f}{3}

C has  u=5\frac{f}{4}

For more information on this visit

brainly.com/question/14468351

3 0
3 years ago
Explain the plate tectonic theory.
Nookie1986 [14]

Answer:

scientific theory describing the large-scale motion of seven large plates and the movements of a larger number of smaller plates of the Earth's lithosphere, since tectonic processes began on Earth between 3.3 and 3.5 billion years ago

Explanation:

3 0
4 years ago
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