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finlep [7]
3 years ago
15

The temperature of a heated object determines how fast the particles vibrate or move within that material. Objects with large te

mperature have particles that are moving very ---Select--- . Particles that are moving ---Select--- will run into each other much more often than particles that are moving ---Select--- and on average will release ---Select--- energy in each collision. Therefore, hot objects should radiate ---Select--- energy than cooler objects.
Physics
1 answer:
gregori [183]3 years ago
8 0

Answer:

The temperature of a heated object determines how fast the particles vibrate or move within that material. Objects with large temperature have particles that are moving very <u><em>rapidly</em></u><u>.</u> Particles that are moving <u><em>rapidly</em></u><em>  </em>will run into each other much more often than particles that are moving <u><em>slower</em></u><em> </em>and on average will release <u><em>more</em></u><em> </em>energy in each collision. Therefore, hot objects should radiate <u><em>more</em></u> energy than cooler objects.

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Answer:

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8 0
3 years ago
An observer in frame S sees lightning simultaneously strike two points 100 m apart. The first strike occurs at xx1 = yy1 = zz1 =
Veseljchak [2.6K]

Answer:

a) 0, = -0.33 us

b) 140m

c) No, The event are not simultaneous i.e they did not occur at the same time, the second even (-0.33 usec) occurs 0.33 usec earlier than the first event.

Explanation:

a)

the lorentz factor expression is written as;

y = 1₀ / √(1 - (v²/c²))

where v  is the relative speed of an observer and c is the speed of light

so we were given that relative speed to be o.7c

therefore

y = 1 / √(1 - ((0.7c)² / c²))

y = 1 / √(1 - (0.49c² / c²))

y = 1 / √(1 - 0.49)

y = 1 / 0.7141

y = 1.4

1 - the coordinates  of the first event, the s' frame of reference is,

x1 ' = y(x1 - vt1) = 0

y1 ' = y1, z1' = z1 and

t1 ' = y [t1 - v/c²x1]

= 0

2 - the coordinates of the second event, the s ' frame of reference is'

x2 ' = y(x2-vt2)

= 1.4(100m - 0)

= 140m

y2 ' = y2, z2 ' = z2

t2 ' = y [ t2 - v/c²x2 ]

= 1.4 [ 0 - 0.7c/c²(100) ]

using speed of light c as 3*10^8

1.4 [ 0 - (0.7*3*10^8) / (3*10^8)²(100) ]

= -0.33 us

b)

distance between

delltaX' = X2' - X1'

= 140m - 0

= 140m

c)

No, The event are not simultaneous i.e they did not occur at the same time.

the second even (-0.33 us) occurs 0.33 us earlier than the first event.

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6 0
3 years ago
If the timber weighs 670 N, calculate its angle of inclination when the water surface is 2.1 m above the pivot. Above what depth
Ymorist [56]

Answer:

Q = 40.1 degrees

Explanation:

Given:

- The weight of the timber W = 670 N

- Water surface level from pivot y = 2.1 m

- The specific density of water Y = 9810 N / m^3

- Dimension of timber = (0.15 x 0.15 x 0.0036) m

Find:

- The angle of inclination Q that the timber makes with the horizontal.

Solution:

- Calculate the Flamboyant Force F_b acting upwards at a distance x along the timber, which is unknown:

                                   F_b = Y * V_timber

                                   F_b = 9810*0.15*0.15*x

                                   F_b = 226.7*x N

- Take static equilibrium conditions for the timber, and take moments about the pivot:

                                   (M)_p = 0

                                   W*0.5*3.6*cos(Q) - x/2 * F_b*cos(Q) = 0

- Plug values in:

                                   670*0.5*3.6 - x^2 * 0.5*226.7 = 0

                                   x^2 = 1206 / 113.35

                                   x = 3.26 m

- Now use the value of x and vertical height y to compute the angle of inclination to be:

                                   sin(Q) = y / x

                                   sin(Q) = 2.1 / 3.26

                                   Q = sin^-1 (0.6441718)

                                   Q = 40.1 degrees

5 0
3 years ago
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