Large Doppler shifts indicate _____. low temperatures low speeds high temperatures high speeds

A 1-mCi source of^60 Co is placed in the center of a cylindrical water-filled tank with an inside diameter of 20 cm and depth of

LenKa [72]

To solve this problem we need the concepts of Energy fluency and Intensity from chemical elements.

The energy fluency is given by the equation

$\Psi=4RcE\pi$

Where

$\Psi =$ The energy fluency

c = Activity of the source

r = distance

E = electric field

In the other hand we have the equation for current in materials, which is given by

$I= I_0 e^{-\mu_{h20}X_{h2o}} e^{-\mu_{Fe}X_{Fe}}$

Then replacing our values we have that

$I = 1*10^{-3} * 3.3*10^{10} * e ^{-0.06*1.1} e^{-0.058*7.861}$

$I = 1.3*10^7 Bq$

We can conclude in this part that 1.3*10^7Bq is the activity coming out of the cylinder.

Now the energy fluency would be,

$\Psi = \frac{cE}{4\pir^2}$

$\Psi = \frac{1.3*10^7*2*1.25}{4\pi*11^2}$

$\Psi = 2.14*10^4 MeV/cm^2.s$

The uncollided flux density at the outer surface of the tank nearest the source is $\Psi = 2.14*10^4 MeV/cm^2.s$

6 0

3 years ago

In Speed Study Number 1, we looked at two cars traveling the same distance at different speeds on city streets. Car "A" traveled

Ganezh [65]

Answer:

230.4 s

Explanation:

The speed of car A is

$v_A = 35 mi/h$

and the distance travelled is

$d = 10 mi$

so the time taken for car A is

$t_A = \frac{d}{v_A}=\frac{10 mi}{35 mi/h}=0.286 h$

The speed of car B is

$v_B = 45 mi/h$

and the distance travelled is

$d = 10 mi$

so the time taken for car B is

$t_B = \frac{d}{v_B}=\frac{10 mi}{45 mi/h}=0.222 h$

So the difference in time is

$\Delta t = t_A - t_B = 0.286 h -0.222 h=0.064 h$

Which corresponds to

$\Delta t = 0.064 h \cdot 3600 s/h = 230.4 s$

so car B arrived 230.4 s before car A.

5 0

3 years ago

Astronaut A can cover 10 meters per minute walking with the heavy shovel. What does

klasskru [66]

Answer:

d) The speed of the astronaut

Explanation:

The sentence describes the speed of the astronaut. This speed value is 10meters per minute.

Now let us understand why;

Speed is the distance divided by time. It is a scalar quantity without regard for direction but it has magnitude.
The value 10meters per minute clearly shows this instance. We do not know the direction the astronaut is moving towards.
Velocity, like speed is the displacement of a body with time. It is a vector quantity and it shows the direction of motion.
For example, 10m/s due west is a velocity value because we know the direction.

Therefore, since there is no directional sense, the value indicates speed.

3 0

3 years ago

A hippo drives 42 km due East. He then turns and drives 28 km at 25° East of South. He turns again and drives 32 km at 40° North

ch4aika [34]

Answer:

a) Please, see the attched figure

b) Total displacement R = (78.3 km; -4.8 km)

c) R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))

d) The hippo is 78.4 km from his starting point.

The total distance traveled is 102 km

Explanation:

a)Please, see the attached figure.

b) The vector A can be expressed as:

A = (magnitude * cos α; magnitude * sin α)

Where

magnitude = 42 km

α= 0

Then,

A = (42 km ; 0) or 42 km i

In the same way, we can proceed with the other vectors:

B = ( Bx ; By)

where

(apply trigonometry of right triangles: sen α = opposite / hypotenuse and

cos α = adjacent / hypotenuse. See the figure to determine which component of vector B is the opposite and adjacent side to α)

Bx = 28 km * sin 25 = 11.8 km

By = 28 km * cos 25 = -25.4 km (it has to be negative since it is directed towards the negative vertical region according to our reference system)

B = (11.8 km; -25.4 km) or 11.8 km i - 25.4 km j

C = (Cx; Cy)

where

Cx = 32 km * cos 40° = 24.5 km

Cy = 32 km * sin 40 = 20.6 km

C = (24.5 km; 20.6 km)

Then:

R = A+B+C = (42 km + 11.8 km + 24.5 km; 0 - 25.4 km + 20.6 km)

= (78.3 km; -4.8 km) or 78.3 km i -4.8 km j

c) R = (78.3 km; -4.8 km)

The magnitude of R is:

$magnitude = \sqrt{(78.3)^{2 }+ (-4.8)^{2}}= 78.4 km$

Using trigonometry, we can calculate the angle:

Knowing that

tan α = opposite / adjacent

and that

opposite = Ry = -4.8 km

adjacent = Rx = 78.3 km

Then:

tan α = -4.8 km / 78.4 km

α = -3.5°

We can now write the vector R in magnitude and direction form:

R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))

d) The displacement of the hipo relative to the starting point is the magnitude of vector R calculated in c):

magnitude R = 78. 4 km

The total distance traveled is the sum of the magnitudes of each vector:

Total distance = 42 km + 28 km + 32 km = 102 km

3 0

3 years ago

Please help!! I will give brainliest!

grandymaker [24]

Answer:

Depends.

Explanation:

Whether the object is going left or right, the speed will stay the same until friction eventually stops it. <em>However, </em>if, for example, we're talking about an object going straight before veering right, then yes, speed <em>does</em> matter. An object will normally have to speed up or slow down momentarily when changing direction to keep itself sustained on the ground.

So, honestly? It really depends on what we're talking about!

Hope this helped!

Source(s) used: None.

7 0

3 years ago