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3 years ago

The model shows the star Wolf 359, the sun, and Star X. It takes 7.8 years for light produced on Wolf 359 to reach the sun.

Physics

1 answer:

saul85 [17]3 years ago

4 0

Answer:

390 light-years

Explanation:

50 x 7.8 =390

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A typical laboratory centrifuge rotates at 3700 rpm . Test tubes have to be placed into a centrifuge very carefully because of t

Alexus [3.1K]

Answer:

Explanation:

acceleration of test tube

= ω² R

= (2πn)² R

= 4π²n²R

n = no of rotation per second

= 3700 / 60

= 61.67

R = .10 m

acceleration

= 4π²n²R

= 4 x 3.14² x 61.67² x .10

= 14999 N Approx

3 0

3 years ago

________ is the average kinetic energy of each atom. radiation temperature potential

SSSSS [86.1K]

It could be radiation because radiation means the emission of energy as electromagnetic wave, but potential fits it best

8 0

3 years ago

A car is accelerating at 30 m/s2, if the car is 400 kg how much force

Verizon [17]

It would be 12,000 because newton’s third 2nd law states F=ma (force=matter x acceleration) so 30x400 would be your force .

please mark brainliest and i hope this helps!

6 0

3 years ago

The leaves of a tree lose water to the atmosphere via the process of transpiration. A particular tree loses water at the rate of

Gnoma [55]

Answer:

The speed of the sap flowing in the vessel is 1.90 mm/s

Explanation:

Given:

The rate of water loss, Q = 3 × 10 ⁻⁸ m³/s

Number of vessels contained, n = 2000

Diameter of the vessel, D = 100 Mu m

thus, the radius of the vessel, r = 50 × 10⁻⁶ m

Now, the rate of flow is given as:

Q = AV .............(1)

where, A is the area of the cross-section

V is the velocity

Total area, A = n × (πr²)

substituting the values in the equation (1), we get

3 × 10 ⁻⁸ m³/s = [2000 × (π × (50 × 10⁻⁶)²)] × V

V = 1.909 × 10⁻³ m/s or 1.90 mm/s

Hence, the speed of the sap flowing in the vessel is 1.90 mm/s

7 0

3 years ago

Uranium-235 decays to thorium-231 with a half-life of 700 million years. When a rock was formed, it contained 6400 million urani

Dahasolnce [82]

Answer:

proof in explanation

Explanation:

First, we will calculate the number of half-lives:

$n = \frac{t}{t_{1/2}}$

where,

n = no. of half-lives = ?

t = total time passed = 2100 million years

$t_{1/2}$ = half-life = 700 million years

Therefore,

$n = \frac{2100\ million\ years}{700\ million\ years}\\\\n = 3$

Now, we will calculate the number of uranium nuclei left ( $n_u$ ):

$n_u = \frac{1}{2^{n} }(total\ nuclei)\\\\n_u = \frac{1}{2^{3} }(6400\ million)\\\\n_u = \frac{1}{8}(6400\ million)\\\\n_u = 800\ million$

and the rest of the uranium nuclei will become thorium nuclei ( $u_{th}$ )

$n_{th} = total\ nuclei - n_u\\n_{th} = 6400\ million-800\ million\\n_{th} = 5600\ million$

dividing both:

$\frac{n_{th}}{n_u}=\frac{5600\ million}{800\ million} \\\\n_{th} = 7n_u$

Hence, it is proven that after 2100 million years there are seven times more thorium nuclei than uranium nuclei in the rock.

6 0

3 years ago