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Harrizon [31]
3 years ago
5

An alpha particle (which has two protons) is sent directly toward a target nucleus with 90 protons. the alpha particle has a kin

etic energy of 0.45 pj when far away. what is the least center-to-center distance that the alpha particle will be from the target nucleus, assuming that the nucleus does not move?
Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
5 0
It does not move because of the sun the sun has no energy.
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Find the magnitude of acceleration (ft/s^2) a person experiences when he or she is texting and driving 58mph, hits a wall, and c
SVEN [57.7K]

Answer:

350 ft/s²

Explanation:

First, convert mph to ft/s.

58 mi/hr × (5280 ft/mi) × (1 hr / 3600 s) = 85.1 ft/s

Given:

v₀ = 85.1 ft/s

v = 0 ft/s

t = 0.24 s

Find: a

v = at + v₀

a = (v − v₀) / t

a = (0 ft/s − 85.1 ft/s) / 0.24 s

a = -354 ft/s²

Rounded to two significant figures, the magnitude of the acceleration is 350 ft/s².

7 0
3 years ago
Let w(x)=3x-7.If w(x)=14, find x
dlinn [17]

Answer:

7

Explanation:

We are given:

    w(x) = 3x   -   7

     w(x)  = 14

The problem here entails us to solve for x;

To solve for x; equate the two expressions:

         

          3x  - 7  = 14

           3x  = 14 + 7

           3x  = 21  

             x = 7

So the value of x  = 7

6 0
3 years ago
All stars go through a lifecycle.
Komok [63]

Mira is much bigger than the Sun.

Only very massive stars will go through a supernova stage, causing the outer layer to explode away and the core to collapse in on itself, becoming very dense.

5 0
2 years ago
Read 2 more answers
An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below
grin007 [14]
<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let P be the pressure at a point.

Let p be the density fluid at a point.

Let v be the velocity of fluid at a point.

Bernoulli's equation states that P+\frac{1}{2}pv^{2}+pgh=constant for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let p_{up} be the pressure of a point just above the wing.

Let p_{do} be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}

So,p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa

Force is given by the product of pressure difference and area.

Given that area is 27ms^{2}.

So,lifting force is 16398.48\times 27=442758.96N

6 0
3 years ago
If 2 grams of element X combine with 4 grams of element Y to form compound XY, how many grams of element Y would combine with 46
Butoxors [25]
92 grams of Y would combine with 46 grams of X
3 0
3 years ago
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