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lubasha [3.4K]
3 years ago
14

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 2.5 x 10-15 m before

interacting. From this information, find the time interval required for the strong interaction to occur.
Physics
1 answer:
Tasya [4]3 years ago
8 0

Explanation:

It is given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, d=2.5\times 10^{-15}\ m

Speed of light, c=3\times 10^{8}\ m\s

Let t is the time interval required for the strong interaction to occur. The speed is given by :

c=\dfrac{d}{t}

t=\dfrac{d}{c}

t=\dfrac{2.5\times 10^{-15}\ m}{3\times 10^{8}\ m/s}

t=8.33\times 10^{-24}\ s

So, the time interval required for the strong interaction to occur is 8.33\times 10^{-24}\ s. Hence, this is the required solution.

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Boyle's Law mainly involves _______.
goblinko [34]
Your answer is B, gases
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Water enters a baseboard radiator at 180 °F and at a flow rate of 2.0 gpm. Assuming the radiator releases heat into the room at
beks73 [17]

Answer:

Temperature of water leaving the radiator = 160°F

Explanation:

Heat released = (ṁcΔT)

Heat released = 20000 btu/hr = 5861.42 W

ṁ = mass flowrate = density × volumetric flow rate

Volumetric flowrate = 2 gallons/min = 0.000126 m³/s; density of water = 1000 kg/m³

ṁ = 1000 × 0.000126 = 0.126 kg/s

c = specific heat capacity for water = 4200 J/kg.K

H = ṁcΔT = 5861.42

ΔT = 5861.42/(0.126 × 4200) = 11.08 K = 11.08°C

And in change in temperature terms,

10°C= 18°F

11.08°C = 11.08 × 18/10 = 20°F

ΔT = T₁ - T₂

20 = 180 - T₂

T₂ = 160°F

8 0
3 years ago
Motion maps for two objects, Y & Z, are shown.
serg [7]

Answer: it’s 3 seconds

Explanation:

5 0
2 years ago
Calculate the average speed of a runner who runs to for 500 meters in 40 second
siniylev [52]

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12.5

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5 0
2 years ago
From an h = 53 feet observation tower on the coast, a Coast Guard officer sights a boat in difficulty. The angle of depression o
maksim [4K]

Answer:

757,93 feets

Explanation:

We can make a right triangle between the boat (A), the Coast Guard officer (B) and the base of the observation tower (C), like in the graph attached. Now, you could also made a rectangle, adding the horizontal at the height of the Coast Guard, starting in B and ending in D, the vertex opossing C.

The angle of depression, its O in the graph.

Now, as we got an rectangle, of course, the segment AD its the same length as CB, and CA, the distance from the boat to shoreline, its the same length as DB.

ADB its an right triangle, with AB, the hypothenuse, and BD and DA, the catheti (or <em>legs</em>).

Now, we know the lenght BC, the height of the tower, 53 feets, so we also know the lenght of DA. DA its the opposite cathetus to the angle O. We wish to know the length AC, equal to the lenght DB, the adjacent cathetus of the angle O.

Know, the trigonometric function that connects the adjacent cathetus with the opossite cathetus its the tangent.

tangent( O ) = \frac{opposite}{adjacent}

We can take that the angle O = 4 °, and knowing that the opossite cathetus its 53 feets, we got:

tangent( 4) = \frac{53 feets}{DB}

DB=  \frac{53 feets}{tangent( 4)}

DB=  757,93 feets

This its equal to the distance from the boat to the shoreline.

4 0
3 years ago
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