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leonid [27]
3 years ago
10

Which type of radiation can reach the furthest? A. Beta B. Gamma C. Alpha

Physics
2 answers:
Mashutka [201]3 years ago
7 0

Answer:

C. Alpha

Explanation:

Gamma radiation, unlike alpha or beta, does not consist of any particles, instead consisting of a photon of energy being emitted from an unstable nucleus. Having no mass or charge, gamma radiation can travel much farther through air than alpha or beta, losing (on average) half its energy for every 500 feet.

kati45 [8]3 years ago
5 0

Answer:

B. Gamma

Explanation:

The gamma radiation can travel much farther through air than alpha or beta.

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The current in an electric hair dryer is 10
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First let's convert the time in seconds:
\Delta t= 5 min= 5 min \cdot (60 s/min)= 300 s

The current is defined as the quantity of charge flowing through a certain section of a circuit per unit of time:
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Using I=10 A, and \Delta t=300 s, we can find the amount of charge flown through the hair dryer in this time:
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A carrot has a volume of 180 cubic centimeters and a mass of 81 grams . Calculate its destiny
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Read 2 more answers
Please help on this one
Sphinxa [80]

Since this is a projectile motion problem, break down each of the five kinematic quantities into x and y components. To find the range, we need to identify the x component of the displacement of the ball.

Let's break them down into components.

                X                                Y

v₁     32 cos50 m/s           32 sin50 m/s

v₂     32 cos50 m/s                    ?

Δd             ?                                0

Δt              ?                                ?

a                0                         -9.8 m/s²


Let's use the following equation of uniform motion for the Y components to solve for time, which we can then use for the X components to find the range.

                           Δdy = v₁yΔt + 0.5ay(Δt)²

                                0 = v₁yΔt + 0.5ay(Δt)²

                                0 = Δt(v₁ + 0.5ayΔt), Δt ≠ 0

                                0 = v₁ + 0.5ayΔt

                                0 = 32sin50m/s + 0.5(-9.8m/s²)Δt

                                0 = 2<u>4</u>.513 m/s - 4.9m/s²Δt

                 -2<u>4</u>.513m/s = -4.9m/s²Δt

-2<u>4</u>.513m/s  ÷ 4.9m/s² = Δt

                          <u>5</u>.00s = Δt


Now lets put our known values into the same kinematic equation, but this time for the x components to solve for range.

Δdₓ = v₁ₓΔt + 0.5(a)(Δt)²

Δdₓ = 32cos50m/s(<u>5</u>.00s) + 0.5(0)(<u>5</u>.00)²

Δdₓ = 32cos50m/s(<u>5</u>.00s)

Δdₓ = 10<u>2</u>.846


Therefore, the answer is A, 102.9m. According to significant digit rules, neither would be correct, but 103m is the closest to 102.9m so I guess that is what it is.






6 0
3 years ago
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