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lidiya [134]
3 years ago
11

3. Name and describe 3 things that can occur when light hits and object. Provide an example of each.

Physics
1 answer:
Katarina [22]3 years ago
6 0
It bounces off one another and goes back in forth
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The half-life of iodine-131 is 8.04 days. It’s decay constant equals
Svetllana [295]

Answer:

Explanation:

we know that half life of an element is

T=0.693/λ

where λ is decay constant in order to find decay constant

λ=0.693/T

λ=0.693/8.04

λ=0.086

7 0
3 years ago
A space probe is launched from Earth headed for deep space. At a distance of 10,000 miles from Earth's center, the gravitational
Pavlova-9 [17]
<span> gravitational force varies based on 1/r^2
when you're double the distance =10,000 to 20,000, the force is 4 times smaller so on and so forth.
</span><span>As force is proportional to 1 / {distance squared}, the force will be 1 / 2^2 (i.e. 1/4) of the force at the reference distance (i.e. 1/4 * 600 = 150 lb)
</span>hope this helps
5 0
3 years ago
Read 2 more answers
a block with length 1.5m width 1m height 0.5m and mass 300kg lays on the table.what is the pressure at the bottom surface of the
Nesterboy [21]

Answer:

your answer will be 320kg that would be the pressure at the bottom surface of the block

6 0
2 years ago
When the leaves are __________________, they fall straight down.
konstantin123 [22]

Answer:Done doing there job, in the winter they have a break

Explanation:

7 0
3 years ago
Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial spe
Lisa [10]

Answer:

1) v₀x = 13.76 m/s

2) v₀y = 19.66 m/s

3) ymax = 21.199 m

4) X = 55.1746 m

5) and 6) y = 18.4 m

Explanation:

1) v₀x = v₀*Cos α = 24 m/s* Cos 55° = 13.76 m/s

2) v₀y = v₀*Sin α = 24 m/s* Sin 55° = 19.66 m/s

3) ymax = y₀ + (v₀y²/(2g)) = 1.5 m + ((19.66 m/s)²/(2*9.81 m/s²)) = 21.199 m

4) We can use this equation

y = y₀ + (tan α)*x – (g / (2* v₀x²))*x²

where y = y₀ = 1.5 m

then

1.5 = 1.5 + tan (55°)*x - (9.81 / (2* (13.76)²))*x²

⇒   0.02588 x² - 1.42815 x = 0

Solving this equation we get

x₁ = 0     and    x₂ = 55.1746 m

The distance between the two girls is 55.1746 m

5) and 6) If   v₀x = 15 m/s = vx   and   ymax = 24 m

y = ?   when x = (xmax/2)

ymax = y₀ + (v₀y²/(2g)) ⇒ v₀y = √(2g*(ymax - y₀))

⇒ v₀y = √(2(9.81 m/s²)(24 m - 1.5 m)) = 21.01 m/s

then we get α' as follows

α' = tan⁻¹(v₀y/v₀x) = tan⁻¹(21.01 m/s/15 m/s) = 54.47°

v₀ = √(v₀x² + v₀y²) = √((15 m/s)² + (21.01 m/s)²) = 25.81 m/s

Now we can apply the equation of the path

y = ymax - ((gx²)/(2v₀²))

⇒  y = 24m - ((9.81)(55.1746/2)²/(2*25.81²))

⇒  y = 18.4 m

7 0
3 years ago
Read 2 more answers
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