3 years ago

Which do you think will penetrate farther into a block of lead, x-rays, or gamma rays

Physics

1 answer:

disa [49]3 years ago

3 0

Gamma rays because it has more penetrating power and frequency but shorter wavelength.

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Suppose you analyze standardized test results for a country and discover almost identical distributions of physics scores for fe

zaharov [31]

Answer:

go ask a teacher

Explanation:

go to school

go find the teacher

ask him/her for help

finish question done :)

7 0

3 years ago

Which is not a common extrusive igneous rock

KonstantinChe [14]

Granite is the answer i think

8 0

3 years ago

A child in danger of drowning in a river is being carried downstream by a current that flows due south uniformly with a speed of

tia_tia [17]

Let the rescue boat starts at an angle theta with the North

now its velocity towards East is given as

$v_x = 24sin\theta$

$v_y = -24cos\theta + 3$

now in some time "t" it will catch the boy

so we will have

$t = \frac{0.5}{24sin\theta}$

also we have

$t = \frac{2}{-24cos\theta + 3}$

now we have

$\frac{2}{-24cos\theta + 3} = \frac{0.5}{24sin\theta}$

$4*24sin\theta = - 24cos\theta + 3$

$96 sin\theta + 24cos\theta = 3$

by solving above we got

$\theta = 164 degree$

3 0

3 years ago

A repeating disturbance that transfers energy through matter or space is called a _____.

Damm [24]

Answer:

Collateral Damage

Explanation:

3 0

3 years ago

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If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the m

Xelga [282]

Answer:

The magnitude of the net force F₁₂₀ on the lid when the air inside the cooker has been heated to 120 °C is $\frac{135.9}{A}N$

Explanation:

Here we have

Initial temperature of air T₁ = 20 °C = ‪293.15 K

Final temperature of air T₁ = 120 °C = 393.15 K

Initial pressure P₁ = 1 atm = ‪101325 Pa

Final pressure P₂ = Required

Area = A

Therefore we have for the pressure cooker, the volume is constant that is does not change

By Chales law

P₁/T₁ = P₂/T₂

P₂ = T₂×P₁/T₁ = 393.15 K× (‪101325 Pa/‪293.15 K) = ‭135,889.22 Pa

∴ P₂ = 135.88922 KPa = 135.9 kPa

Where Force = $\frac{Pressure}{Area}$ we have

Force = $F_{120}=\frac{135.9}{A}N$ .

4 0

3 years ago