Answer:
Explanation:
27dB = 2.7 B
So I / I₀ = 10⁻²°⁷ , I₀ is intensity of main sound and I is intensity after reduction.
= 1.99 X 10⁻³
So intensity will reduce by 1.99 X 10⁻³ .
Well it's energy comes from the sun since the sun gives it sunlight to help it grow
Answer:
²₁H + ³₂He —> ⁴₂He + ¹₁H
Explanation:
From the question given above,
²₁H + ³₂He —> __ + ¹₁H
Let ⁿₐX be the unknown.
Thus the equation becomes:
²₁H + ³₂He —> ⁿₐX + ¹₁H
We shall determine, n, a and X. This can be obtained as follow:
For n:
2 + 3 = n + 1
5 = n + 1
Collect like terms
n = 5 – 1
n = 4
For a:
1 + 2 = a + 1
3 = a + 1
Collect like terms
a = 3 – 1
a = 2
For X:
n = 4
a = 2
X =?
ⁿₐX => ⁴₂X => ⁴₂He
Thus, the balanced equation is
²₁H + ³₂He —> ⁴₂He + ¹₁H

Actually Welcome to the Concept of the Projectile Motion.
Since, here given that, vertical velocity= 50m/s
we know that u*sin(theta) = vertical velocity
so the time taken to reach the maximum height or the time of Ascent is equal to
T = Usin(theta) ÷ g, here g = 9.8 m/s^2
so we get as,
T = 50/9.8
T = 5.10 seconds
thus the time taken to reach max height is 5.10 seconds.
Answer:
The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²
Explanation:
Thickness of the wall is L= 20cm = 0.2m
Thermal conductivity of the wall is K = 2.79 W/m·K
Temperature at the left side surface is T₁ = 50°C
Temperature of the air is T = 22°C
Convection heat transfer coefficient is h = 15 W/m2·K
Heat conduction process through wall is equal to the heat convection process so

Expression for the heat conduction process is

Expression for the heat convection process is

Substitute the expressions of conduction and convection in equation above


Substitute the values in above equation

Now heat flux through the wall can be calculated as

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²