It is defined by their wavelength. Different colors have different wavelengths. For example, radio waves have a really long wavelength, whereas gamma-rays have a very short wavelength.
Answer:
(a) charge q=5.33 nC
(b) charge density σ=10.62 nC/m²
Explanation:
Given data
radius r=0.20 m
potential V=240 V
coulombs constant k=9×10⁹Nm²/C²
To find
(a) charge q
(b) charge density σ
Solution
For (a) charge q
As
For (b) charge density
As charge density σ is given as:
σ=q/(4πR²)
σ=(5.333×10⁻⁹) / (4π×(0.20)²)
σ=10.62 nC/m²
Answer:
Explanation:
By energy conservation we know that spring energy is converted into kinetic energy of the block
so we will have
so we will have
now we will have same thing for another mass 4m which moves out with speed 5v
so we have
now from above two equations we have
so we have
Answer:Greenhouse effect
Explanation:
Short wavelength radiation from the sun passes through a planet's atmosphere but some of the outgoing long-wavelength energy is absorbed and radiated again this process is known as Green house effect which increases the temperature of earth and as a result causes more to be trapped inside earth atmosphere.
Answer:
1.84 m
Explanation:
For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.
So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,
a = g + rω²
= 9.8 m/s² + 1.10 m × (18.85 rad/s)²
= 9.8 m/s² + 390.85 m/s²
= 400.65 m/s²
Now, using v² = u² + 2a(h₂ - h₁) where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.
v² = u² + 2a(h₂ - h₁)
So, v² - u² = 2a(h₂ - h₁)
h₂ - h₁ = (v² - u²)/2a
h₂ = h₁ + (v² - u²)/2a
substituting the values of the variables into the equation, we have
h₂ = 1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)
h₂ = 1.3 m + [-430.15 (m/s)²]/-801.3 m/s²
h₂ = 1.3 m + 0.54 m
h₂ = 1.84 m
