The other types of EM radiation<span> that make up the electromagnetic spectrum are microwaves, </span>infrared light<span>, ultraviolet </span>light, X-rays and gamma-rays. ... Inspace<span>, </span>infrared light<span> helps us map the dust between stars. Visible: Our eyes detect </span>visible light. Fireflies,light<span> bulbs, and stars all emit </span>visible light<span>.</span>
Explanation:
The first ionization energy varies in a predictable way across the periodic table. The ionization energy decreases from top to bottom in groups, and increases from left to right across a period. Thus, helium has the largest first ionization energy, while francium has one of the lowest.
Answer:
Stop watch
Stop clock
Explanation:
Absolute time is time measured in definite periods such as minutes, days, and years. It can be measured uising stop watch or stop clock
Answer:
0.247 J = 247 mJ
Explanation:
From the principle of conservation of energy, the workdone by the applied force, W = kinetic energy change + electric potential energy change.
So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁) where m = mass of particle = 5.4 × 10⁻² kg, q = charge of particle = 5.10 × 10⁻⁵ C, v₁ = initial speed of particle = 2.00 m/s, v₂ = final speed of particle = 3.00 m/s, V₁ = potential at surface A = 5650 V, V₂ = potential at surface B = 7850 V.
So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁)
= 1/2 × 5.4 × 10⁻²kg × ((3m/s)² - (2 m/s)²) + 5.10 × 10⁻⁵ C(7850 - 5650)
= 0.135 J + 0.11220 J
= 0.2472 J
≅ 0.247 J = 247 mJ
Answer:
the impulse experienced by the egg is 0.053 kgm/s.
Explanation:
Given;
mass of the egg, m = 60 g = 0.06 kg
initial velocity of the egg, u = 6 m/s
height moved by the egg, h = 50 cm = 0.5 m
Determine the final velocity of the egg as it moves upward;
v² = u² + 2(-g)h
v² = u² - 2gh
where;
v is the final velocity
-g is negative acceleration due gravity as it moves upward
v² = 6² - 2(9.8 x 0.5)
v² = 26.2
v = √26.2
v = 5.12 m/s
The impulse applied to the egg is the change in linear momentum;
J = ΔP
ΔP = mu - mv
ΔP = m(u - v)
ΔP = 0.06(6 - 5.12)
ΔP = 0.053 kgm/s
Therefore, the impulse experienced by the egg is 0.053 kgm/s.
