Answer:
In the clarification portion elsewhere here, the definition of the concern is mentioned.
Explanation:
So like optical telescopes capture light waves, introduce it to concentrate, enhance it, as well as make it usable through different instruments via study, so radio telescopes accumulate weak signal light waves, introduce that one to focus, enhance it, as well as make this information available during research. To research naturally produced radio illumination from stars, galaxies, dark matter, as well as other natural phenomena, we utilize telescopes.
Optical telescopes detect space-borne visible light. There are some drawbacks of optical telescopes mostly on the surface:
Mostly at night would they have been seen. Unless the weather gets cloudy, bad, or gloomy, they shouldn't be seen.Although radio telescopes monitor space-coming radio waves. Those other telescopes, when they are already typically very massive as well as costly, have such an improvement surrounded by optical telescopes. They should be included in poor weather and, when they travel through the surrounding air, the radio waves aren't obscured by clouds. Throughout the afternoon and also some at night, radio telescopes are sometimes used.
In a head-on collision, a ball of mass 0.3 kg travelling with velocity 2.8 m/s in the positive x-direction hits a stationary second ball of mass 0.4 kg. What is the velocity of the 0.3 kg ball after the collision? Assume collision is elastic.
Answer:
The final velocity of the first ball (0.3 kg) is 0.4 m/s in negative x direction.
Explanation:
Given;
mass of the first object, m₁ = 0.3 kg
initial velocity of the first ball, u₁ = 2.8 m/s
mass of the second ball, m₂ = 0.4 kg
initial velocity of the second ball, u₂ = 0
let the final velocity of the first ball, = v₁
let the final velocity of the second ball, = v₂
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(0.3 x 2.8) + (0.4 x 0) = 0.3v₁ + 0.4v₂
0.84 = 0.3v₁ + 0.4v₂
2.8 = v₁ + 1.333v₂ -------equation (1)
Apply one-direction velocity;
u₁ + v₁ = u₂ + v₂
2.8 + v₁ = 0 + v₂
v₂ = 2.8 + v₁
substitute the value of v₂ into equation (1)
2.8 = v₁ + 1.333v₂
2.8 = v₁ + 1.333(2.8 + v₁)
2.8 = v₁ + 3.732 + 1.333v₁
2.8 - 3.732 = v₁ + 1.333v₁
-0.932 = 2.333v₁
v₁ = -0.932 / 2.333
v₁ = -0.4 m/s
Therefore, the final velocity of the first ball (0.3 kg) is 0.4 m/s in negative x direction.
A bicycle starts at 2.5m/s and accelerates along a straight path to a speed of 12.5m/s in a time of 4.5 seconds. What is the bicyclist’s acceleration to the nearest tenth of a m/s^2 ?
Answer:
The bicyclist's acceleration is 2.2m/s^2
Explanation:
Given
[tex]u = 2.5m/s[/tex] ---- Initial Velocity
[tex]v = 12.5m/s[/tex] ---- Final Velocity
[tex]t = 4.5s[/tex] ---- Time
Required
Determine the acceleration
This will be solved using the first equation of motion
[tex]v = u + at[/tex]
Substitute values for v, u and a
[tex]12.5 = 2.5 + a * 4.5[/tex]
[tex]12.5 = 2.5 + 4.5a[/tex]
Collect Like Terms
[tex]4.5a = 12.5 - 2.5[/tex]
[tex]4.5a = 10.0[/tex]
Solve for a
[tex]a = 10.0/4.5[/tex]
[tex]a = 2.2m/s^2[/tex] ---- (approximated)
Hence, the bicyclist's acceleration is 2.2m/s^2