You can only see stars whose peak intensity of radiation is in the visible band. True False

Answers

Answer 1

Answer: That's false. You can only see a star that radiates SOMETHING in the visible band, but it doesn't have to be the peak.

Answer 2

Answer:

Answer: FALSE

Explanation:

i tested it, i got it right on my quiz

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Less force is required to poke through an eggshell from the _____.a. outside in
b. Force is the same in both directions.
c. inside out

Answers

I believe the answer is A. hope this helps :)

when a beam of light passes at an oblique angle into a material of lower optical density, the angle of incidence is ?

Answers

Penn Foster Students: less than the angle of refraction

What is the polarity of each of the earths magnetic poles ? Explain you answer

Answers

Answer:

When you put un-like poles together (South facing North) you can feel magnetic attraction. In the Northern Hemisphere, your compass needle points North, but if you think about it for a moment, you will discover that the magnetic pole in the Earth's Northern Hemisphere has to be a South polarity.

True or False Another name for constructive interference is intense interference

Answers

The answer to this statement is FALSE

The momentum of a bald eagle in flight is calculated to be 345. The mass of the eagle is 5.0 kg. What is the magnitude of the velocity of the eagle?

Answers

Answer : v = 69 m/s.

Explanation :

It is given that,

The momentum of a bald eagle in flight, $p = 345\ Kgm/s$

Mass of the eagle, m = 5 kg

We know that the momentum of any object is defined as the product of mass and the velocity with which it is moving.

$p=mv$

$v=(p)/(m)$

$v=(345\ kgm/s)/(5\ kg)$

$v=69\ m/s$

The velocity of the eagle is 69 m/s.

Hence, this is the required solution.

The formula to find the magnitude of velocity is:

▲v= ▲M/m

▲v=Velocity Change
▲M=Momentum Change
m=Mass

Plug in the information;

▲v=345/5

The answer is 69

a balloon is ascending at 3.0 m/s at a height of 20.0 m above ground when a package is released. the time taken, in the absence of air resistance, for the package to reach the ground is:

Answers

Answer:

Approximately $2.35\; {\rm s}$ , assuming that $g = 9.81\; {\rm m\cdot s^(-2)}$ .

Explanation:

Under the assumptions, the package would start with an initial upward velocity of $u = 3.0\; {\rm m\cdot s^(-1)}$ and accelerate downward at a constant $a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}$ (negative because acceleration points downward.)

Right before landing, the package would be $20.0\; {\rm m}$ below where it was released. Hence, the displacement of the package at that moment would be $x = (-20.0)\; {\rm m}$ (negative since this position is below the initial position.)

The duration of the motion can be found in the following steps:

Apply the SUVAT equation $v^(2) - u^(2) = 2\, a\, x$ to find velocity $v$ right before landing.
Divide the change in velocity $(v - u)$ by acceleration to find the duration of the motion.

Rearrange the SUVAT equation $v^(2) - u^(2) = 2\, a\, x$ to find $v$ , the velocity of the package right before reaching the ground. Notice that because the package would be travelling downward, the value of $v\!$ should be negative.

$\begin{aligned} v &= -\sqrt{u^(2) + 2\, a\, x} \n &= -\sqrt{(3.0)^(2) + 2\, (-9.81)\, (-20.0)}\; {\rm m} \n &\approx (-20.035)\; {\rm m}\end{aligned}$ .

Subtract the initial velocity from the new value to find the change in velocity. Divide this change in velocity by acceleration (rate of change in velocity) to find the duration of the motion:

$\begin{aligned}t &= (v - u)/(a) \n &\approx ((-20.035) - (3.0))/((-9.81))\; {\rm s} \n &\approx 2.35\; {\rm s}\end{aligned}$ .

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An automobile having a mass of 2000 kg deflects its suspension springs 0.02 m under static conditions. Determine the nafural frequency of the automobile in the vertical direction by assuming damping to be negligible.

A bumper car with a mass of 200 kilograms traveling at a velocity of 3.5 meters/second collides with a stationary bumper car with a mass of 250 kilograms. If the two bumper cars stick and move together at a velocity of 2.3 meters/second, what is the kinetic energy of the system after the collision