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3 years ago

Than a cool

1 answer:

7 0

Alnilam is the brightest star

Absolute brightness is how bright the star actually is. Apparent brightness is how bright the star looks.

Factors that influence how bright a star looks is distance. The fact that Alnilam has the same apparent brightness of the other two stars even though it is farther away, means that it is brighter.

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An object moving with a constant

jeka57 [31]

Answer:

Acceleration:

${ \tt{a = \frac{v - u}{t} }} \\ { \tt{a = \frac{20 - 10}{5} }} \\ { \tt{a = 2 \: m {s}^{ - 2} }}$

From third equation:

${ \bf{ {v}^{2} = {u}^{2} + 2as}} \\ { \tt{s = \frac{ {20}^{2} - {10}^{2} }{2 \times 2} }} \\ = { \tt{s = 75 \: m}}$

5 0

2 years ago

Read 2 more answers

Difference between of echo and reverberation

KiRa [710]

Answer:

A difference between of echo and reverberation is described below in details.

Explanation:

Here's a piece of immediate information: An echo is an individual consideration of a soundwave off a horizon exterior. Reverberation is the consideration of sound waves generated by the superposition of the before-mentioned echoes. ... A reverberation can happen when a sound wave is displayed off a nearby covering.

5 0

2 years ago

Can someone please explain number 8?

guajiro [1.7K]

Answer: Ax=(Vx-Vox)/(T)

Vx=Vox+Ax*T

Solving for Ax in terms of Vx, Vox, T

Vx-Vox=Ax*t

Ax=(Vx-Vox)/(T)

This is saying the acceleration in the x-direction can be found by taking the difference between the finial and initial Velocity in x-direction and dividing it by the Total Time.

Any questions please feel free to ask. Thanks

8 0

2 years ago

In one experiment, the students allow the block to oscillate after stretching the spring a distance A. If the potential energy s

atroni [7]

Answer:

K = m g (A - A2)

Explanation:

In a block spring system the total energy is the sum of the potential energy plus the kinetic energy, for maximum elongation all the energy is potential

Em = U₀ = m g A

For when the system is at an ele

Elongation A2 less than A, energy has two parts

Em = K + U₂

K = Em –U₂

We substitute

K = m g A - m gA2

K = m g (A - A2)

3 0

3 years ago

Read 2 more answers

A small object carrying a charge of -2.50 nc is acted upon by a downward force of 18.0 nn when placed at a certain point in an e

Vesna [10]

Missing question in the text:
"A.What are the magnitude and direction of the electric field at the point in question?

B.<span>What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?"</span>

<span>Solution:

A) A charge q </span>under an electric field of intensity E will experience a force F equal to:

$F=qE$

In our problem we have $q=-2.5 nC=-2.5\cdot 10^{-9} C$ and $F=18 nN = 18 \cdot 10^{-9} N$ , so we can find the magnitude of the electric field:

$E= \frac{F}{q}= \frac{18\cdot 10^{-9}N}{2.5\cdot 10^{-9}C}=7.2 V/m$

The charge is negative, therefore it moves against the direction of the field lines. If the force is pushing down the charge, then the electric field lines go upward.

B) The proton charge is equal to

$e=1.6\cdot 10^{-19} C$

Therefore, the magnitude of the force acting on the proton will be

$F=eE=1.6\cdot 10^{-19} C \cdot 7.2 V/m=1.15 \cdot 10^{-18} N$

And since the proton has positive charge, the verse of the force is the same as the verse of the field, so upward.

7 0

2 years ago