1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
777dan777 [17]
3 years ago
10

Maria listed four important points about Einstein’s special theory of relativity. 1) applies to any object moving more slowly th

an the speed of light
2) applies to objects moving in a straight path
3) describes the relationship between time and space
4) applies to uniformly moving objects? Which best describes how Maria can correct her error?

ANSWER A (change the first point to “applies to objects moving almost at the same speed of light”)
Physics
2 answers:
DiKsa [7]3 years ago
6 0
<span> applies to any object moving more slowly than the speed of light</span>
kobusy [5.1K]3 years ago
4 0

Answer:

(Change the first point to “applies to objects moving almost at the speed of light.” ) or A

Explanation:

This is the Answer on Edge

You might be interested in
What is the atmospheric pressure of the U.S.
rusak2 [61]
When you say atmospheric pressure it is a standard on all places on earth. The value of standard atmospheric pressure is 1 Bar approximately or 14.7 lb per square inch. This pressure is an attribute of free molecules on the air which is not bound the accumulation of the pressure of this in average is the atmospheric pressure.
4 0
3 years ago
What is the formula to calculate liquid pressure?/
Mice21 [21]

1. set up the equation

2. multiply the variables

3. analyze your results

7 0
4 years ago
Read 2 more answers
Potassium-41 has 19 protons. How many neutrons does this isotope have ?<br>​
AysviL [449]

Answer:

20 neutrons

Explanation:

This will get you the mass number of the most abundant potassium isotope. You now know that the most abundant isotope of potassium contains 20 neutrons. All isotopes of potassium will have the same number of protons inside their nucleus, i.e. 19 protons.

I hope this helps you.

3 0
3 years ago
A 90.0-kg fullback running east with a speed of 5.00 m/s is tackled by a 95.0-kg opponent running north with a speed of 3.00 m/s
V125BC [204]

Answer:

a) Please see below as the answer is self-explanatory.

b) 2.88 m/s

c) 785. 8 J

d) It is expended like thermal energy, due to internal friction.

Explanation:

a) In a tackle, both players keep emmeshed each other, so it is a perfectly inelastic collision; Immediately after the tackle, both masses behave like they were only one.

b) Assuming no external forces act during the collision, total momentum must be conserved.

As momentum is a vector, the conservation principle must be met by all vector components at the same time.

In our case, as the players move in directions mutually perpendicular, we can decompose the momentum vector along both directions, taking into account that after the collision, the momentum vector will have components along both directions.

So, if we call the W-E axis our X-axis (being the direction towards east as the positive one) , and to the S-N axis our Y -axis (being the northward direction the positive one), we can write the following equations:

pₓ₀ = pₓf ⇒ m₁*v₁ = (m₁+m₂)*vf*cosθ

py₀ = pyf ⇒ m₂*v₂ = (m₁+m₂)*vf*sin θ

where θ, is the angle that both players take regarding the x-axis after the collision (north of east).

Replacing by the values, we have the following equations:

vf*cosθ = (90.0 kg*5.00 m/s) / (90.0 kg + 95.0 kg) = 2.43 m/s (1)

vf*sin θ = (95.0 kg* 3.00 m/s) / (90.0 kg + 95.0 kg) = 1.54 m/s (2)

Dividing both sides:

sin θ / cos θ = tan θ = 1.54 / 2.43 = 0.634

⇒ arc tan (0.634) = 32.3º

Replacing in (1) we have:

vf = 2.43 m/s / cos 32.3º = 2.43 m/s / 0.845 = 2.88 m/s

c) As the collision happens in one dimension, all mechanical energy, before and after the collision, is just the kinetic energy of the players.

Before the collision:

K₀ = 1/2*m₁*v₁₀² + 1/2 m₂*v₂₀²

= 1/2*( ( 90.0) kg*(5.0)²(m/s)² + (95.0)kg*(3.0)(m/s)²) = 1,553 J

After the collision:

Kf = 1/2 *(m₁+ 767.2 Jm₂)*vf² = 1/2*185 kg*(2.88)²(m/s)²= 767.2 J

The mechanical energy lost during the collision is just the difference between the final and initial kinetic energy:

ΔK = Kf - K₀ = 767.2 - 1,553 J = -785.8 J

So, the magnitude of the energy lost during the collision is 785.8 J.

d) This energy is lost during the collision as thermal energy, due to the internal friction between both players.

7 0
4 years ago
When training to increase physical fitness, adaptation occurs....
user100 [1]
I believe A.. training never gives instant results and is never a quick or short process !
3 0
4 years ago
Read 2 more answers
Other questions:
  • lipids are stored until they are needed by cells. there are molecules in the cells that break the lipids down and convert them i
    8·1 answer
  • True or false a reference point is the object to which we compare another object's motion
    14·2 answers
  • How to remember physics derivation easier? <br>​
    12·1 answer
  • What is the wavelength of an fm radio signal that has a frequency of 91.7 mhz?
    14·1 answer
  • Examine the lightbulbs in the circuit below. Write a sentence explaining what would happen if lightbulb A burned out. Repeat thi
    11·1 answer
  • two balls each with with a mass of 0.5 kg, collide on a pool table. is the law of conservation of momentum satisfied in this col
    13·2 answers
  • Which of these will have the same units if acceleration is multiplied by time? Force Time velocity Pressure
    8·2 answers
  • How are sports related to the concepts of force and motion?
    14·1 answer
  • The pressure exerted by a phonograph needle on a record is surprisingly large, due to the very small width of the needle. If the
    8·1 answer
  • 1) Differences between Magnetic Induction and Demagnetization
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!