WHAT ARE THE NECESSARY CONDITIONS TO CREATE A WAVE?

1. Define dark matter and dark energy, and clearly distinguish between them. What types of observations have led scientists to p

Vlada [557]

Explanation:

Dark Energy is a property of the entire universe. And the universe is roughly made up of 68% of this dark energy whereas dark matter makes up 27% of the universe matter.

The influence of the dark matter shows in the individual galaxies whereas dark energy acts only on the entire universe. dark matter is responsible for the gravity effect on the plants and the heavenly bodies which produces gravity force. But the dark energy is known as the anti gravity force as it is responsible for the expansion of the universe. Dark matter attracts and produces an attractive force while dark energy repels and produces a repulsive force.

Scientist believe that they exists to account for the very fact that gravity holds all the heavenly bodies and the galaxies together. They are even responsible for the fluctuation in Cosmic Microwave background. Researchers and scientist firmly believe that all the visible matter that exists may not have enough gravitational pull to hold everything together in the galaxies. And also the expanding universe strongly proves that dark energy exists in the universe.

7 0

3 years ago

Use Newton's laws to explain why a falling object dropped from a 57m tower accelerates initially but then reaches constant veloc

snow_lady [41]

Answer:

At the point of dropping the object, by Newton's first law due to gravitational force $F_g$ = m × g, accelerates

By Newton's Second law the object reaches impacts on the air with the gravitational force resulting in changing momentum of m×(Final Velocity - Initial Velocity)

As the velocity increases, the rate of change of momentum becomes equivalent to the gravitational force and by Newton's third law, the action action and reaction are equal and opposite hence they cancel each other out

The body then moves at a constant uniform motion down according to Newton's first law

Explanation:

At the point the object of mass, m, is dropped from the height of the tower, the only force acting on the object is the gravitational force such that the object has an acceleration which is the acceleration due to gravity, g, and the gravitational force is therefore = m × g

As the speed of the object increases while the object is falling with the gravitational acceleration the rate at which the object cuts through layers of air which (by Newton's first law of motion, are at rest ) has some buoyancy effect also increases therefore, the object is constantly increasingly changing the momentum of the air which by Newton's second law results, at an high enough velocity, and by Newton's third law, in a force equal to the applied gravitational force

Therefore, the force of the air drag becomes equal to the gravitational force, cancelling each other out and the object then moves according to Newton;s first law, in uniform motion of a constant speed while still falling down.

5 0

3 years ago

Two metra trains approach each other on separate but parallel tracks. one has a speed of 90 km/hr, the other 80 km/hr. initially

Gennadij [26K]

The trains take <u>57.4 s</u> to pass each other.

Two trains A and B move towards each other. Let A move along the positive x axis and B along the negative x axis.

therefore,

$v_A=90 km/h\\ v_B=-80 km/h$

The relative velocity of the train A with respect to B is given by,

$v_A_B=v_A-v_B\\ =(90km/h)-(-80km/h)\\ =170km/h$

If the train B is assumed to be at rest, the train A would appear to move towards it with a speed of 170 km/h.

The trains are a distance d = 2.71 km apart.

Since speed is the distance traveled per unit time, the time taken by the trains to cross each other is given by,

$t= \frac{d}{v_A_B}$

Substitute 2.71 km for d and 170 km/h for $v_A_B$

$t= \frac{d}{v_A_B}\\ =\frac{2.71 km}{170 km/h} \\ =0.01594 h$

Express the time in seconds.

$t=(0.01594h)(3600s/h)=57.39s$

Thus, the trains cross each other in <u>57.4 s</u>.

6 0

3 years ago

A 2.30-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so it is free to swing about that end. a solid sph

Marina86 [1]

Solution:

initial sphere mvr = final sphere mvr + Iω
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m²
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω
where: ω = 2.87 rad/s

So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)²
E = 12.64 J becomes PE = mgh, so
12.64 J = 2.3 kg * 9.8m/s² * h
h = 0.29 m

h = L(1 - cosΘ) → where here L is the distance to the CM
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ
Θ = arccos((1-0.29)/1) = 44.77 º

8 0

3 years ago

If two charged objects in a laboratory are brought to a distance of 0.22 meters away from each other. What is

zysi [14]

Answer:

$q_2=2.47\times 10^{-4}\ C$

Explanation:

The charge on one object, $q_1=9.9\times 10^{-5}\ C$

The distance between the charges, r = 0.22 m

The force between the charges, F = 4,550 N

Let q₂ is the charge on the other sphere. The electrostatic force between two charges is given by the formula as follows :

$F=\dfrac{kq_1q_2}{r^2}\\\\q_2=\dfrac{Fr^2}{kq_1}\\\\q_2=\dfrac{4550\times (0.22) ^2}{9\times 10^9\times 9.9\times 10^{-5}}\\\\q_2=2.47\times 10^{-4}\ C$

So, the charge on the other sphere is $2.47\times 10^{-4}\ C$ .

7 0

3 years ago