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2 years ago

What happens to momentum during a collision?...i give brainliest

Physics

1 answer:

Vlada [557]2 years ago

7 0

Answer:

Momentum is the product of a moving object's mass and velocity . ... When two objects collide the total momentum before the collision is equal to the total momentum after the collision (in the absence of external forces). This is the law of conservation of momentum. It is true for all collisions.

Explanation:

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Examples of energy balance: physical education

mart [117]

Answer:children burn calories to being a student

Explanation:That mean when a children getting ready to go to high school

7 0

2 years ago

An 84.0 kg sprinter starts a race with an acceleration of 1.76 m/s2. If the sprinter accelerates at that rate for 11 m, and then

gulaghasi [49]

Answer:

t=17.838s

Explanation:

The displacement is divided in two sections, the first is a section with constant acceleration, and the second one with constant velocity. Let's consider the first:

The acceleration is, by definition:

$a=\frac{dv}{dt}=1.76$

So, the velocity can be obtained by integrating this expression:

$v=1.76t$

The velocity is, by definition: $v=\frac{dx}{dt}$ , so

$dx=1.76tdt\\x=1.76\frac{t^{2}}{2}$ .

Do x=11 in order to find the time spent.

$11=1.76\frac{t^2}{2}\\ t^2=\frac{2*11}{76} \\t=\sqrt{12.5}=3.5355s$

At this time the velocity is: $v=1.76t=1.76*3.5355s=6.2225\frac{m}{s}$

This velocity remains constant in the section 2, so for that section the movement equation is:

$x=v*t\\t=\frac{x}{v}$

The left distance is 89 meters, and the velocity is $6.2225\frac{m}{s}$ , so:

$t=\frac{89}{6.2225}=14.303s$

So, the total time is 14.303+3.5355s=17.838s

7 0

2 years ago

What is the approximate size of the Earth's magnetic field? (dont ask me to specify thats what the question is and im as confuse

Olegator [25]

Answer:

The Earth's magnetic field intensity is roughly between 25,000 - 65,000 nT (.25 -.65 gauss).

Explanation:

To measure the Earth's magnetism in any place, we must measure the direction and intensity of the field. The Earth's magnetic field is described by seven parameters. These are declination (D), inclination (I), horizontal intensity (H), the north (X), and east (Y) components of the horizontal intensity, vertical intensity (Z), and total intensity (F). The parameters describing the direction of the magnetic field are declination (D) and inclination (I). D and I are measured in units of degrees, positive east for D and positive down for me. The intensity of the total field (F) is described by the horizontal component (H), vertical component (Z), and the north (X) and east (Y) components of the horizontal intensity. These components may be measured in units of gauss but are generally reported in nanoTesla (1nT * 100,000 = 1 gauss). The Earth's magnetic field intensity is roughly between 25,000 - 65,000 nT (.25 - .65 gauss). Magnetic declination is the angle between magnetic north and true north. D is considered positive when the angle measured is east of true north and negative when west. The magnetic inclination is the angle between the horizontal plane and the total field vector, measured positive into Earth. In older literature, the term “magnetic elements” is often referred to as D, I, and H.

8 0

2 years ago

Do atoms have full outer energy level combine with other atoms

Pavlova-9 [17]

All of the Noble Gases, which are on the right side of the periodic table, have a full outer energy level. The elements that are Noble Gases are the following: Neon Argon Krypton Xenon Radon Ununoctium.

Hope this helps.

8 0

2 years ago

Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible co

Illusion [34]

Answer:

q = 8.61 10⁻¹¹ m

charge does not depend on the distance between the two ships.

it is a very small charge value so it should be easy to create in each one

Explanation:

In this exercise we have two forces in balance: the electric force and the gravitational force

F_e -F_g = 0

F_e = F_g

Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.

Let's write Coulomb's law and gravitational attraction

$k \frac{q_1q_2}{r^2} = G \frac{m_1m_2}{r^2}$

In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.

k q² = G m²

q = $\sqrt{ \frac{G}{k} }$ m

we substitute

q = $\sqrt{ \frac{ 6.67 \ 10^{-11}}{8.99 \ 10^{9}} }$ m

q = $\sqrt{0.7419 \ 10^{-20}}$ m

q = 0.861 10⁻¹⁰ m

q = 8.61 10⁻¹¹ m

This amount of charge does not depend on the distance between the two ships.

It is also proportional to the mass of the ships with the proportionality factor found.

Suppose the ships have a mass of m = 1000 kg, let's find the cargo

q = 8.61 10⁻¹¹ 10³

q = 8.61 10⁻⁸ C

this is a very small charge value so it should be easy to create in each one

6 0

2 years ago