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2 years ago

What is the acceleration of a Porsche that can go from 15 mi/hr to 75 mi/hr in 4 seconds?

Physics

2 answers:

Elodia [21]2 years ago

8 0

Hi there!

Acceleration = change in velocity / change in time = Δv/Δt

Thus:

a = (75 - 15)/4 = 60/4 = 15 mi/hr²

AveGali [126]2 years ago

6 0

Equation you need to use is a=Vf-Vi/t
Vi=initial velocity
Vf= Final Velocity
Write out the variables and their numbers
Vf=75 mi/hr Vi=15 mi/hr t=4 seconds
Plug it all in
75-15/4
60/4
a=15 mi/hr^2

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Continuing in an existing state. Resistance to change.

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Which of the following is the best definition of an element?

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Answer:

A. chemical substance whose atoms all have the same number of protons

Explanation:

An element is a substance which contains identical atoms that have the same number of protons in the nucleus.

Elements are arranged in the periodic table according to their atomic number (= number of protons): so atoms of different elements have a different number of protons in their nuclei.

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3 years ago

What pushes against gravity in: a main sequence star, a white dwarf, a neutron star, and a black hole? electron degeneracy, neut

Inga [223]

Answer:

heat pressure, electron degeneracy, neutron degeneracy, and nothing

Explanation:

Main Sequence Star: It is a star in which nuclear fusion is happening in the core of the star. Hydrogen molecules fuse together to generate Helium. This nuclear fusion generates outward gas pressure and radiation pressure which balances the inward gravity thus creating an equilibrium which keeps the stars in shape.

White dwarf: It is the end stage of a medium sized star like the Sun. Outer layers of the star are thrown in the form a shell/bubble leaving a small and dense core in the center called as white dwarf. This core consists of carbon and oxygen. Nuclear fusion doesn't occur in the core of white dwarfs. The inward gravity is balanced by the electron degeneracy pressure. Thus these stars will keep on radiating the remaining heat and will turn in to a black dwarf at the end.

Neutron Star: This is the end stage of a supermassive star (1-3 times the mass of the Sun). At the last stage of the life the core collapses. In these stars the inward gravity is so huge that the pressure overcomes the electron degeneracy pressure and crushes together the electron and proton to form neutron. The neutron then stops the collapse and balances the inward gravity.

Black Hole: This is the end stage of a hyper massive stars weighing more than 3 times the mass of the Sun. The inward gravitational force is so huge that even the neutrons are not able to stop the collapse the core. thus the mass of the star collapses into a very small area of immense gravity. There is nothing that can balance this inward gravity.

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2 years ago

A 0.500 kg bullet is fired from a gun at 25.0 m/s, how much kinetic energy does it have?

slamgirl [31]

Considering the definition of kinetic energy, the bullet has a kinetic energy of 156.25 J.

<h3>Kinetic energy</h3>

Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and in a rest position, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless there is a change in speed or the body returns to its rest state by applying a force to it.

The kinetic energy is represented by the following expression:

Ec= ½ mv²

Where:

Ec is the kinetic energy, which is measured in Joules (J).
m is the mass measured in kilograms (kg).
v is the speed measured in meters over seconds (m/s).

<h3>Kinetic energy of a bullet</h3>

In this case, you know:

m= 0.500 kg
v= 25 m/s

Replacing in the definition of kinetic energy:

Ec= ½ ×0.500 kg× (25 m/s)²

Solving:

Finally, the bullet has a kinetic energy of 156.25 J.

Learn more about kinetic energy:

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1 year ago

You have been hired as a technical consultant for an early-morning cartoon series for children to make sure that the science is

katen-ka-za [31]

The initial potential energy of the wagon containing gold boxes will enable

it roll down the hill when cut loose.

The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

$K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}$

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

$\dfrac{1}{2} \times 166 \times v^2 = 17457.0912$

$v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5$

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

$\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2) }}$

Which gives;

$\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66) } \approx 8.1$

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

$Time = \dfrac{Distance}{Velocity}$

$\mathrm{The \ time \ the\ Lone \ Ranger \ and \ Tonto \ have, \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s$

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

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2 years ago