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3 years ago

8

You should always wear your seatbelt just in case the car comes to an abrupt stop. The seatbelt will hold you in place so that y

our body does not continue moving when the car stops.
What property of matter will keep your body in motion when the car comes to a halt?

Group of answer choices

inertia

mass

force

acceleration

2 answers:

JulsSmile [24]3 years ago

5 0

Answer:

inertia

first

Explanation

BigorU [14]3 years ago

4 0

Answer:

inertia

Explanation:

The property of matter that will keep the body in motion when the car comes to a halt is the inertia force.

Inertia is the ability of a body to remain in static position. It is the tendency to remain in a stable condition where there is no motion.

Newton's first law is the law of inertia and it states that a body remain in a state of rest or of uniform motion unless acted upon by an external force.
The ability to remain in state of rest by a body is predicated on the force of inertia.

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A runner of mass 80 kg is moving at 8.0 m/s. Calculate her kinetic energy.

lana [24]

Answer:

2560J

Explanation:

By definition the kinetic energy can be calculated in the following way:

K = (mv²)/2 = 80kg·(8.0m/s)²/2 = 2560 J

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2 years ago

VARVARA [1.3K]

Answer:

the correct one is D,

Ultraviolet, x-ray, gamma ray

Explanation:

Electromagnetism radiation are waves of energy that is expressed by the Planck relationship

E = h f

where h is the plank constant and f the frequency of the radiation.

Also the speed of light is

c = λ f

we substitute

E = h c /λ

therefore to damage the cells of the body radiation of appreciable energy is needed

microwave radiation has an energy of 10⁻⁵ eV

infrared radiation E = 10⁻² eV

visible radiation E = 1 to 3 eV

radiation Uv E = 3 to 6 eV

X-ray E = 10 eV

gamma rays E = 10 5 eV

therefore we see that the high energy radiation is gamma rays, x-rays and ultraviolet light.

When checking the answers, the correct one is D

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2 years ago

The _____ of a mechanical wave is a direct measure of its energy.

KiRa [710]

I think it's amplitude

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3 years ago

Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o

polet [3.4K]

Answer:

a. charge C experiences the greatest net force, and charge B receives the smallest net force

b. ratio=9

Explanation:

<u>Electrostatic Force</u>

Two point-charges $q_1$ and $q_2$ separated a distance d will exert a force on each other of a magnitude given by the Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{r^2}$

Where k is the proportional constant of value

$k=9*10^9\ N.m^2/c^2$

The diagram provided in the question shows four identical charges (let's assume their value is Q) separated by identical distance (of value d). The force between the charges next to others is

$\displaystyle F_1=\frac{k\ Q\ Q}{d^2}$

$\displaystyle F_1=\frac{k\ Q^2}{d^2}$

The force between charges separated 2d is

$\displaystyle F_2=\frac{k\ Q^2}{(2d)^2}$

$\displaystyle F_2=\frac{k\ Q^2}{4d^2}$

And the force between the charges A and D is

$\displaystyle F_3=\frac{k\ Q^2}{(3d)^2}$

$\displaystyle F_3=\frac{k\ Q^2}{9d^2}$

Now, let's analyze each charge and the force applied to them by the others

Let's recall equally signed charges repel each other and differently signed charges attrach each other

Charge A. It receives force to the left from B and C and to the right from D

$\displaystyle F_A=-F_1-F_2+F_3=-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}$

$\displaystyle F_A=\frac{k\ Q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

$\displaystyle F_A=-\frac{41}{36}F_1$

Charge B. It receives force to the right from A and D and to the left from C

$\displaystyle F_B=F_1-F_1+F_2=\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{4d^2}$

$\displaystyle F_B=\frac{1}{4}F_1$

Charge C. It receives forces to the right from all charges.

$\displaystyle F_C=F_2+F_1+F_1=\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{d^2}$

$\displaystyle F_C=\frac{9}{4}F_1$

Charge D. It receives forces to the left from all charges

$\displaystyle F_D=-F_3-F_2-F_1=-\frac{k\ Q^2}{9d^2}-\frac{k\ Q^2}{4d^2}-\frac{k\ Q^2}{d^2}$

$\displaystyle F_D=-\frac{49}{36}F_1$

Comparing the magnitudes of each force is just a matter of computing the fractions

$\displaystyle \frac{41}{36}=1.13,\ \frac{1}{4}=0.25,\ \frac{9}{4}=2.25,\ \frac{49}{36}=1.36$

a.

We can see the charge C experiences the greatest net force, and charge B receives the smallest net force

b.

The ratio of the greatest to the smallest net force is

$\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}=9$

The greatest force is 9 times the smallest net force

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Answer:

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