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3 years ago

Rachel is in a car that was struck by lightning. is she safe?

Physics

2 answers:

ArbitrLikvidat [17]3 years ago

8 0

NO.

Lightning is dangerous

Also a car has many parts that are electrical and they can explode and cause other damage.

The car will explode (very likely) if the person is not injured immediately

Eduardwww [97]3 years ago

5 0

yes she is very safe inside

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what are the various ways in which lithospheric plates interact with each other as they move around on a dynamic earth??

Ivan

<span>The plates make up Earth's outer shell, called the lithosphere. (This includes the crust and uppermost part of the mantle.) Churning currents in the molten rocks below propel them along like a jumble of conveyor belts in disrepair. Most geologic activity stems from the interplay where the plates meet or divide.</span>

3 0

3 years ago

Please help!! giving a lot of points

Readme [11.4K]

Question 1.

mass = 4500 kg
potential energy (p.e) = 67500 J

now, we know :

=》

$p.e = mgh$

=》

$67500 = 4500 \times 10 \times h$

=》

$67500 = 45000 \times h$

=》

$h = \dfrac{67500}{45000}$

=》

$h = 1.5 \: m$

note : if we take acceleration due to gravity as 9.8, then height = 1.53 m

Question 2.

mass = 4500 kg
kinetic energy = 63000 j

we know,

=》

$k.e = \dfrac{1}{2} mv {}^{2}$

=》

$63000 = \dfrac{1}{2} \times 4500 \times {v}^{2}$

=》

${v}^{2} = \dfrac{63000 \times 2}{4500}$

=》

${v}^{2} = 28$

=》

$v = \sqrt{28}$

=》

$v = 2 \sqrt{7} \: \: ms {}^{ - 1}$

=》

$5.29 \: \: ms {}^{ - 1}$

7 0

3 years ago

A satellite is in circular orbit at an altitude of 1500 km above the surface of a nonrotating planet with an orbital speed of 9.

Ksju [112]

To solve this problem we will use the Newtonian theory about the speed of a body in space for which the speed of a body in the orbit of a planet is summarized as:

$v = \sqrt{\frac{2GM}{R}}$

Where,

G = Gravitational Universal Constant

M = Mass of Planet

r = Radius of the planet ('h' would be the orbit from the surface)

The escape velocity is

$v = 14.9km/h = 14900m/s$

Through this equation we can find the mass of the Planet in function of the distance, therefore

$M = \frac{v^2R}{2G}$

$M = \frac{14900^2R}{2(6.67*10^{-11})}$

$M = 16.64*10^{17}R$

The orbital velocity is

$v_o = \sqrt{\frac{GM}{R+h}}$

$9200^2 = \frac{(6.67*10^{-11})(16.64*10^{17})R}{R+1500*10^3}$

$11.1*10^7R = (R+15000*10^3)(9200)^2$

$2.64*10^7R = 12.69*10^{13}$

$R = 4.81*10^6m$

The time period of revolution is,

$T = \frac{2\pi(R+h)}{v_o}$

$T = \frac{2\pi(4.81*10^6+1.5*10^6)}{9200}$

$T = 4307s$

$T = 72min = 1hour12min$

Therefore the orbital period of the satellite is closes to 1 hour and 12 min

3 0

3 years ago

According to Coulomb’s Law, what happens to the force when the distance increase between 2 particles?

ohaa [14]

Answer:

The size of the force varies inversely as the square of the distance between the two charges. Therefore, if the distance between the two charges is doubled, <u>the attraction or repulsion becomes weaker</u>, decreasing to one-fourth of the original value.

Explanation:

Coulomb’s law, mathematical description of the electric force between charged objects. Formulated by the 18th-century French physicist Charles-Augustin de Coulomb, it is analogous to Isaac Newton’s law of gravity.

Both gravitational and electric forces decrease with the square of the distance between the objects, and both forces act along a line between them. In Coulomb’s law, however, the magnitude and sign of the electric force are determined by the electric charge, rather than the mass, of an object. Thus, charge determines how electromagnetism influences the motion of charged objects. Charge is a basic property of matter. Every constituent of matter has an electric charge with a value that can be positive, negative, or zero.

Coulomb's Law says that the force between 2 charges is proportional to the product of the quantities of charge on each and inversely proportional to the square of the distance between them. The formula for Coulomb's Law is $F=k\frac{q_{1}q_{2} }{r^{2} }$ .

$F$ is the force.

$k$ is the Coulomb's constant ( $8.987*10^{9} \frac{Nm^{2} }{C^{2} }$ ).

$q_{1}$ is the electric charge of object 1.

$q_{2}$ is the electric charge of object 2.

$r$ is the distance between the two charges.

Electric force is inversely proportional to ( $r^{2}$ ) instead of ( $r$ ). As the distance between charges increases, the electric force decreases by a factor of $\frac{1}{r^{2} }$ .

8 0

3 years ago

A fighter jet accelerates from 10 m/s to 75 m/s in 7.0 s. The acceleration of the jet is

liraira [26]

Acceleration = final velocity - inital / time
a = 75-10 / 7
a = 65 / 7
a = 9.29 m/s^2

4 0

3 years ago