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2 years ago

Hi so this is 15 points :D

Physics

1 answer:

mario62 [17]2 years ago

8 0

Answer:

A. The core makes up the majority of Earth's volume.

Explanation:

Although the core and mantle are about equal in thickness, the core actually forms only 15% of earth's volume whereas the mantle takes up 84%

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suppose a car manufacturer tested its cars for front end collsion by hauling them up on a crane and dropping them from a certain

IRINA_888 [86]

Initial height: 66.5 m

Explanation:

The problem can be solved by using the principle of conservation of energy.

If we neglect air resistance, the total mechanical energy of the car is conserved during the fall, therefore we can write:

$K_i + U_i = K_f + U_f$

where :

$K_i = 0$ is the kinetic energy of the car at the top (it starts from rest)

$U_i = mgh$ is the gravitational potential energy of the car at the top, with:

m = the mass of the car

g = the acceleration of gravity

h = the heigth of the car

$K_f = \frac{1}{2}mv^2$ is the kinetic energy of the car just before hitting the ground, with

v = 130 km/h final speed of the car

$U_f = 0$ is the gravitational potential energy of the car at the bottom

Re-arranging the equation, we find

$mgh=\frac{1}{2}mv^2$

and we have:

$g=9.8 m/s^2\\v = 130 km/h = 36.1 m/s$

Solving for h, we find the initial height of the car:

$h=\frac{v^2}{2g}=\frac{36.1^2}{2(9.8)}=66.5 m$

Learn more about kinetic energy and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

5 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS AND I NEED ALL CORRECT ANSWERS ONLY!!!

Dominik [7]

the purpose of fuses and circuit breakers is (first answer)

7 0

3 years ago

Read 2 more answers

Can someone solve this problem and explain to me how you got it

evablogger [386]

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒ $F\alpha\frac{q1*q2}{r^{2}}$

∴ $F=k\frac{q1*q2}{r^{2}}$

where $F$ is the force of attraction or repulsion

$k$ is Coulumb's constant= $9*10^{9}Nm^{2}C^{-2}$

$q1$ and $q2$ are the magnitude of the charges

$r$ is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question5:

Given: q1=0.041 C, q2=0.029 C, r=12 m

therefore by Coulumb's law,

$F=9*10^{9}*\frac{0.041*0.029}{12^{2}}$

$F=74312.5N$

Question6:

Given: q1= $3*10^{-18}C$ , r=5 m, F= $4*10^{-11}N$

therefore by Coulumb's law,

$4*10^{-11}=9*10^{9}*\frac{3*10^{-18}*q2}{5^{2}}$

⇒ $q2=\frac{4*10^{-11}*25}{9*10^{9}*3*10^{-18}} \\=0.37C$

4 0

3 years ago

What two characteristics are used to classify air masses

IgorLugansk [536]

Answer: Temperature and humidity are the two characteristics used to classify air masses.

Explanation:

7 0

3 years ago

A 110-g bullet is fired from a rifle having a barrel 0.636 m long. Choose the origin to be at the location where the bullet begi

astraxan [27]

Data:
mass = 110-g bullet
d = 0.636 m
Force = 13500 + 11000x - 25750x^2, newtons.

a) Work, W

W = ∫( F* )(dx) =∫[13500+ 11000x - 25750x^2] (dx) =

W = 13500x + 5500x^2 - 8583.33 x^3 ] from 0 to 0.636 =

W = 8602.6 joule

b) x= 1.02 m

W = 13500x + 5500x^2 - 8583.33 x^3 ] from 0 to 1.02

W = 10383.5

c) %

[W in b / W in a] = 10383.5 / 8602.6 = 1.21 => W in b is 21% more than work in a.

7 0

3 years ago