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3 years ago

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Damian is texting while driving and ends up slamming into the back end of the car in front of him, which then strikes the car in

front of it. By texting while driving, Damian has created a

1 answer:

Sphinxa [80]3 years ago

3 0

The answer for this question is negative externality

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How do i solve this?

kenny6666 [7]

Answer:

hmmm i dont know....

Explanation:

i just wanted free point. TANKS YOU SIR!!

7 0

3 years ago

A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.

mash [69]

a) The acceleration of gravity is $4.96 m/s^2$

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

a)

The acceleration of gravity for an object near a planet is given by

$g=\frac{GM}{(R+h)^2}$

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2$

b)

The critical velocity for a satellite orbiting around a planet is given by

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

where we have again:

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s$

Converting into km/h,

$v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h$

c)

The period of the orbit is given by the circumference of the orbit divided by the velocity:

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

where

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

v = 6668 m/s

Substituting,

$T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s$

d)

One day consists of:

$t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s$

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

$n=\frac{t}{T}=\frac{86400}{8452}=10.2$

e)

The escape velocity for an object in the gravitational field of a planet is given by

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

where here we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

Substituting, we find

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s$

f)

We can apply again the formula to find the escape velocity for the rocket:

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

Where this time we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=0$ , because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

6 0

3 years ago

PLEASE HELPPPPP

Ganezh [65]

The only graph that accurately depict the given motion is graph D.

The given parameters;

initial position of the man = 0

direction of the man's first displacement = backward
time of first motion, t₁ = 6 seconds
velocity of this first displacement = v₁
time without any motion (<em>zero movement</em>) = 6 seconds
direction of the second displacement = forward
velocity of second displacement = 2v₁

Let the acceleration of the first displacement = a

Acceleration of the second displacement = 2a

From the given graphs we can eliminate every graph without initial decrease or motion towards the negative direction.

The only options with initial motion towards the negative direction are;

<em>graph B</em>, and

<em>graph D</em>.

The difference between graph B and D;

in graph B there is a uniform motion for 6 seconds

in graph D there is no motion for 6 seconds (<em>this is obvious as the line fall directly on top of the horizontal axis maintaining a value of zero for 6 seconds</em>).

Thus, the only graph that accurately depict the given motion is graph D.

Learn more here: brainly.com/question/21095906

5 0

2 years ago

When you're asked to calculate moments and torque, what's the main difference in calculation? Also what is a couple?

leva [86]

Answer:

a) Explanation below. b) Explanation below

Explanation:

Torque is defined as the product of a force by a radius, while momentum is defined as the product of force by a distance. Mathematically we would have

T = F * r

M = F * d

where:

T = torque = [N*m]

M = moment = [N*m]

F = force =[N]

d = distance [m]

r = radius [m]

Although they have the same units, the difference between them is the application. For the case of torque this is always applied in parts that are in rotation, such as the shafts of cars, the shafts of pumps, torque in gears and etc. While the moment can be applied to a body without the need for it to rotate.

A couple, is as its name suggests a couple of forces of equal magnitude but opposite sense and do not share a line of action. A body under the action of a couple of forces tends to rotate the body without moving it from one point to another.

7 0

3 years ago

In a clock, the average speed is maximum for the tip of<br />a) Seconds hand b) Hours hand c) Minutes hand

Simora [160]

A, hope this helped! I didn’t really get it but I think it’s correct?

3 0

3 years ago

Read 2 more answers