Calculate the kinetic energy of a 100.0-kg meteor approaching the Earth at a speed of 10.0 km/s. Remember that 1 km = 1000 m.



5.00 × 109 J


10 × 107 J


50.0 × 108 J


1.00 × 108 J

Answers

Answer 1
Answer: 5.0 x 10^9 J
Calculate The Kinetic Energy Of A 100.0-kg Meteor Approaching The Earth At A Speed Of 10.0 Km/s. Remember

Related Questions

A swimmer heading directly across a river that is 200 m wide reaches the opposite bank in 6 min 40 s. During this swim, she is swept downstream 480 m. How fast can she swim in still water

Answers

Answer:

The speed of the swimmer in stil water is 0.5 m/s

Explanation:

Given;

total time taken to swim across = 6 mins 40 s = (6 x 60s) + 40 s = 400 s

width of the river, = 200 m

Please find the image attached for explanation.

jjjjjjjjjjjjalkshadl,wnkdalsssssicj

Answers

Intramolecular Forces:-

An intramolecular force (or primary forces) is any force that binds together the atoms making up a molecule or compound, not to be confused with intermolecular forces, which are the forces present between molecules.

Which of the following elements has the largest atomic radius?
Silicon
Aluminum
Sulfur
Phosphorous

Answers

aluminum has the largest atomic radius

Answer:

francium

Atomic radii vary in a predictable way across the periodic table. As can be seen in the figures below, the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. Thus, helium is the smallest element, and francium is the largest.

A pilot wishes to fly from Bayfield to Kitchener, a distance of 100 km on a bearing of 105°. The speed of the plane in still air is 240 km/h. A 20 km/h wind is blowing on a bearing of 210°.
Remembering that she must fly on a bearing of 105° relative to the ground (i.e. the resultant must be on that bearing), find (6 marks)
a) the heading she should take to reach her destination.
b) how long the trip will take.

Answers

Here is are by step :) hope this helps

First things first, you want the resultant to be in the direction of the bearing of 105 °. The distance required is irrelevant for question 1.

You want the addition of the wind speed and the planes velocity to equal some resultant in the direction one 105 °. Draw a parallelogram (vector addition) of the resultant (the 100km in direction 105°) and the wind speed. The diagonal that results from the vector addition will equal 240. You only know the angle between the resultant and the wind speed right now (105°).

https://www.geogebra.org/geometry/rbejhvta

What is the angle between the resultant (the direction we want to travel) and the direction the plane will actually go when wind speed is taken into accound???

Law of sines.

You said you have it, so we will continue under the assumption you got 4.617°.

now, take into account that the plane is accelerating 4.617° off of the 105° bearing in order to travel on the 105° bearing when wind speed is taken into account... what heading is the plane taking? 105° - 4.617° = 100.4°

There is part 1.

now you simply need to find the actual speed she is travelling along the bearing 105°. (she is traveling 240km/h with a 100.4° heading, find the correct speed for the resultant).

EDIT: The representation says 105.1°, it is supposed to say 105°

A constant force moves an object along the line segment from to . Find the work done if the distance is measured in meters and the force in newtons.

Answers

This question is incomplete, the complete question is;

Flag

A constant force F = 6i+8j-6k moves an object along a straight line from point (6, 0, -10) to point (-6, 7, 2).

Find the work done if the distance is measured in meters and the magnitude of the force is measured in newtons.

Answer:

the work done is -88 J

Explanation:

Given the data in the question;

we know that;

Work done = F × S

where constant force F = ( 6i + 8j - 6k )

S = ( -6i + 7j + 2k ) - ( 6i + 0j - 10k )

S = ( (-6i - 6i) + (7j - 0j) + ( 2k - ( -10k) ) )

S = ( -12I + 7j + 12k )

so

Work force = ( 6i + 8j - 6k ) × ( -12I + 7j + 12k )

Work force = ( 6 × -12 ) + ( 8 × 7 ) + ( -6 × 12 )

Work force = -72 + 56 - 72

Work force = -88 J

Therefore, the work done is -88 J


During the fission reaction shown, how did the target nucleus change ?

Answers

Answer:

A. The target nucleus split into two nuclei, each with fewer nucleons than the original.

Explanation:

Answer: A. The target nucleus split into two nuclei, each with fewer nucleons than the original.

Explanation:

Help!!
A table is pushed across the floor for a distance of 32 m with a force of 320 N in 150 seconds. How much power was used?
A.70.2W
B.68.3W
C.56.7W
D.49.8W

Answers

Compute the work done on the table:

W = Fd = (320 N) (32 m) = 10,240 J

Divide this by the given time duration to get the power output:

P = W/∆t = (10,240 J) / (150 s) ≈ 63.3 W

What is the friction force on a box that has a mass of 15kg as it slides across the floor. The coefficient of friction of the not very clean floor is 0.25
please explain everything including formula used ​

Answers

Answer:

36.75 N

Explanation:

Applying

F = mgμ................. Equation 1

Where F = Friction force on the box, m = mass of the box, g = acceleration due to gravity of the box, μ = coefficient of static friction

From the question,

Given: m = 15 kg, μ = 0.25

Constant: g = 9.8 m/s²

Substitute these values into equation 1

F = 15(9.8)(0.25)

F = 36.75 N

Hence the friction force on the box is 36.75 N

An empty 12,954 kg railroad car, traveling at a speed of 28 m/s strikes a partially filled 17,616 kg railroad car moving in the same direction at a speed of 5 m/s. What is the total momentum of the two railroad cars AFTER the collision?

Answers

Answer:

450792 kgm/s

Explanation:

by conservation of momentum,

total momentum AFTER collision = total momentum BEFORE collision

                                                       =mv+m'v'

                                                       =12954×28+17616×5

                                                       =450792 kgm/s

PLEASE ANSWR 1ST AND I WILL MARK U BRAINLIEST
Two statements are given- one labeled Assertion (A) and the other labeled Reason ®. Select
the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:
a. Both A and R are true, and R is correct explanation of the assertion.
b. Both A and R are true, but R is not the correct explanation of the assertion.
c. A is true, but R is false.
d. A is false, but R is true.
Assertion: An object has a negative acceleration.
Reason: The velocity of an object decreases in the same direction.

Answers

Answer:

Where is the R statement?

A hoop rolls with constant velocity and without sliding along level ground. Its rotational kinetic energy is:______a- half its translational kinetic energyb- the same as its translational kinetic energyc- twice its translational kinetic energyd- four times its translational kinetic energy

Answers

Answer:

The same as its translational KE.

The easy way to do this is to make up numbers and use them.

So, I'll say m=2 and r=3. I will also say v=3 .

Rot. Inertia of a hoop is mr^2. So the rot KE is: 1/2 (mr^2)(w^2)

note: (1/2*I*w^2)

Translational kinetic energy is basically normal KE, so 1/2(m)(v^2)

Now, lets plug our made up values in:

Rot Ke : 1/2 (9*2)(3/3) *note w = v/r

Tran Ke: 1/2(2)(9)

Rot Ke: 9

Tran Ke: 9

9=9, same.

differentiate between step up and step down transformer​

Answers

Answer:

The main difference between the step-up and step-down transformer is that the step-up transformer increases the output voltage, while the step-down transformer reduces the output voltage.

A giant chorus of 1000 vocalists is singing the same note. Suddenly, 999 vocalists stop,leaving one soloist. By how many decibels does the sound intensity level decrease? Explain.

Answers

Answer:

The decrease in decibels is 0.1 dB.

Explanation:

Let the intensity of one chorus is Io.

let the intensity of 1000 vocalist is dB.

The intensity of 1000 vocalist is 1000 Io.

[tex]dB = 10log\frac{1000Io}{Io}=30[/tex]..... (1)

let the intensity of 999 vocalist is dB'.

[tex]dB' = 10log\frac{999Io}{Io}=29.9[/tex]..... (2)

So, the change is

= dB - dB' = 30 - 29.9 = 0.1 dB

explain why this is important for life on earth​

Answers

Answer:

THIS is very important.

Explanation:

If the word "this" was never invented, people wouldn't have a way to tell another person what is directly in front of them.

BTW this is a joke.

Find the ratio of the diameter of copper to iron wire, if they have the same resistance per unit length (as they might in household wiring). dCu dFe =

Answers

Answer:

The ratio of diameter of copper wire to the iron wire is 0.42.

Explanation:

length of both the wires is same as L and resistance is R.

resistivity of copper = 1.7 x 10^-8 ohm m

resistivity of iron = 9.7 x 10^-8 ohm m

Let the diameter of copper is d and for iron is d'.

The formula of the resistance is

[tex]R = \rho\times\frac{L}{A}\\\\R = \rho\times\frac{4L}{\pi d^2}.... (1)\\And\\ R = \rho'\times\frac{4L}{\pi d'^2}.... (2)\\\\comparing (1) by (2)\\\\1.7\times10^{-8}\times\frac{4L}{\pi d^2}=9.7\times10^{-8}\times\frac{4L}{\pi d'^2}\\\\d : d' = 0.42[/tex]

A spring scale hung from the ceiling stretches by 6.1cm when a 2.0kg mass is hung from it. The 2.0kg mass is removed and replaced with a 2.8kg mass.What is the stretch of the spring?

Answers

2.0kg and the mass of sometimes makes it hard 2.8kg

A 25 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic friction is 0.25, and the coefficient of static friction is 0.35. As the angle is increased, find the minimum angle at which the box starts to slip.


24.5°


18.5°


21.3°


19.3°

Answers

Up until the moment the box starts to slip, the static friction is maximized with magnitude f, so that by Newton's second law,

• the net force acting on the box parallel to the ramp is

F = mg sin(α) - f = 0

where mg sin(α) is the magnitude of the parallel component of the box's weight; and

• the net force acting perpendicular to the ramp is

F = n - mg cos(α) = 0

where n is the magnitude of the normal force and mg cos(α) is the magnitude of the perpendicular component of weight.

From the second equation we have

n = mg cos(α)

and f = µn = µmg cos(α), where µ is the coefficient of static friction. Substituting these into the first equation gives us

mg sin(α) = µmg cos(α)   ==>   µ = tan(α)   ==>   α = arctan(0.35) ≈ 19.3°

According to Coulomb's law, if the separation between two particles of the same charge increases four times, the potential energy of the two particles:__________.
a. is one-sixteenth as high as it was before the separation.
b. is four times as high as it was before the distance separation.
c. is one-fourth as high as it was before the separation.
d. does not change.

Answers

Answer:

c.   V = k Q1 * Q2 / R1      potential energy of Q1 and Q2 separated by R

V2 / V1 = (R1 / R2) = 1/4

V2 = V1 / 4

According to Coulomb's law, if the separation between two particles of the same charge increases four times, the potential energy of the two particles is is one-fourth as high as it was before the separation. The correct option is c.

What is Coulomb's law?

Coulomb's law states that the force of attraction or repulsion among both two charged bodies is directly related to the sum of their charges and inversely proportional to the square of the distance between them.

Coulomb force is the attraction or repulsion of particles or objects due to their electric charge. It is also known as electrostatic force or Coulomb interaction.

It denotes the electric force's inverse square dependence. It can also be used to accurately provide relatively simple derivations of Gauss' law for general cases.

Coulomb's law states that if the separation between two particles of the same charge increases four times, the potential energy of the two particles is one-fourth of what it was before the separation.

Thus, the correct option is c.

For more details regarding Coulomb's law, visit:

https://brainly.com/question/506926

#SPJ2

Problem

A charged particle is moving in the presence of uniform magnetic field. The mass of the particle

is m = 10−6 kg its charge is Q = 10−5 C and the magnetic field vector is B~ = (1T, 0, 0). At the

beginning the velocity vector of the particle is ~v0 = (12 m/s, 0, 5 m/s).

a.) How large will the x component of the velocity of the particle be in t = 2 s?

b.) Where will the particle be in t = 3.14 s?

c.) How large will the magnitude of the velocity be in t = 2.5 s?​

Answers

Answer:

Answer is a I checked the work

A train moves with a uniform velocity of 36km/hr 10sec. calculate the distance travelled​

Answers

Given:-

Speed = 36 km/hr

converting speed into m/s

Speed = 36*5/18

Speed = 10 m/s

t = 10 sec  

By using the Formula

Distance = Speed * time

D = 10*10

D = 100 m

Hope it helps....

A truck is hauling a 300-kg log out of a ditch using a winch attached to the back of the truck. Knowing the winch applies a constant force of 2850 N and the coefficient of kinetic friction between the ground and the log is 0.45, determine the time for the log to reach a speed of 0.5 m/s

Answers

Answer:

0.1 s

Explanation:

The net force on the log is F - f = ma where F = force due to winch = 2850 N, f = kinetic frictional force = μmg where μ = coefficient of kinetic friction between log and ground = 0.45, m = mass of log = 300 kg and g = acceleration due to gravity = 9.8 m/s² and a = acceleration of log

So F - f = ma

F - μmg = ma

F/m - μg = a

So, substituting the values of the variables into the equation, we have

a = F/m - μg

a = 2850 N/300 kg - 0.45 × 9.8 m/s²

a = 9.5 m/s² - 4.41 m/s²

a = 5.09 m/s²

Since acceleration, a = (v - u)/t where u = initial velocity of log = 0 m/s (since it was a rest before being pulled out of the ditch), v = final velocity of log = 0.5 m/s and t = time taken for the log to reach a speed of 0.5 m/s.

So, making t subject of the formula, we have

t = (v - u)/a

substituting the values of the variables into the equation, we have

t = (v - u)/a

t = (0.5 m/s - 0 m/s)/5.09 m/s²

t = 0.5 m/s ÷ 5.09 m/s²

t = 0.098 s

t ≅ 0.1 s

how to reduce fluid friction

Answers

Answer:

By making the object sharp pointed

Two thin conducting plates, each 56.0 cm on a side, are situated parallel to one another and 7.0 mm apart. If 10^-10 electrons are moved from one plate to the other, what is the electric field between the plates?

Answers

Answer:

[tex]E=576.5V/m[/tex]

Explanation:

From the question we are told that:

Length [tex]l=56.0cm=0.56m[/tex]

Distance apart [tex]d=7.0mm=0.007m[/tex]

Electron Transferred [tex]n=10^{-10}[/tex]

Therefore

Total Charge

Since Charge on each electron is

[tex]e=1.602*10^{-19}[/tex]

Therefore

[tex]T=1.602*10^{-19} *10^{10}[/tex]

[tex]T=1.602*10^{-9}[/tex]

Generally the equation for Charge density is mathematically given by

[tex]\rho=T/A[/tex]

Where

Area

[tex]A=0.56*0.56[/tex]

[tex]A=0.3136[/tex]

Therefore

[tex]\rho=1.602*10^{-9}/0.3136[/tex]

[tex]\rho=5.10*10^{-9}[/tex]

Generally the equation for Electric Field in the capacitor is mathematically given by

[tex]E=\frac{\rho}{e_0}[/tex]

[tex]E=\frac{5.10*10^{-9}}{8.85x10{-12}}[/tex]

[tex]E=576.5V/m[/tex]

1. Convert the following length into meters
a. 123.50mm
b. 560cm
c. 100dm
d. 125.89km​

Answers

I don’t really know this either

you are given a set of facts regarding a lens : object heigh, and dostance to objects. Given this jnformation, how can you tell if you're dealing with a concave or convex lens

Answers

Answer:

concave curves inward like an hourglass and convex is an outward curve like a football

Explanation:

hope this helps

The cycle is a process that returns to its beginning, but it does not repeat
itself.
True
False

Answers

False. It does repeat itself

if p=2i+4j+3k and q=I+5j-2k,find P×q.

Answers

Answer:

[tex]p\times q=-23i+7j+6k[/tex]

Explanation:

We are given that

p=2i+4j+3k

q=i+5j-2k

We have to find pxq

We know that

[tex]p\times q=\begin{vmatrix}  i&j  &k\\  2&4  & 3\\  1& 5 & -2\end{vmatrix}[/tex]

[tex]p\times q=i(-8-15)-j(-4-3)+k(10-4)[/tex]

[tex]p\times q=-23i+7j+6k[/tex]

Hence,[tex]p\times q=-23i+7j+6k[/tex]

A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. If the ball hits the ground 6.0 seconds later, approximately how high is the cliff? ​

Answers

Answer:

144 meters

Explanation:

it takes 6 seconds to hit the ground right and the ball lays off 24 m per second .

so by the time the ball hits the ground 6 seconds passed. so that means the cliff is 6.0×24=144

An electric field is given by in units of N/C when is in meters. What is the potential difference from point B at (0,7) m to point A at (7,0) m

Answers

Complete Question

An electric field is given by E= (5x, 0, 0) in units of N/C when x is in meters. What is the potential difference VA - VB from point B at (0,7) m to point A at (7,0) m? V

Answer:

[tex]V_A-V_B=-122.5V[/tex]

Explanation:

From the question we are told that:

Position of point [tex]B=(0,7) m[/tex]

Position of point [tex]A=(7,0) m[/tex]

Generally the equation for pd across the points is mathematically given by

 [tex]V_A-V_B=-\int_0^7(5x)d[/tex]

 [tex]V_A-V_B=-[\frac{5x^2}{2}]_0^7[/tex]

 [tex]V_A-V_B=-[\frac{5(7)^2}{2}][/tex]

 [tex]V_A-V_B=-122.5V[/tex]

A 8.37*10^-5 F capacitor has 2.15*10^-4 C of charge on its plates. How much energy is stored on the capacitor

Answers

Answer:

2.76*10^-4

Explanation:

Trust me :)

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