how are s waves and p waves simuliar?
A.they shake the ground
B.they travel through liquids
C. they arrive at the same time
D.they shake the ground from side to side

Answers

Answer 1

Answer:

A

Explanation:

hope this helps

Answer 2
The answer is A thank me later

Related Questions

the radius of the earth social

Answers

6,371km is the radius of the earth

1. According to its computer, a rocket launched
traveled 1200 m, had an average speed of 100.0
m/s. How-long did the trip take?

Answers

Answer:

I think it's = 12 seconds

Explanation:

the formula for speed is:

speed=[tex]\frac{distance}{time}[/tex] SO, time is equal to:

time=[tex]\frac{distance}{speed}[/tex]

(sub the numbers in the formula)

distance=1200m, speed=100m/s

time=[tex]\frac{1200}{100}[/tex]

=12 seconds

A diffusion couple, made by welding a thin onecentimeter square slab of pure metal A to a similar slab of pure metal B, was given a diffusion anneal at an elevated temperature and then cooled to room temperature. On chemically analyzing successive layers of the specimen, cut parallel to the weld interface, it was observed that, at one position, over a distance of 5000 nm, the atom fraction of metal A, NA, changed from 0.30 to 0.35. Assume that the number of atoms per m3 of both pure metals is 9 x 10^28. First determine the concentration gradient dnA/dx. Then if the diffusion coefficient, at the point in question and annealing temperature, was 2 10^-14 m^2/s.

Required:
Determine the number of A atoms per second that would pass through this cross-section at the annealing temperature.

Answers

Answer:

The value  is    [tex]H  =  18*10^{2} \  Atom / sec  [/tex]

Explanation:

From the question we are told that

  The atom fraction of metal A at point G is [tex] A  =  0.30 \ m[/tex]

   The atom fraction of metal  A at a distance 5000nm from G is  [tex]A_2 = 0.35[/tex]

   The number of atoms per [tex]m^3[/tex] is    [tex]N_h =  9 * 10^{28}[/tex]

    The diffusion coefficient is  [tex]D =   2* 10^{-14 } m^2/s[/tex]

Generally of the concentration of atoms of metal A at G is  

       [tex] N_A = A * N_h [/tex]

=>    [tex] N_A =  0.3  * 9 * 10^{28}[/tex]

=>     [tex] N_A =   2.7 * 10^{28} 2.7 atoms/m^3[/tex]

Generally of the concentration of atoms of metal A at a distance 5000nm from G is  

       [tex]D =  0.35 *9 * 10^{28}[/tex]

=>     [tex]D =  3.15 * 10^{28} \  atoms / m^3[/tex]

The concentration gradient is mathematically represented as

   [tex]\frac{dN_A}{dx}  =  \frac{(3.15 - 2.7) * 10^{28} }{5000nm - 0 }[/tex]

=> [tex]\frac{dN_A}{dx}  =  \frac{(3.15 - 2.7) * 10^{28} }{[5000 *10^{-9}] - 0 }[/tex]  

=>   [tex]\frac{dN_A}{dx}  = 9 * 10^{20} / m^4[/tex]  

Generally the flux of the atoms per unit  area according to Fick's Law  is mathematically represented as

       [tex]J =  -D* \frac{d N_A}{dx}[/tex]

=>    [tex]J =  -2* 10^{-14 * 9 * 10^{20} [/tex]

=>    [tex] J =  18*10^{6}\   atoms\ crossing\ /m^2 s  [/tex]

Generally if the cross-section area is [tex] a  =  1 cm^2 =  10^{-4} \  m^2[/tex]

Generally the number of atom crossing the above area  per second is mathematically is  

      [tex]H  =  18*10^{6}    *  10^{-4} [/tex]

=>    [tex]H  =  18*10^{2} \  Atom / sec  [/tex]

An object, initially at rest, is subject to an acceleration of 45 m/s^2. How long will it take that object to travel 1000m? Round to one decimal place.

Answers

Answer:

6.7 seconds

Explanation:

d=(1/2)at^2

equation

1000=(1/2)45t^2.

substitute

2000=45t^2.

multiply by 2 for both sides

44.44=t^2.

divide both sides by 45

6.7=t

take the square root of both sides

5. In Investigation 2, if everything stays the same, except the diameter of the outer ring is doubled, how does the electric field change?

Answers

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

There is a change in the electric field by this factor [tex]\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]

Explanation:

From the question we are told that

  The electric field is  [tex]E(r)_1 =  [\frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r}[/tex]

    Now when the outer diameter is doubled, the radius(b) is also doubled

So

    [tex]E(r)_2 =  [\frac{V_o}{ln(2b) -ln(a)} ] * \frac{1}{r}[/tex]

Now

      [tex]\frac{E(r)_2}{E(r)_1}  =  \frac{\frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r}}{\frac{V_o}{ln(2b) -ln(a)} ] * \frac{1}{r}}[/tex]

=>   [tex]\frac{E(r)_2}{E(r)_1}  =  \frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r} * \frac{ ln(2b) -ln(a)}{V_o} ] * \frac{r}{1}[/tex]

=>  [tex]\frac{E(r)_2}{E(r)_1}  =\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]

[tex]=> E(r)_2 =\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]} }{E(r)_1}[/tex]

Here we see that the electric field changes by the factor [tex]\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]

The car alarm on a stationary car emits sound waves with a frequency of 450 Hz. If you are moving away from the stationary car at 20 m/s, what alarm frequency do you perceive

Answers

Answer:

Explanation:

The frequency of wave is directly proportional to velocity

f = kV

k = f/V

f1/V1 = f2/V2

Given

f1 = 450Hz

V1 = 343m/s

f2 = ?

V2 = 20m/s

Substitute into the formula

450/343 = f2/20

Cross multiply

343f2 = 450×20

343f2 = 9000

f2 = 9000/343

f2 = 26.24Hz

A fountain shoots a jetof water straight up. The nozzle is 1 cm in diameter and the speed of the water exiting the nozzle is 30 m/s. What is the force exerted by the water jet

Answers

Answer:

Explanation:

mass of water coming out per second = A x v where A is area of cross section of the nozzle and v is velocity of water

A = 3.14 x .005²

= 785 x 10⁻⁷ m²

mass of water coming out per second = 785 x 10⁻⁷ x 30 = 23.55 x 10⁻⁴ kg

momentum of this mass = 23.55 x 10⁻⁴ x 30 = 706.5 x 10⁻⁴ kg m /s .

Rate of change of momentum  = 706.5 x 10⁻⁴

Let force be F

F - mg = 706.5 x 10⁻⁴

F = mg + 706.5 x 10⁻⁴

F =  23.55 x 10⁻⁴  x 9.8 + 706.5 x 10⁻⁴

= 937.3 x 10⁻⁴ N .

calculating light in physics

Answers

Formula: c = f where:
c = the speed of light = 300,000 km/s or 3.0 x 108 m/s.
= the wavelength of light, usually measured in meters or Ångströms (1 Å = 10-10 m)
f = the frequency at which light waves pass by, measured in units of per seconds (1/s).

Victor has 100 trading cards, each with a distinct power level between 101 and 200 inclusive. Whenever he heads out, he always randomly selects 21 cards to bring with him just in case he meets a fellow collector. Prove that no matter which 21 cards he brings, Victor will always be able to select 4 of those cards that exhibit the following property:
Let the average power level of all 4 cards bep. The cards can be split into two pairs, each of which also has an average power level of p.

Answers

Answer:

Victor will always be able to select 4 of those cards with the following property

Explanation:

Number of trading cards = 100

victor selects 21 cards

let the 4 cards be labelled : A,B,C and D

The average power level of : A,B,C,D = ( A + B + C + D )/ 4 = P

let the two pairs be : ( A + B ) and ( C + D )

note: average power of each pair = P  and this shows that

( A + B ) = ( C + D ) for Victor to select 4 cards out of the 21 cards that exhibit the same property

we have to check out the possible choices of two cards out of 21 cards yield distinct sums.

= C(21,2)=(21x20)/2 = 210.

from the question the number of distinct sums that can be created using 101  through 200 is < 210 .

hence it is impossible to get 210 distinct sums therefore Victor will always be able to select 4 of those cards

correct me if im wrong

Answers

Your answer is correct. No problem and Have a nice day

A car with a mass of 1500kg is traveling at a speed of 30m/s. What force must be applied to stop the car in 3 seconds?

Answers

Answer:

The answer is 15,000 N

Explanation:

To find the force given the mass , velocity and time can be found by using the formula

[tex]f = \frac{m \times v}{t} \\ [/tex]

where

m is the mass

v is the velocity

t is the time

From the question

m = 1500 kg

v = 30 m/s

t = 3 s

We have

[tex]f = \frac{1500 \times 30}{3} = \frac{45000}{3} \\ [/tex]

We have the final answer as

15,000 N

Hope this helps you

A ball is thrown at 20 m/s from the ground upwards at an angle of elevation of 30°. How far away does it land? 35.35 m

Answers

Answer:

35.35 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 20 m/s

Angle of projection (θ) = 30°

Acceleration due to gravity (g) = 9.8 m/s²

Range (R) =.?

The range (i.e how far away) of the ball can be obtained as follow:

R = u² Sine 2θ /g

R = 20² Sine (2×30) / 9.8

R = 400 Sine 60 / 9.8

R = (400 × 0866) / 9.8

R = 346.4 / 9.8

R = 35.35 m

Therefore, the range (i.e how far away) of the ball is 35.35 m

How far will a 600 kg boat travel in 12 s if there is a constant 900 N force on it and it starts from rest?

Answers

Answer:

108 metres

Explanation:

Given

[tex]Force = 900N[/tex]

[tex]Mass = 600kg[/tex]

[tex]Time = 12s[/tex]

Required

Determine the distance moved

First, we need to determine the acceleration.

[tex]Force = Mass * Acceleration[/tex]

[tex]900N = 600kg * a[/tex]

Solve for a

[tex]a = 900/600[/tex]

[tex]a = 1.5m/s^2[/tex]

Next, we determine the distance using:

[tex]S = ut + \frac{1}{2}at^2[/tex]

Since it starts from rest,

[tex]u = 0[/tex]

[tex]t = 12[/tex]

[tex]a = 1.5[/tex]

So:

[tex]S = 0 * 12 + \frac{1}{2} * 1.5 * 12^2[/tex]

[tex]S = \frac{1}{2} * 1.5 * 144[/tex]

[tex]S = \frac{1}{2} * 216[/tex]

[tex]S = 108m[/tex]

In a spy movie, the hero, James, stands on a scale that is positioned horizontally on the floor. It registers his weight as 810 N . Unknown to our hero, the floor is actually a trap door, and when the door suddenly disappears, James and the scale fall at the acceleration of gravity, down towards an unknown fate. As James falls, he looks at the scale to see his weight. What does he see

Answers

Answer:

His weight would be zero on the scale i.e he is weightless at that instance.

Explanation:

weight = mg

where m is the mass of the object, and g is the acceleration of gravity.

⇒ 810 = mg

During free fall, the weight of an object can be determined by:

W = mg - ma (provided that acceleration of gravity is greater than acceleration of the object)

where a is the acceleration of the object.

But since James fall at the acceleration of gravity, then:

g = a

mg = ma = 810 N

So that;

W = 810 - 810

    = 0 N

Therefore though the weight of James is 810 N, but the scale reads 0 N. this condition is referred to as weightlessness.

how much power is used if it takes frank (a 450 N boy ) 3 seconds to run 2 meters ?

Answers

Answer:

300

Explanation:

450Newton × 2Meter ÷ 3sec

A cyclist and his bicycle have a combined mass of 88 kg and a combined
weight of 862.4 N. The cyclist accelerates at 1.2 m/s2. After 2 seconds he
reaches a speed of 2.4 m/s. What is his momentum at this point?
A. 36.7 kg m/s
B. 359.3 kg:m/s
C. 105.6 kg-m/s
D. 211.2 kg:m/s

Answers

The cyclist accelerates at 1.2 m / s² after 2 seconds he reaches a speed of 2.4 m / s, then the momentum at this point would be 211.2 kg-m/s, therefore the correct answer is option D.

What is momentum?

It can be defined as the product of the mass and the speed of the particle, it represents the combined effect of mass and the speed of any particle, and the momentum of any particle is expressed in Kg m/s unit.

As given in the problem a cyclist and his bicycle has a combined mass of 88 kg and a combined weight of 862.4 N. The cyclist accelerates at 1.2 m/s2. After 2 seconds he reaches a speed of 2.4 m/s.

The momentum of the cyclist = 88 × 2.4

                                                 = 211.0 kgm/s

Thus, the momentum of the cyclist would be  211.0 kgm/s.

To learn more about momentum from here, refer to the link given below;

brainly.com/question/17662202

#SPJ2

Matching type. Send help please. ASAP!

Answers

Answer:

46-D

47-C

48-F

49-A

50-B

I am not very sure I am right about those answers though.

Why do you feel that you are being thrown upward out of your seat when going over an arced hump on a roller coaster

Answers

Answer: The options are not given.

Here are the options.

a) There is an additional force lifting up on you.

(b) At the top you continue going straight and the seat moves out from under you.

(c) You press on the seat less than when the coaster is at rest.Thus the seat presses less on you. (

d) Both b and c are correct.

(e) a, b, and c are correct.

The correct option Is D.

B.At the top you continue going straight and the seat moves out from under you. C.At the same time, you press on the seat less than when the coaster is at rest because the normal force expirienced will be less.

Explanation:

At the top you continue going straight and the seat moves out from under you.At the same time, you press on the seat less than when the coaster is at rest because the normal force expirienced will be less because it is as a result of a phenomenon called Weightlessness. This occur when there is no force or little force is acting on your body. At the top you continue going straight and the seat moves out from under you because there is no force acting on your body and when the body is in free fall i.e acceleration due to gravity , the person is not supported by any thing at.

That is the scenarion that occur...

One student runs with a velocity of +10 m/s while a second student runs with a velocity of –10 m/s. Which student has the faster velocity? Why?

Answers

Answer:

The one with the faster velocity is the one with a velocity of -10m/s

What is the approximate weight of a 400 kg object?

Answers

Answer:

881.84905 LBS

Explanation:

ThErE :p

3922.66 newtons.

This is an exact amount, to get newtons form kg, multiply by 9.8, or in this case, 10.

This gives you 4000 newtons

why do some athletes get injuries before and after the game?

Answers

Answer:

they don't strech so they tear a muscle when they perform

Explanation:

What is the force of a 12 kg object that is accelerating 6 m/s

Answers

We are given:

Mass of object (m) = 12 kg

acceleration (a) = 6 m/s²

Solving for the Force:

From newton's second law of motion:

F = ma

replacing the variables

F = 12*6

F = 72N

A car with a mass of 2,000 kg travels a distance of 400m as it moves from one stoplight to the next. At its fastest , the car travels 18m/s. What is the kinetic energy at this point ?

Answers

Answer: KE= 324,000

Explanation: I hope that this helps! -_-

Answer:

324,000

Explanation:

The volume of water in a water bottle, is about 398
g
cm
km/hr
Kg
g/mL
ml
km
m/s

Answers

Answer:

milliliters (ml)

Explanation:

millileters is the correct measurement for liquids

a motor boat is traveling at 25 knots towards 340 degree on a river flowing at 20 knots towards 175 degrees. What is the actual speed of the boat as seen by a helicopter piolet hovering above?

Answers

Answer:

Vbx = 25 * cos 340 = 23.5 knot        x-component of boats speed

Vrx = 20 cos 175 = -19.9 knot        x-component of rivers speed

Vx = 3.58 knot      x-component of boat and river speed

Vby = 25 sin 340 = -8.55 knot    y-component of boat speed

Vry = 20 sin 175 = 1.74 knot    y-component of river speed

Vy = -6.81 knot     y-component of boat and river speed

V = (Vx^2 + Vb^2)^1/2 = (3.58^2 + 6.81^2)^1/2 = 7.69 knots

A hiker walks 11 km due north from camp and then turns and walks 11 km due east.
What is the total distance walked by the hiker?
What is the displacement (on a straight line) of the hiker from the camp?
please answer all questions

Answers

The total distance is gonna be 11+11=22km.

The displacement is gonna be square root of 11^2+11^2 which is 15.556km.

I hope this helps

if a ramp is 12 meters long has a mechanical advantage of 6 whats its height brainly HELPP!

Answers

Answer:

h = 2 m

Explanation:

Mechanical advantage of a ramp is given by :

MA = length of incline/height of incline

Length of the ramp, l = 12 m

MA = 6

We need to find the height of the ramp.

So,

height of ramp = length of incline/MA

h = 12/6

h = 2 m

So, the height of the ramp is 2 m.

A person has a mass of 1000g and an acceleration of 20 m/s/s. What is the force on the person

Answers

Answer:

20000

Explanation:

Newtons Second law states that the force acting on an object is equal to its mass times its acceleration, f=ma. To solve for force, plug in your values for m and a, and then solve. f = (1000)*(20) = 20000

A car traveling at 32.4 m/s skids to a stop in 4.55 s. Determine the skidding distance of the car (assume uniform acceleration).

Answers

Answer:

x = 73.71 [m]

Explanation:

In order to solve this problem we must use two formulas of kinematics. It is important to make it clear that these formulate are for uniformly accelerated motion, i.e. with constant acceleration.

[tex]v_{f }= v_{i}-(a*t)[/tex]

where:

Vf = fnal velocity = 0

Vi = initial velocity = 32.4 [m/s]

t = time = 4,55 [s]

a = acceleration or desacceleration [m/s^2]

0 = 32.4 - (a*4.55)

a = 7.12 [m/s^2]

Note: it is important to clarify that the negative sign in the above equation is because the car stops and decreases its speed to zero, thus its final velocity is equal to zero.

Now using the following equation:

[tex]x = x_{o} + (v_{i}*t)-(\frac{1}{2} )*a*t^{2}[/tex]

where:

xo = initial distance = 0

x = final distance [m]

Therefore we have:

x = 0 + (32.4*4.55) - (0.5*7.12*4.55^2)

x = 73.71 [m]

A pair of glasses are dropped from the top of a 32.0 m high stadium. A pen is dropped 2.00 s later. How high above the ground is the pen when the glasses hit the ground? (Disregard air resistance. a = -g = -9.81 m/s2.)

Answers

Answer:

Explanation:

the distance have the following relation:

d = (1/2)gt2

D=32.0 m

t =√ (2D/g) = √(2*32.0m/9.8m/s2) = 2.56s

it take 2.56s from the glasses to hit the ground

when the glasses hit the ground, the pen only travel Δt =2.56s - 2.00s = 0.56s

x = (1/2)g(Δt)2 = 0.5*9.8m/s2*(0.56s)2 = 1.54 m

the pen only travel 1.54m

so the pen is above the ground 32.0m - 1.54m = 30.46m

The pen is 30.46m above the ground. when the glasses hit the ground. It is represented by x.

What is the height?

Height is a numerical representation of the distance between two objects or locations on the vertical axis.

The height can refer to a physical length or an estimate based on other factors in physics or common use. |

The given data in the problem is;

h is the height from the top of a stadium = 32.0 m

t is the time period when the pen is dropped later =  2.00 s

x is the height above the ground

a is the air resistance. a = -g = -9.81 m/s²

From the second equation of motion;

[tex]\rm H =ut+\frac{1}{2} gt^2 \\\\ \rm H =\frac{1}{2} gt^2 \\\\ \rm t = \sqrt{\frac{2H}{g} } \\\\ \rm t = \sqrt{\frac{2 \times32.0 }{9.81} } \\\\ \rm t =2.56\ sec[/tex]

When the glasses fall to the ground, the pen only travels a short distance;

[tex]\rm \triangle t = 2.56 -2.00 \\\\ \rm \triangle t = 0.56 \ sec[/tex]

So the pen travel the distance;

[tex]\rm h= \frac{1}{2} g \triangle t^2 \\\\ \rm h= \frac{1}{2} \times 9.81 (0.56)^2 \\\\ h=1.54 \ m[/tex]

The pen above the ground is found as;

[tex]\rm x = H-h \\\\ \rm x = 32.0-1.54 \\\\ \rm x =30.46 \ m[/tex]

Hence the pen is 30.46m above the ground. when the glasses hit the ground.

To learn more about the height refer to the link;

https://brainly.com/question/10726356

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