Which are electromagnetic waves? check all that apply.earthquake waves
infrared waves
ocean waves
radio waves
untraviolet waves

Answers

Answer 1
Answer:

Since electromagnetic waves do not require a medium for their transmission, the electromagnetic waves are radio waves, ultraviolet waves and infrared waves.

What are electromagnetic waves?

Electromagnetic waves or radiations are waves which occur as a result of the interaction between the electric and magnetic fields.

Electromagnetic waves do not require a material medium for their transmission and as such can travel through a vacuum.

Some examples of electromagnetic waves are radio waves, ultraviolet waves, microwaves, infrared waves etc.

Therefore, the electromagnetic waves are radio waves, ultraviolet waves and infrared waves.

Learn more about electromagnetic waves at: brainly.com/question/25847009

Answer 2
Answer: The electromagnetic waves are:
Radio waves
Ultraviolet waves
And Infrared waves
Hope i helped
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You have devised an experiment to measure the kinetic coefficient of friction between a ramp and block. You place the block on the ramp at an angle high enough that it starts sliding. You measure the time it takes to fall down a known distance. The time it takes to fall down the ramp starting from a standstill is 0.5 sec, ???? = 1 kg, θ = 45o, and the distance it falls, L, is 0.5 m. What is µk? (8 pts)

A lead ball is dropped into a lake from a diving board 6.10 mm above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 4.50 ss after it is released. How deep is the lake?

Answers

Answer:

D=1.54489 m

Explanation:

Given data

S=6.10 mm= 0.0061 m

To find

Depth of lake

Solution

To find the depth of lake first we need to find the initial time ball takes to hit the water.To get the value of time use below equation

S=v_(1)t+(1/2)gt^(2) \n 0.0061m=(0m/s)t+(1/2)(9.8m/s^(2) )t^(2)\n t^(2)=(0.0061m)/(4.9m/s^(2) )\n  t=\sqrt{1.245*10^(-3) }\n t=0.035s

So ball takes 0.035sec to hit the water

As we have found time Now we need to find the final velocity of ball when it enters the lake.So final velocity is given as

v_(f)=v_(i)+gt\nv_(f)=0+(9.8m/s^(2) )(0.035s)\n v_(f)=0.346m/s

Since there are (4.50-0.035) seconds left for (ball) it to reach the bottom of the lake

So the depth of lake given as:

D=|vt|\nD=|0.346m/s*4.465s|\nD=1.54489m

Answer: d = 1.54m

The depth of the lake is 1.54m

Explanation:

The final velocity of the ball just before it hit the water can be derived using the equation below;

v^2 = u^2 + 2as ......1

Where ;

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled.

Since the initial velocity is zero, and the acceleration is due to gravity, the equation becomes:

v^2 = 2gs

v = √2gs ......2

g = 9.8m/s^2

s = 6.10mm = 0.0061m

substituting into equation 2

v = √(2 × 9.8× 0.0061)

v = 0.346m/s

The time taken for the ball to hit water from the time of release can be given as:

d = ut + 0.5gt^2

Since u = 0

d = 0.5gt^2

Making t the subject of formula.

t = √(2d/g)

t = √( 2×0.0061/9.8)

t = 0.035s

The time taken for the ball to reach the bottom of the lake from the when it hits water is:

t2 = 4.5s - 0.035s = 4.465s

And since the ball falls for 4.465s to the bottom of the lake at the same velocity as v = 0.346m/s. The depth of the lake can be calculated as;

depth d = velocity × time = 0.346m/s × 4.465s

d = 1.54m

The depth of the lake is 1.54m

Tarik winds a small paper tube uniformly with 163 turns of thin wire to form a solenoid. The tube's diameter is 6.13 mm and its length is 2.49 cm . What is the inductance, in microhenrys, of Tarik's solenoid?

Answers

Answer:

The  inductance is L  =  40\mu H

Explanation:

From the question we are told that

    The number of turns is  N  =  163 \ turns

    The  diameter is  D  =  6.13 \ mm  =  6.13 *10^(-3) \ m

    The  length is  l  =  2.49 \ cm  =  0.0249 \ m

     

The radius is evaluated as r =  (d)/(2)

substituting values

        r =  (6.13 *10^(-3))/(2)

       r =  3.065 *10^(-3) \  m

The  inductance of the Tarik's solenoid is mathematically represented as

            L  =  (\mu_o * N^2  *  A )/(l )

Here \mu_o is the permeability of free space with value  

        \mu_o  =  4\pi *10^(-7) \ N/A^2

A is the area which is mathematically evaluated as

         A  =  \pi r^2

substituting values

       A  =  3.142 * [ 3.065*10^(-3)]^2

       A  =  2.952*10^(-5) \ m^2

substituting values into formula for L  

      L  =  ( 4\pi *10^(-7) * [163]^2  *  2.952*10^(-5) )/(0.0249 )

     L  =  40\mu H

     

A large convex lens stands on the floor. The lens is 180 cm tall, so the principal axis is 90 cm above the floor. A student holds a flashlight 120 cm off the ground, shining straight ahead (parallel to the floor) and passing through the lens. The light is bent and intersects the principal axis 60 cm behind the lens. Then the student moves the flashlight 30 cm higher (now 150 cm off the ground), also shining straight ahead through the lens. How far away from the lens will the light intersect the principal axis now?A. 30 cm
B. 60 cm
C. 75 cm
D. 90 cm

Answers

B. 60 cm 

  All parallel light rays are bent through the focal point of a convex lens, so the rays from the flashlight 150 cm above the floor must go through the same point on the principal axis as the rays from the flashlight 120 cm above the floor. The location of the focal point does not change when the position of the object is moved either vertically or horizontally.

A ball with a mass of 170 g which contains 3.80×108 excess electrons is dropped into a vertical shaft with a height of 145 m . At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has a magnitude of 0.250 T and direction from east to west.A)If air resistance is negligibly small, find the magnitude of the force that this magnetic field exerts on the ball just as it enters the field.

Use 1.602×10−19 C for the magnitude of the charge on an electron.

B)Find the direction of the force that this magnetic field exerts on the ball just as it enters the field.

a-from north to south

b-from south to north

Answers

Answer:

A. F=6.65*10^{-10}N

B. south - north

Explanation:

A) We use the Lorentz force

F = qv X B

|F| = qvB

to calculate the magnitude of the force we need the speed of the of the ball.

v_(f)^(2)=v_(0)^(2)+2gy\nv_(f)=\sqrt{0+2(9.8(m)/(s^(2)))(145m)}=53.31(m)/(s)

and by replacing in the formula for the magnitude of the force we have (taking into account the excess of electrons)

F=(3.8*10^(8))(1.602*10^(-19)C)(53.31(m)/(s))(0.205T)=6.65*10^(-10)N

B)

b.  south - north (by the rigth hand rule)

I hope this is usefull for you

regards

Select True or False for the following statements about Heisenberg's Uncertainty Principle. True False It is not possible to measure simultaneously the x and y positions of a particle exactly.True False It is not possible to measure simultaneously the x and y momentum components of a particle exactly.
True False It is not possible to measure simultaneously the z position and the z momentum component of a particle exactly.

Answers

Answer:

Statement 1) False

Statement 2) False

Statement 3) True

Explanation:

The uncertainty principle states that " in a physical system certain quantities cannot be measured with random precision no matter whatever the least count of the instrument is" or we can say while measuring simultaneously the position and momentum of a particle the error involved is

P\cdot\delta x\geq (h)/(4\pi )

Thus if we measure x component of momentum of a particle with 100% precision we cannot measure it's position 100% accurately as the error will be always there.

Statement 1 is false since measurement of x and y positions has no relation to uncertainty.

Statement 2 is false as both the momentum components can be measured with 100% precision.

Statement 3 is true as as demanded by uncertainty principle since they are along same co-ordinates.

A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. One person hits the water 5.00 m from the end of the slide in a time of 0.504 s after leaving the slide. Ignore friction and air resistance. Find the height H.

Answers

Answer:

4.93 m

Explanation:

According to the question, the computation of the height is shown below:

But before that first we need to find out the speed which is shown below:

As we know that

Speed = (Distance)/(Time)

Speed = (5)/(0.504)

= 9.92 m/s

Now

v^2 - u^2 = 2* g* h

9.92^2 = 2* 9.98 * h

98.4064 = 19.96 × height

So, the height is 4.93 m

We simply applied the above formulas so that the height i.e H could arrive

Final answer:

The height of the water slide is 5.04 meters.

Explanation:

The problem described in this question involves a water slide, where swimmers start from rest at the top and leave the slide traveling horizontally. To determine the height of the slide, we can use the equations of motion in the horizontal direction. The horizontal displacement (x) is given as 5.00 m and the time (t) is given as 0.504 s. Assuming no friction or air resistance, we can use the equation x = v*t, where v is the horizontal velocity. Rearranging the equation, we can solve for v, which is equal to x/t. Substituting the given values, we have v = 5.00 m / 0.504 s = 9.92 m/s. The horizontal velocity (v) is constant throughout the motion, so we can use the equation v = sqrt(2*g*H), where g is the acceleration due to gravity (9.8 m/s^2) and H is the height of the slide. Rearranging the equation, we can solve for H, which is equal to v^2 / (2*g). Substituting the known values, we have H = (9.92 m/s)^2 / (2*9.8 m/s^2) = 5.04 m.