What do radio waves and gamma rays have in common?

They are both electromagnetic waves.
They are both low frequency waves.
They can only travel in a vacuum.
They both are part of the visible light spectrum.

Answers

Answer 1
They are both electromagnetic waves.
Answer 2

Answer:

Both Magnetic

Explanation:


Related Questions

If a wave has a frequency of 5 hertz, the period for one wave cycle will be

Answers

Answer:

0.2 seconds

Explanation:

f=1/t

1/5=0.2

40 POINTS!!! PLEASE HELPP!!!
Please select the word from the list that best fits the definition

One form of energy changes to one other form

(A.) single transformation
(B.) multiple transformation

Answers

Answer:

A

Explanation:

if it was B it would say from one to another to another

Answer A



Single transformation

How do our attitudes help us organize our reality?

Answers

Answer:

If we are upset, often our mind cannot think straight. we need to take deep breaths, and focus on reality. When we are upset we do things out of anger.

Explanation:

Hope this helps!

our attitude towards our peers depicts how we will be treated by others

ian pushed a piano across the room with the correct amount of force. Which of newton’s laws is this?
Newton’s 1st, 2nd, 3rd, or 1st and 2nd law?

Answers

Answer:

1st and 2nd

Explanation:

A train has a mass of 100000 kilograms.if its kinetic energy is 3mj what velocity is it travelling at​

Answers

Answer:

V = 7.75 meters per seconds.

Explanation:

Given the following data;

Mass = 100000 Kilograms

Kinetic energy = 3 Megajoules = 3 * 10⁶ Joules

To find the velocity of the train;

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

[tex] K.E = \frac{1}{2}MV^{2}[/tex]

Where;

K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

Substituting into the formula, we have;

[tex] 3000000 = \frac{1}{2}*100000*V^{2}[/tex]

Cross-multiplying, we have;

[tex] 6000000 = 100000V^{2}[/tex]

[tex] V^{2} = \frac{6000000}{100000} [/tex]

[tex] V^{2} = 60 [/tex]

Taking the square root of both sides, we have;

[tex] V = \sqrt {60} [/tex]

V = 7.75 m/s

convert 1 day into seconds.(you will need to show your work to receive credit.) please help thank you​

Answers

Answer:

the answer is 86400seconds.

Explanation:

1day= 24 hours

24hours to seconds =

24×60×60= 86400 seconds.

( 60 second = 1 minutes)

( 60 minutes = 1 hour)

3) 4 electrons are placed - one electron per corner - at the corners of a square of side 1 meter. One fixed proton is placed in the middle of the square.The 4 electrons are held in place by some mechanism. The 4 electrons are released by the mechanism at the same time. They move and reach the corners of a square of side 0.8 meters, and keep on moving . Find the velocity of each electron at the corners of the square of side 0.8 meters.

Answers

Explanation:

3

i believe that they are all going at 3.2 meters each, I did 4 times 0.8

The velocity of each electron at the corners of the square is 15.92 m/s.

The given parameters;

charge of electron, q = 1.6 x 10⁻¹⁹ Clength of the square, L = 0.8 m

The diagonal length of the square is calculated as;

[tex]d^2 = 0.8^2 + 0.8^2\\\\d = \sqrt{0.8^2 + 0.8^2} \\\\d = 1.13 \ m[/tex]

The distance of each corner charge and the middle charge is calculated as;

[tex]r = \frac{1.13}{2} \\\\r = 0.565 \ m[/tex]

The force between each corner charge and the middle charge is calculated as;

[tex]F= \frac{kq^2}{r^2}[/tex]

The centripetal force on each charge moving around the square is calculated as;

[tex]F = \frac{mv^2}{r}[/tex]

solve the forces together;

[tex]\frac{mv^2}{r} = \frac{kq^2}{r} \\\\v^2 = \frac{kq^2}{m} \\\\v = \sqrt{ \frac{kq^2}{m} } \\\\v = \sqrt{ \frac{(9\times 10^9) \times (1.602\times 10^{-19})^2}{9.11 \times 10^{-31}} } \\\\v = 15.92 \ m/s[/tex]

Thus, the velocity of each electron at the corners of the square is 15.92 m/s.

Learn more here:https://brainly.com/question/13219544

Help please & actually answer thank youu :)

Answers

Answer:

highest KE and lowest GPE = Dhighest GPE and lowest KE = Asome KE and some PE = B

PLEASE HELP! I'LL GIVE BRAINLEST​

Answers

Answer:

Weight = 8.162 Newton.

Explanation:

Given the following data;

Mass = 2.2 kg

Acceleration due to gravity = 3.71 N/kg

To find the weight of the textbook;

Weight = mass * acceleration due to gravity

Weight = 2.2 * 3.71

Weight = 8.162 N

Therefore, the weight of the science textbook in mars is 8.162 Newton.

Please help The position of masses 4kg, 6kg, 7kg, 10kg, 2kg, and 12kg are (-1,1), (4,2), (-3,-2), (5,-4), (-2,4) and (3,-5) respectively. Determine the position of the center of mass of this system?​

Answers

Answer:

(1.9756, -2.1951)

Explanation:

The center of mass equation is: [tex]x_{cm}[/tex] = [tex]\frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5} + m_{6}x_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}[/tex], where m represents the masses and x represents the position.

In order to find the coordinates of the center of mass, we need to use this equation for both the x-values and the y-values.

x-values:

[tex]x_{cm}[/tex] = [tex]\frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5} + m_{6}x_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}[/tex] = [tex]\frac{4(-1)+6(4)+7(-3)+10(5)+2(-2)+12(3)}{4+6+7+10+2+12}[/tex] = [tex]\frac{(-4)+(24)+(-21)+(50)+(-4)+(36)}{41}[/tex] = [tex]\frac{81}{41}[/tex] = 1.9756

y-values:

[tex]y_{cm}[/tex] = [tex]\frac{m_{1}y_{1} + m_{2}y_{2} + m_{3}y_{3} + m_{4}y_{4} + m_{5}y_{5} + m_{6}y_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}[/tex] = [tex]\frac{4(1)+6(2)+7(-2)+10(-4)+2(4)+12(-5)}{4+6+7+10+2+12}[/tex] = [tex]\frac{(4)+(12)+(-14)+(-40)+(8)+(-60)}{41}[/tex] = [tex]\frac{-90}{41}[/tex] = -2.1951

center of mass:

(1.9756, -2.1951)

scholastic science world
CHECK FOR UNDERSTANDING: VIDEO GAME TYCOONS

Answers

I dont see it ;-; REEEEEEEE

PLEASE HELPPPPPP!ASAPPPPPPP​

Answers

Answer:

Altitude is the height of landmass calculated from sea level. No it has no effect on weight.

Explanation:

The altitude at sea level is always assumed to be 0°, thus as you move away from altidude is calculated in increments while moving into the sea, the vice versa is true.

The wireless system is operating at 2GHz. A base station and a mobile unit are separated by 16km. The maximum gain of the transmitting antenna at the base station is 20dB. The input power to the transmitter is 100W, and the power received by the antenna is 5nW. The antennas are aligned, and there are no re ections or loses. What will be the received power if the distance between the mobile unit and the base station increases to 20km?

Answers

If everything else remains constant, then the received signal power is simply an inverse-square function of its distance from the transmitter.

That is, down 6dB when distance is doubled etc.

If distance increases from 16 to 20, then received power decreases by the factor of (16/20)^2.

That's (0.8)^2 = 0.64

New receive power is 5x0.64 = 3.2 nW

=. ===== ==========

Here's how I do it at my job:

Initial RSL = 5nW ~ - 53 dBm

loss = 20log(20/16)= 20log(1.25)~1.94dB

New RSL = - 54.94 dBm or ~ 3.2 nW .

Niobium metal becomes a superconductor when cooled below 9K. Itssuperconductivity is destroyed when the surface magnetic fieldexceeds 0.100 T. Determine the maximum current a 2.00-mm-diameterniobium wire can carry and remain superconducting, in the absenceof any external magnetic field.

Answers

Answer:

the maximum current is 500 A

Explanation:

Given the data in the question;

the B field magnitude on the surface of the wire is;

B = μ₀i / 2πr

we are to determine the maximum current so we rearrange to find i

B2πr = μ₀i

i = B2πr / μ₀

given that;

diameter d = 2 mm = 0.002 m

radius = 0.002 / 2 = 0.001 m

B = 0.100 T

we know that permeability; μ₀ = 4π × 10⁻⁷ Tm/A

so we substitute

i = (0.100)(2π×0.001 ) / 4π × 10⁻⁷

i = 500 A

Therefore, the maximum current is 500 A

PLEASE HELP! I'LL GIVE BRAINLEST​

Answers

Neil is right as you don’t lose mass on the moon. The gravity is just stronger in earth so you weigh more on earth.

A 3 kg block collides with a massless spring of spring constant 98 N/m attached to a wall. The speed of the block was observed to be 1.5 m/s at the moment of collision. The acceleration of gravity is 9.8 m/s 2 . How far does the spring compress if the surface on which the mass moves is frictionless

Answers

Answer: 0.83 m

Explanation:

Given

mass of the block is m=3 kg

spring constant k=98 N/m

The Speed at the time of collision is v=1.5 m/s

Here, the kinetic energy of the block is converted into Elastic potential energy

[tex]\Rightarrow \dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2 \\\\\Rightarrow 3\times 1.5^2=98\times x^2\\\\\Rightarrow x^2=0.6887\\\\\Rightarrow x=0.829\approx 0.83\ m[/tex]

Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. Suppose the mass of a certain spherical neutron star is twice the mass of the Sun and its radius is 13.0 km. Determine the greatest possible angular speed the neutron star can have so that the matter at its surface on the equator is just held in orbit by the gravitational force. (The mass of the Sun is 1.99 1030 kg.)

Answers

Answer:

The required angular speed the neutron star is 10992.32 rad/s

Explanation:

Given the data in the question;

mass of the sun M[tex]_S[/tex] = 1.99 × 10³⁰ kg

Mass of the neutron star

M[tex]_N[/tex] = 2( M[tex]_S[/tex] )

M[tex]_N[/tex] = 2( 1.99 × 10³⁰ kg )

M[tex]_N[/tex] = ( 3.98 × 10³⁰ kg )

Radius of neutron star R[tex]_N[/tex] = 13.0 km = 13 × 10³ m

Now, let mass of a small object on the neutron star be m

angular speed be ω[tex]_N[/tex].

During rotational motion, the gravitational force on the object supplies the necessary centripetal force.

GmM[tex]_N[/tex] = / R[tex]_N[/tex]² = mR[tex]_N[/tex]ω[tex]_N[/tex]²

ω[tex]_N[/tex]² = GM[tex]_N[/tex] = / R[tex]_N[/tex]³

ω[tex]_N[/tex] = √(GM[tex]_N[/tex] = / R[tex]_N[/tex]³)

we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²

we substitute

ω[tex]_N[/tex] = √( (  6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)

ω[tex]_N[/tex] = √( 2.65466 × 10²⁰ / 2.197 × 10¹²

ω[tex]_N[/tex] = √ 120831133.3636777

ω[tex]_N[/tex] = 10992.32 rad/s

Therefore, The required angular speed the neutron star is 10992.32 rad/s

Plz help, will Mark brainliest. A 20.0 Ohm and 60.0 Ohm resistor are connected in series to a 9.00 V battery. How much current flows out of the battery?
(Unit = A)

Answers

Answer: 0.1125A

Explanation:

We need to know the total resistance which will be:

R = R1+R2

where,

R1 = 20 ohm

R2 = 60 ohm

Therefore, R = 20 + 60 = 80 ohm

We should note that from Ohm's law

v = iR

i = v/R

where,

i = 9/80 = 0.1125 A

Therefore, the current that flows out of the battery is 0.1125A

Which of the following has no energy?"

Amoving ball
Awrecking ball
Non of the above

Answers

The answer is none of the above. Everything has energy! Good luck!

define the term change and state one negative change you may encounter as a student or as an employee in the future​

Answers

Answer:

hey iam bored pls join

ID: 724 645 3790

Password: 1111

The term change refers to an alteration of present conditions.

What is change?

The term change refers to an alteration of present conditions. When a change occurs, things cease to be the way they were. As a student, the complexity of academic work increases with time. This is a change that has taken place.

As an employee, you are given real world tasks to accomplish rather than mere mental exercises like when you were a student, this is change that has taken place.

Learn more about change: https://brainly.com/question/11739819

Energy associated with moving objects or that could move later is?

Answers

Answer: Mechanical Energy

Explanation: Mechanical energy can be either kinetic energy (energy of motion) or potential energy (stored energy of position)

Hope this helped! :)

If a 5-L balloon at 25 degrees celsius were gently heated to 30 degrees celsius, what new volume would the balloon have? Show all work for credi

Answers

Answer: 5.08 L.

Explation down below

32. Assume the helium-neon lasers commonly used in student physics laboratories have power outputs of 0.500 mW. (a) If such a laser beam is projected onto a circular spot 1.00 mm in diameter, what is its intensity

Answers

Answer:

636.6 W/m²

Explanation:

From the given information:

The area of the circular spot can be calculated as:

A = πr²

A = π(0.5 × 10⁻³ m)²

A = 7.85 × 10⁻⁷ m²

The intensity can be determined by using the formula:

[tex]I = \dfrac{P}{A}[/tex]

[tex]I = \Big ( \dfrac{0.500 \ mW}{\pi (0.5 \times 10^{-3} \ m )^2} \Big) \Big( \dfrac{10^{-3} \ W}{1\ mW} \Big)[/tex]

[tex]I = \dfrac{0.500 \times 10^{-3} }{\pi(0.25 \times 10^{-6} )}[/tex]

I = 636.6 W/m²


can someone please help me !!!!

Answers

Answer:

it's A subduction, deep water trench


A fisherman sitting on the end of a pier notices that 6 wave crests pass him in 3 seconds. What is the frequency of the waves?
9.0 Hz
18.0 Hz.
20 Hz
4.5 Hz

Answers

Do you know the muffin man?

The frequency of the waves will be 18 Hz. Then the correct option is B.

What is the frequency?

The frequency of a repeated event is its number of instances per unit of time. It differs from angular frequency and is sometimes referred to as temporal frequency for clarification. One event occurs per second when measuring frequency in hertz.

Frequency is the number that represents the number of oscillations or intervals each second. The hertz is the SI unit for frequency (Hz). One cycle per second equals one hertz.

The number of finished waves produced each second is considered the wave frequency.

Six wave crests pass a fisherman sitting at the end of a pier in the span of three seconds.

The frequency of the waves is given as,

f = 6 x 3

f = 18 Hz

The waves will have a frequency of 18 Hz. Then, choice B is the best one.

More about the frequency link is given below.

https://brainly.com/question/29739263

#SPJ6


PLEASE HELP !!!!!!!!!!

Answers

Answer:

1

Explanation:

Half Time At 3:00 the hour hand and the minute hand of a clock point in directions that are 90.0
apart. What is the first time after 3:00 that the angle between the two hands has decreased by half to 45.0?

Answers

The first time after 3:00 that the angle between the two hands has decreased by half to 45.0 is 3:08:11.

Angular velocity of the hour hand

The angular velocity of the hour hand for 12 hours time is calculated as follows;

[tex]\omega_h = \frac{2\pi}{12(3600 \ s)}[/tex]

Angular velocity of the minute hand

The angular velocity of the minute hand for 12 hours time is calculated as follows;

[tex]\omega _m = \frac{2\pi}{3600}[/tex]

Difference in the angular velocity

[tex]\Delta \omega = \frac{2\pi }{3600} - \frac{2\pi }{12(3600)} = \frac{11(2\pi)}{12(3600)} = \frac{22\pi}{12 (3600)}[/tex]

First angle between the two hands in radians

[tex]45^0 = \frac{45}{180} \pi= \frac{\pi}{4}[/tex]

Time after 3:00

The time after 3:00 is calculated as follows;

[tex]t =( \frac{\pi}{4}) /\frac{22\pi}{12(3600)} \\\\t = \frac{\pi }{4} \times \frac{12(3600)}{22\pi} \\\\t = 490.91 \ s\\\\t = 8 \min, \ 10.91 \ s[/tex]

[tex]t \approx 8\min, 11 \ s[/tex]

Thus, the first time after 3:00 that the angle between the two hands has decreased by half to 45.0 is 3:08:11.

Learn more time in angular difference here: https://brainly.com/question/7815

You are trying to catch the mutated mouse and you have a rope
that both you and the mouse are pulling with a force of 500 Newtons,
but the rope does not move.
How much work is done?


PLS ANSWER ASAP! WILL MARK AS BRAINLYIST!!!!!
time left (5:00)!!

Answers

Answer:

none no work cuz no motion

Explanation:

GOOD LUCK

a
Ten 2.2 v cells each having an internal
resistance of o.1ohms in are connected in series
to a load of 21ohm. Determine
pd at the battery terminals​

Answers

Answer:

22Volts

Explanation:

The pd at the terminal is known as the emf

Since there are Ten 2.2V cells

Terminal voltage = number of cells * pd of one cell

Terminal voltage = 10 * 2.2

Terminal voltage = 22V

Hence the pd at the battery terminals​ is 22Volts

You are holding a finishing sander with your right hand. THe sander has a flywheel which spins counterclockwise as seen from behind the handle. You are sanding a wall in front of you. as you turn the sander towards the right, you feel a tendency in the sander to...
a. pull away from you
b. turn towards the left
c. turn downward
d. turn upward
e. push toward you

Answers

Answer:

c. turn downward

Explanation:

From the information given:

To find the tendency of the sander;

We need to apply the right-hand rule torque; whereby we consider the direction of the flywheel, the direction at which the torque is acting, and the movement of the sander toward the right.

Since the flywheel of the sander is in counterclockwise movement, hence the torque direction will be outward placing on the wall. However, provided that the movement of the sander is toward the right, then there exists an opposite force that turns downward which showcases the tendency in the sander is downward.

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