Match each type of wave made during an earthquake to its wave form.

Answers

Answer 1

Answer:

I see anything so I can answer you .


Related Questions

the region where a magnet force is strongest is at the​

Answers

Answer:

Magnetic poles

Explanation:

Magnetic poles are the regions where the magnet's force is strongest, one is the north pole and one is the south pole.

The region were there are the strongest magnetic felts are in the North Pole and South Pole

Why are protists difficult classify?

Answers

Answer:

Protists are difficult to characterize because of the great diversity of the kingdom. These organisms vary in body form, nutrition, and reproduction. They may be unicellular, colonial, or multicellular.

Answer:

There are many varying characteristics and exceptions to each type of protist.

They have been previously categorized based on what they are not

Recent studies show that protists have not descended from one common ancestor.

Explanation:

its right i took test :)

A red pool ball is rolling directly east before it collides with the
white cue ball moving directly north. Due to conservation of
momentum the total momentum of both objects after the
collision would be in which direction?

Answers

Answer: North East

Explanation: Trust me, I was just doing this on the Ck-12 and this the answer I choose and It said I'm correct.

The gravitational potential energy of an object is defined as the energy it has due to its position in a gravitational field. A ball with a weight of 50 N is lifted to a height of 1 meter. Which graph correctly represents the change in gravitational potential energy (shaded in blue) as it is lifted to this height?

Answers

Answer:

athletic

Explanation:

because internet system has been down since we were in few days

A hair dryer draws a current of 12.8 A.

(a)How many minutes does it take for
6.8 x 10° C of charge to pass through the
hair dryer? The fundamental charge is
1.602 x 10-19 C.
Answer in units of min.

(b)How many electrons does this amount of
charge represent?
Answer in units of electrons.

Answers

Answer:

(a) 8.85×10⁻³ minutes

(b) 4.24×10¹⁹ electrons

Explanation:

(a) Using,

Q = it............................. Equation 1

Where Q = quantity of charge, i = current, t = time.

Make t the subject of the equation

t = Q/i............................. Equation 2

Given: Q = 6.8×10⁰ C, i = 12.8 A

Substitute these values into equation 2

t = 6.8×10⁰/12.8

t = 8.85×10⁻³ minutes

(b) n = Q/(1.602×10⁻¹⁹)................. Equation 3

Where n = number of electrons.

Given: Q = 6.8×10⁰ C

Substitute into equation 2

n = 6.8×10⁰/1.602×10⁻¹⁹

n = 4.24×10¹⁹ electrons

(a) The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes

(b) Amount of the electrons in the charge will be  [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons

What will be the time of the charge and number of the electrons in the charge ?

As we know Q = IT

Where Q = quantity of charge, i = current, T = time.

From the above equation

                    T= Q/I.

Given: Q = [tex]6.8\times\d10^{0}[/tex] C, i = 12.8 A

Substitute these values  

T=  [tex]6.8[/tex]×[tex]\d10^{0}[/tex] /12.8

T =  [tex]8.85[/tex]×[tex]\d10^{-3}[/tex] minutes

Now the number of the electrons present in the charge will be

n = Q/( [tex]1.602[/tex]×[tex]\d10^{-19}[/tex])

Where n = number of electrons.

Given: Q = [tex]6.8\times\d10^{0}[/tex] C

Substitute Value of Q  

n =  [tex]6.8\times\d10^{0}[/tex]/ [tex]1.602\times\d10^{-19}[/tex]

n = [tex]4.24\times\d10^{19}[/tex] electrons

Thus

(a)The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes

(b)Amount of the electrons in the charge will be  [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons

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A force of 10 N is applied at right angles to the handle of a spanner, 0.5 m from the centre of a nut. The
moment on the nut is:
20 Nm
50 Nm
5 Nm

Answers

Explanation:

the movement of the nut is 20Nm

Q1what is pinhole camera?​

Answers

A pinhole camera is a simple camera without a lens but with a tiny aperture (the so-called pinhole)—effectively a light-proof box with a small hole in one side. Light from a scene passes through the aperture and projects an inverted image on the opposite side of the box, which is known as the camera obscura effect.

A 50 N force causes a spring to compress 0.09 m. What is the spring constant? What is the potential energy of the spring?​

Answers

Initial considerationsApplied force = 50NDistance to equilibrium position = 0.09mSpring constant = ?

Resolution

using hooke's relation

F = K . d

50N = k . 0.09m

k = 50N / 0.09m

k = 5555.56 N/m

Calculating the potential energy of the spring

Ep = 1/2 k . x²

Ep = 1/2 (5555.56 N/m) (0.09m)²

Ep = 22.5 Joules

Answer

the spring constant? =

k =  5555.56 N/m

potential energy of the spring?​

Ep = 22.5 Joules

The Potential energy of the spring is 2.25 J

What is the Potential energy of spring?

This is the energy stored in spring due to its elastic properties.

To calculate the potential energy of the spring, we use the formula below.

Formula:

E = Fe/2................ Equation 1

Where:

E = Potential energy of the springF = Force applied to the springe = compression.

From the question,

Given:

F = 50 Ne = 0.09 m

Substitute these values into equation 1

E = 50(0.09)/2E = 2.25 J.

Hence, The Potential energy of the spring is 2.25 J

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please answer asap! will give brainliest if correct! ^^

A train of waves is moving at a speed of 30 m/s. The frequency of the waves is 10 Hz. What is the wavelength?
a. 300 m
b. 30 m
c. 3 m
d. 0.1 m

Answers

Answer:

the answer is c, 3m......

When a train is moving at the speed of 30 m/s. The wavelength of the wave is 3 m.

What is wavelength?

Wavelength is the characteristic of the wave which represents the distance between the adjacent crest or trough of the wave.

Given is the speed of the train of waves 30 m/s and frequency of the wave is 10 Hz, then the wavelength is related to speed and wavelength as

v = fλ

30 = 10 × λ

λ = 3 m

Thus, the wavelength of the wave is 3 m.

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1. A horizontal force of 50 N is applied to push a desk 40 m across a
warehouse floor. How much work is done?
2000 J
1000 J
3000 J
2001 J

Answers

Work = N × m = 50 x 40 = 2000 J

When grip strength increases:

a. action potential voltage increases.
b. action potential frequency decreases.
c. action potential frequency increases.
d. action potential frequency increases.
e. the number of active motor units increases.

Answers

Answer:

e. the number of active motor units increases.

Explanation:

There is a direct relationship between the number of active motor units and the grip strength in a given scenario. For example, increase in the grip strength leads to increase in the number of active motor units. In the other-hand, the decrease in grip strength leads to the decrease in the number of active motor units.

Diffraction occurs for all types of waves, including sound waves.

a. True
b. False

Answers

Answer:

a. True

Explanation:

Sound are mechanical waves that are highly dependent on matter for their propagation and transmission.

Sound travels faster through solids than it does through either liquids or gases. A student could verify this statement by measuring the time required for sound to travel a set distance through a solid, a liquid, and a gas.

Mathematically, the speed of a sound is given by the formula:

[tex] Speed = wavelength * frequency [/tex]

Generally, the frequency of a sound wave determines the pitch of the sound that would be heard.

Diffraction occurs for all types of waves, including sound waves.

What is the weight of a 48kg rock?

Answers

Answer:

48kg

Explanation:

3. Materials that lets electricity to pass​

Answers

Answer:

materials that allow electricity to pass through them are called conductors, some examples of conductors are many metals, such as copper, iron and steel.

A falling 0.60 kg object experiences a frictional force due to air resistance of 1.5 N. What is the object's acceleration?​

Answers

Answer:

7.5 m/s².

Explanation:

From the question given above, the following data were:

Mass (m) of object = 0.6 Kg

Force of friction (Fբ) = 1.5 N

Acceleration (a) =?

Next, we shall determine the force of gravity on the object. This can be obtained as follow:

Mass (m) of object = 0.6 Kg

Acceleration due to gravity (g) = 10 m/s²

Force of gravity (F₉) =?

F₉ = mg

F₉ = 0.6 × 10

F₉ = 6 N

Next, we shall determine the net force acting on the object. This can be obtained as follow:

Force of friction (Fբ) = 1.5 N

Force of gravity (F₉) = 6 N

Net force (Fₙ) =?

Fₙ = F₉ – Fբ

Fₙ = 6 – 1.5

Fₙ = 4.5 N

Finally, we shall determine the acceleration of the object. This can be obtained as follow:

Mass (m) of object = 0.6 Kg

Net force (Fₙ) = 4.5 N

Acceleration (a) of object =?

Fₙ = ma

4.5 = 0.6 × a

Divide both side by 0.6

a = 4.5 / 0.6

a = 7.5 m/s²

Therefore, the acceleration of the object is 7.5 m/s²

(will give brainliest to whoever is first and gives reason) A mass is spun in a circle with a frequency of 40Hz. What is the period of its rotation?

Answers

Answer:

[tex]\huge\boxed{T = 0.025\ seconds}[/tex]

Explanation:

Given:

Frequency = f = 40 Hz

Required:

Time period = T = ?

Formula:

[tex]\sf T = 1 / f[/tex]

Solution:

T = 1 / 40

T = 0.025 seconds

[tex]\rule[225]{225}{2}[/tex]

Hope this helped!

~AH1807


1. What types of natural phenomena could serve as time standards?

Answers

In practice, something that follows a very predictable pattern can be used as a time standard. This include things like radioactive decay, planetary orbit, and the speed of light, among others.

Answer:

The movement of Sun and moon

Explanation:

When the sun rise.it is am and when it sets .it is pm.

A ball rolls along the floor with a constant velocity of 12 m/s. How far will it have
gone after 6 seconds?

Answers

Answer:

72m

12m x 6s=72m

there ya go

You would do 12x6 to get the velocity. After 6 seconds it would be 72m/s

A string is wound tightly around a wheel. When the end of the string is pulled through a distance of 10 cm, the wheel rotates through 5 revolutions. What is the radius of the wheel

Answers

Answer:

0.318cm

Explanation:

The computation of the radius of the wheel is shown below:

As we know that

One revolution of the wheel = 2πr

Here r denotes the radius of the wheel

Now

5 revolutions of the wheel would be

= 2πr ×  5

= 10πr

So,

10πr = 10cm

Thus r = 1 ÷ π

= 0.318cm

The radius of the wheel will be r=0.318 cm

What will be the radius of the wheel?

The radius is defined as the distance between the center of circle and the outer layer of the circle.

It is given in the question that

Revolutions of the wheel = 5

The length of the string =10 cm

Now the radius will be calculated as

One revolution of the wheel

= [tex]2\pi r[/tex]

Now 5 revolutions of the wheel would be

=  [tex]2\pi r\times 5=10\pi\ r[/tex]

Since the string has the length of 10 cm then

[tex]10\pi\ r=10[/tex]

[tex]r=\dfrac{10}{10\pi}[/tex]

[tex]r=0.318\ cm[/tex]

Thus the radius of the wheel will be r=0.318 cm

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If you swing an object on the end of a string around a circle, the string pulls on the object to keep it moving in a circle. What is the name of this force?
A. inertial
B. centripetal
C. resistance
D. gravitational

Answers

Answer:

B

Explanation:

The centripetal force keeps an object moving in a circular path. Therefore option (B) is correct.

What is centripetal force?

A centripetal force can be described as a force that makes a body follow a curved path and its direction is orthogonal to the motion of the body. Gravity offers the centripetal force causing astronomical orbits.

The centripetal force is directed perpendicular to the direction of the displacement of an object. It always acts towards the center of the circle on an object moving in a circular path. For example, When spinning a ball on a string, the tension on the rope pulls the object toward the center.

The Centripetal Force can be described as the product of mass and velocity squared, divided by the radius.

F = mv²/r

Where F is the Centripetal force, m is the mass, r is the radius of the circle and v is the velocity of the object.

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In high air pressure the molecules are

A-Warm and moving fast
b-Close together and moving slowly
c-far apart and moving slowly
d-hot and moving rapidly

Answers

Aaaaa is the answer

The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 cm from the axis of rotation at the knee. How much force must the knee extensors exert to produce an angular acceleration at the knee of 1 rad/s2 , given a mass of the lower leg and foot of 4.5 kg, and a radius of gyration of 23 cm

Answers

Answer:

the knee extensors must exert 15.87 N

Explanation:

Given the data in the question;

mass m = 4.5 kg

radius of gyration k = 23 cm = 0.23 m

angle ∅ = 30°

∝ = 1 rad/s²

distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m

using the expression;

ζ = I∝

ζ = mk²∝

we substitute

ζ = 4.5 × (0.23)² × 1

ζ  = 0.23805 N-m

so

from; ζ = rFsin∅

F = ζ / rsin∅

we substitute

F = 0.23805 / (0.03 × sin( 30 ° )

F = 0.23805 / (0.03 × 0.5)

F F = 0.23805 / 0.015

F = 15.87 N

Therefore, the knee extensors must exert 15.87 N

A 3.0-kilogram object initially at rest explodes and splits into three fragments. One fragment has a mass of 0.50 kg and flies off along the negative x axis at a speed of 2.8 m/s, and another has a mass of 1.3 kg and flies off along the negative y axis at a speed of 1.5 m/s.

Required:
What are the speed and direction of the third fragment?

Answers

Answer:

The speed and direction of the third fragment are 2 meters per second and 54.316º, respectively.

Explanation:

The object is not under the influence of any external force, meaning that Principle of Momentum Conservation to calculation of the velocity of the third fragment:

[tex](m_{1}+m_{2}+m_{3})\cdot \vec {v}_{o} = m_{1}\cdot \vec v_{1} + m_{2}\cdot \vec v_{2} + m_{3}\cdot \vec v_{3}[/tex] (1)

Where:

[tex]m_{1}[/tex], [tex]m_{2}[/tex], [tex]m_{3}[/tex] - Masses of the first, second and third fragments, in kilograms.

[tex]\vec v_{o}[/tex] - Initial velocity of the object, in meters per second.

[tex]\vec v_{1}[/tex], [tex]\vec v_{2}[/tex], [tex]\vec v_{3}[/tex] - Velocities of the first, second and third fragments, in meters per second.

If we know that [tex]m_{1} = 0.5\,kg[/tex], [tex]m_{2} = 1.3\,kg[/tex], [tex]m_{3} = 1.2\,kg[/tex], [tex]\vec v_{o} = (0,0)\,\left[\frac{m}{s} \right][/tex], [tex]\vec v_{1} = \left(-2.8, 0\right)\,\left[\frac{m}{s} \right][/tex] and [tex]\vec v_{2} = \left(0,-1.5\right)\,\left[\frac{m}{s} \right][/tex], the velocity of the third fragment is:

[tex](-1.4,0) + (0,-1.95) + 1.2\cdot \vec v_{3} = (0,0)[/tex]

[tex]1.2\cdot \vec v_{3} = (1.4,1.95)[/tex]

[tex]\vec v_{3} = (1.167, 1.625)\,\left[\frac{m}{s} \right][/tex]

The speed of the third fragment is the magnitude of the result found above:

[tex]v_{3} = 2\,\frac{m}{s}[/tex]

And the direction of the third fragment is:

[tex]\theta_{3} = \tan^{-1} \left(\frac{1.625}{1.167}\right)[/tex]

[tex]\theta_{3} \approx 54.316^{\circ}[/tex]

The speed and direction of the third fragment are 2 meters per second and 54.316º, respectively.

5. How much heat is needed to warm .052 kg of gold from 30°C to 120°C? Note: Gold has a specific heat of 136

J/kg °C

Answers

Answer:

Q = 636.48 J

Explanation:

Given that,

The mass of gold, m = 0.052 kg

The temperature increase from 30°C to 120°C.

The specific heat of gold is 136  J/kg °C.

We need to find the heat needed to warm the gold. The formula for heat needed is given by :

[tex]Q=mc\Delta T\\\\Q=0.052\times 136\times (120-30)\\\\Q=636.48\ J[/tex]

So, 636.48 J of heat is needed to warm gold.

1 point
3. A 75 N box requires 250 J of work to move to a shelf. How high is the
shelf?
O 3.3 m
3.25m
0 3.9m
O 3.6m

Answers

O 3.3m your welcome ussugdh

a place where two bones come together is known as an

Answers

Answer:

a place where two bones come together is known as a join

Answer:

Hey mate.....

Explanation:

This is ur answer....

Joints

Joints – A place in the body where bones come together. Non-Moveable Joints (sometimes called fixed or fibrous) – A place in the body where two or more bones come together but do not move.

Hope it helps!

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Explain how you think an asteroid impact could affect the tilt of Earth’s axis. Explain how this effect would change Earth’s seasons and why. Support your explanation with scientific reasoning and evidence.

Answers

Answer:

An asteroid impact could affect the tilt of the Earth due to the force it applies onto the planet. This would change Earth's seasons due to the fact that Earth's tilt causes seasons.

g Two long parallel wires are a center-to-center distance of 2.50 cm apart and carry equal anti-parallel currents of 2.70 A. Find the magnitude of the magnetic field at the point P which is equidistant from the wires. (R

Answers

Answer:

864 mT

Explanation:

The magnetic field due to a long straight wire B = μ₀i/2πR where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, i = current in wire, and R = distance from center of wire to point of magnetic field.

The magnitude of magnetic field due to the first wire carrying current i = 2.70 A at distance R which is mid-point between the wires is B = μ₀i/2πR.

Since the other wire also carries the same current at distance R, the magnitude of the magnetic field is B = μ₀i/2πR.

The resultant magnetic field at B is B' = B + B = 2B = 2(μ₀i/2πR) = μ₀i/πR

Now R = 2.50 cm/2 = 1.25 cm = 1.25 × 10⁻² m and i = 2.70 A.

Substituting these into B' = μ₀i/πR, we have

B' = 4π × 10⁻⁷ H/m × 2.70 A/π(1.25 × 10⁻² m)

B = 10.8/1.25 × 10⁻⁵ T

B = 8.64 × 10⁻⁵ T

B = 864 × 10⁻³ T

B = 864 mT

This question involves the concept of the magnetic field due to two current-carrying wires in the same direction, parallel to each other.

The magnitude of the magnetic field at the point P, which is equidistant from the wires is "8.64 x 10⁻⁵ T".

The following formula is used to find the magnetic field at the center distance between two parallel current-carrying wires in the same direction:

[tex]B = \frac{\mu_oI_1}{2\pi r}+\frac{\mu_oI_2}{2\pi r}\\\\But,\ I_1=I_2=I\\\\B = \frac{\mu_oI}{\pi r}[/tex]

where,

B = magnetic field at required point = ?

μ₀ = permeability of free space = 4π x 10⁻⁷ H/m

I = current = 2.7 A

r = distance from wires to the point = 2.5 cm/2 = 1.25 cm = 0.0125 m

Therefore,

[tex]B=\frac{(4\pi\ x\ 10^{-7}\ H/m)(2.7\ A)}{\pi (0.0125\ m)}[/tex]

B = 8.64 x 10⁻⁵ T

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why do you have to soak the leaf. in boiling water after heating it in alcohol​

Answers

Answer:

The hot water kills the leaf and the alcohol breaks down the chlorophyll, taking the green color out of the leaf.

Explanation:

Soaking the leaf in alcohol helps to remove the chlorophyll in the leaves and make it easy to identify the starch present in the them.

What is starch test in leaf ?

The starch in plants formed in their leaves by photosynthesis can be identified using some analytical teste. For this the leaves have to placed inside a dark room for one day.

After that soak the leaf in hot water after heating it in alcohol. The alcohol degrades the chlorophyll, removing the leaf's green hue, while the hot water kills the leaf. One of the leaves will become blue-black and the other will turn reddish-brown when you apply iodine to them.

Iodine is an indicator that changes color from blue to black when starch is present. The leaf that was in the sun changes from green to blue-black, showing that it has been doing photosynthesis and making starch.

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A small town has decided to forego the use of electrical power and send energy through town via mechanical waves on ropes. They use rope with a mass per length of 1.50 kg/m under 6000 N tension. If they are limited to a wave amplitude of 0.500 m, what must be the frequency of waves necessary to transmit power at the average rate of 2.00 kW

Answers

Answer:

the required frequency of waves is 2.066 Hz

Explanation:

Given the data in the question;

μ = 1.50 kg/m

T = 6000 N

Amplitude A = 0.500 m

P = 2.00 kW = 2000 W

we know that, the average power transmit through the rope can be expressed as;

p = [tex]\frac{1}{2}[/tex]vμω²A²

p = [tex]\frac{1}{2}[/tex]√(T/μ)μω²A²

so we solve for ω

ω² = 2P / √(T/μ)μA²

we substitute

ω² = 2(2000) / √(6000/1.5)(1.5)(0.500)²

ω² = 4000 / 23.71708

ω² = 168.65

(2πf)² = ω²

so

(2πf)² = 168.65

4π²f² = 168.65

f² = 168.65 / 4π²

f² = 4.27195

f = √4.27195

f = 2.066 Hz

Therefore, the required frequency of waves is 2.066 Hz

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