Giant waves produced by gravity are called what?

Answers

Answer 1

Answer: Either a tsunami or an up welling!

Hope this helped!

Related Questions

The amplitude of a sound wave determines itsloudness pitch frequency wavelength
What do tendons connect skeletal muscles to?
If a droplet has three extra electrons, what is it’s measured charge?
A cubic centimeter can be expressed as a.cm3 or cc b. ccm c. ccm3 or d.ct
Forces contribute to the net force on a car rolling down a ramp

The potential energy of a 40 kg cannon ball is 14000 J. How high was the cannon ball to have this much potential energy?

Answers

From the information given, cannon ball weighs 40 kg and has a potential energy of 14000 J.

We need to find its height.

We will use the formula P.E = mgh

Therefore h = P.E / mg

where P.E is the potential energy,

m is mass in kg,

g is acceleration due to gravity (9.8 m/s²)

h is the height of the object's displacement in meters.

h = P.E. / mg

h = 14000 / 40 × 9.8

h = 14000 / 392

h = 35.7

Therefore the canon ball was 35.7 meters high.

Final answer:

The height of the cannon ball to have 14000 J of potential energy is approximately 357.14 meters. This is calculated using the formula for potential energy: PE = mgh, and solving for 'h'.

Explanation:

The potential energy (PE) of an object is given by the formula PE = mgh, where 'm' is the mass of the object, 'g' is the acceleration due to gravity (standard approximate value is 9.8 m/s² on Earth), and 'h' is the height. In your question, we want to find the height 'h'. Given that the potential energy is 14000 J and the mass of the cannon ball is 40 kg, we can rearrange the formula to solve for 'h': h = PE / (m*g).

So, inputting the given values, h = 14000 J / (40 kg * 9.8 m/s²). Solving this, we find that the height is approximately 357.14 meters. This means the cannon ball was at around this height to have 14000 J of potential energy.

Learn more about Potential Energy here:

brainly.com/question/24284560

#SPJ11

What is one way humans can help a prairie ecosystem after a wildfire? A. bring water to the plants in the ecosystem B. remove seeds from the ecosystem C. spray DDT over the ecoystem D. build roads through the ecoystem

Answers

Answer:

Explanation:

Answer:

(A) Bring water to the plants in the ecosystem.

How are intensity sound and energy related

Answers

sound is a form of energy in that it consists fluctuations of air pressure . The speed of the fluctuations is measured in cycles per second or Hertz (HZ)

Intensity is how large the fluctuations are, also known as amplitude and for the sound the unit is decibels of sonic pressure level (dB SPL)

What is the purpose of the universal gravitation constant in Newton's law of universal gravitation

Answers

The formula for the force of gravity is

   Force = G m₁ m₂ / R² .

If the formula were

      Force = m₁ m₂ / R² , without any 'G',

then in SI units, a couple of 10-kilogram blocks, 1 meter apart,
would attract each other with forces of 100 newtons ... about
23 pounds !

And in English units, a couple of 12-pound bowling balls
1 foot apart would attract each other with forces of 144 pounds !

As I'm sure you've noticed, they don't.

The forces of gravity are proportional to the product of the masses
divided by the square of the distance, but they're not equal to that
expression. In order to make an equation from a proportion, you
need a constant in there, and that's what 'G' is for.

Answer:

Since I don't know what that is here you go.

Law of universal gravitation: all objects attract one another with a gravitational force dependent on their masses and the distance between them

Explanation:

which is the of these their statements correctly describes a key difference between aerobic activity and anaerobic activity

Answers

aerobic uses oxygen to liberate heat while anaerobic is in the absence of oxygen.

3. A block of mass m1=1.5 kg on an inclined plane of an angle of 12° is connected by a cord over a mass-less, frictionless pulley to a second block of a mass m2= 1.2 kg hanging vertically. If the hanging block falls 0.92min 1.23 s, what is the coefficient of friction between m1 and the inclined plane?

Please show work

Answers

Answer:

$\mu=0.377$

Explanation:

we need to start by drawing the free body diagram for each of the masses in the system. Please see attached image for reference.

We have identified in green the forces on the blocks due to acceleration of gravity ( $w_1$ and $w_2$ ) which equal the product of the block's mass times "g".

On the second block ( $m_2$ ), there are just two forces acting: the block's weight ( $m_2\,*\,g$ ) and the tension (T) of the string. We know that this block is being accelerated since it has fallen 0.92 m in 1.23 seconds. We can find its acceleration with this information, and then use it to find the value of the string's tension (T). We would need both these values to set the systems of equations for block 1 in order to find the requested coefficient of friction.

To find the acceleration of block 2 (which by the way is the same acceleration that block 1 has since the string doesn't stretch) we use kinematics of an accelerated object, making use of the info on distance it fell (0.92 m) in the given time (1.23 s):

$x_f-x_i=v_i\,t-(1)/(2) a\,t^2$ and assume there was no initial velocity imparted to the block:

$x_f-x_i=v_i\,t-(1)/(2) a\,t^2\n-0.92\,m=0\,-(1)/(2) a\,(1.23)^2\na=(0.92\,*\,2)/(1.23^2) \na=1.216 \,(m)/(s^2)$

Now we use Newton's second law in block 2, stating that the net force in the block equals the block's mass times its acceleration:

$F_(net)=m_2\,a\nw_2-T=m_2\,a\nm_2\,g-T=m_2\,a\nm_2\,g-m_2\,a=T\nm_2\,(g-a)=T\n1.2\,(9.8-1.216)\,N=T\nT=10.3008\,N$

We can round this tension (T) value to 10.3 N to make our calculations easier.

Now, with the info obtained with block 2 (a - 1.216 $(m)/(s^2)$ , and T = 10.3 N), we can set Newton's second law equations for block 1.

To make our study easier, we study forces in a coordinate system with the x-axis parallel to the inclined plane, and the y-axis perpendicular to it. This way, the motion in the y axis is driven by the y-component of mass' 1 weight (weight1 times cos(12) -represented with a thin grey trace in the image) and the normal force (n picture in blue in the image) exerted by the plane on the block. We know there is no acceleration or movement of the block in this direction (the normal and the x-component of the weight cancel each other out), so we can determine the value of the normal force (n):

$n-m_1\,g\,cos(12^o)=0\nn=m_1\,g\,cos(12^o)\nn=1.5\,*\,9.8\,cos(12^o)\nn=14.38\,N$

Now we can set the more complex Newton's second law for the net force acting on the x-axis for this block. Pointing towards the pulley (direction of the resultant acceleration $a$ ), we have the string's tension (T). Pointing in the opposite direction we have two forces: the force of friction (f ) with the plane, and the x-axis component of the block's weight (weight1 times sin(12)):

$F_(net)=m_1\,a\nT-f-w_1\,sin(12)=m_1\,a\nT-w_1\,sin(12)-m_1\,a=f\nf=[10.3-1.5\,*\,9.8\,sin(12)-1.5\,*1.216]\,N\nf=5.42\,N$

And now, we recall that the force of friction equals the product of the coefficient of friction (our unknown $\mu$ ) times the magnitude of the normal force (14.38 N):

$f=\mu\,n\n5.42\,N=\mu\,*\,14.38\,N\n\mu=(5.42)/(14.38)\n\mu=0.377$

with no units.

Other Questions

A sample of dna conains 500 adenine bases. how many thymine bases does it contain

________________________ from the wind transfers to the water as the water moves toward land.A. EnergyB. MoistureC. Vibrations

What type of soil can become as hard as stone when dries out

Which waves are longitude waves? Check all that apply. 1.sound waves 2.ocean waves 3.ultraviolet waves 4.earthquake P-waves 5.earthquake S-waves 6.radio waves