Name the plate boundary responsible for this.<br><br> Please help!!

Julli [10]

An acrostic poem for transformation simply refers to those simple poems conveying transformation messages in which the first letter of each line forms a word or phrase vertically.

A poem can be defined as a a piece of writing in which the expression of feelings and ideas is given intensity by particular attention to diction.

So therefore, an acrostic poem for transformation simply refers to those simple poems conveying transformation messages in which the first letter of each line forms a word or phrase vertically.

Complete question:

What do you understand by acrostic poem for transformation?

Learn more about poem:

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5 0

1 year ago

TECHNOLOGY A Lamborghini Aventador accelerates from rest to 100 km / h in 2.9 s. The mass of the car is 1575 kg Calculate the po

DerKrebs [107]

At 100 km/hr, the car's kinetic energy is

KE = (1/2) (mass) (speed)²

KE = (1/2) (1575 kg) ( [100 km/hr] x [1000 m/km] x [1 hr/3600 sec] )²

KE = (787.5 kg) (27.78 m/s)²

KE = 607,639 Joules

In order to deliver this energy in 2.9 seconds, the engine must supply

(607,639 J / 2.9 sec) = 209,531 watts

<em>Power = 281 HP</em>

6 0

3 years ago

Runaway truck ramps are common on mountainous highways in case the brakes fail on large trucks. If a

dusya [7]

Answer:

$W=-21,870,000\ J$

Explanation:

<u>Work and Kinetic Energy </u>

The work an object does due to its motion is equal to the change of its kinetic energy. Being ko and k1 the initial and final kinetic energy respectively and m the mass of the object, then

$W=\Delta k=k_1-k_0$

Since

$\displaystyle k=\frac{mv^2}{2}$

We have

$\displaystyle W=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}$

The truck has a mass of 60,000 kg and is moving at 27 m/s. The runaway truck ramp must stop the truck, so the final speed is 0. Thus

$\displaystyle W=\frac{(60,000)0^2}{2}-\frac{(60,000)(27)^2}{2}$

$W=0-21870000\ J$

$\boxed{W=-21,870,000\ J}$

3 0

3 years ago

At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet

expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

also the PE is given as function of position as

$PE = \frac{1}{2}m\omega^2x^2$

now it is given that

KE = PE

now we will have

$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

$A^2 - x^2 = x^2$

$2x^2 = A^2$

$x = \frac{A}{\sqrt2}$

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

now to find the fraction of kinetic energy

$f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}$

$f = \frac{A^2 - (\frac{A}{3})^2}{A^2}$

$f_k = \frac{8}{9}$

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

$f_{PE} = 1 - f_k$

$f_{PE} = 1 - (8/9) = 1/9$

7 0

3 years ago

A one-piece cylinder has a core section protruding from the larger drum and is free to rotate around its central axis. A rope wr

PilotLPTM [1.2K]

Answer:

Magnitude the net torque about its axis of rotation is 2.41 Nm

Solution:

As per the question:

The radius of the wrapped rope around the drum, r = 1.33 m

Force applied to the right side of the drum, F = 4.35 N

The radius of the rope wrapped around the core, r' = 0.51 m

Force on the cylinder in the downward direction, F' = 6.62 N

Now, the magnitude of the net torque is given by:

$\tau_{net} = \tau + \tau'$

where

$\tau$ = Torque due to Force, F

$\tau'$ = Torque due to Force, F'

$tau = F\times r$

$tau' = F'\times r'$

Now,

$\tau_{net} = - F\times r + F'\times r'$

$\tau_{net} = - 4.35\times 1.33 + 6.62\times 0.51 = - 2.41\ Nm$

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

$|\tau_{net}| = 2.41\ Nm$

3 0

2 years ago

Name the plate boundary responsible for this. Please help!!