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3 years ago

the three major scales used to measure earthquakes are Mercalli scale Richard scale and magnitude scale

1 answer:

3 0

False; the three major scales used to measure earthquakes are the Mercalli Scale, the Richter Scale, and the Magnitude Scale. I hope this helps!

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Complete the ray diagram and label incident ray, refracted ray, angle of incidence, and angle of refraction

Novay_Z [31]

Answer:

Solution

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(a) The labelled diagram is shown.

(b) The refractive index of diamond is 2.42. Refractive index of diamond is the ratio of the speed of light in air to the speed of light in diamond.i.e.,

μ=

Speedoflightindiamond

Speedoflightinair

and, the ratio of these velocities is 2.42. i.e., This means that the speed of light in diamond will reduce by a factor of 2.42 as compared to its speed in air. In other words, the speed of light in diamond is

1/2.42

times the speed of light in vacuum.

Explanation:

a) Draw and label the diagram given :

(i) Incident ray

(ii) Refracted ray

(iii) Emergent ray

(iv) Angle of reflection

(v) Angle of deviation

(v) Angle of emergence

(b) The refractive index of diamond is 2.42. What is the meaning of this statement in relation to speed of light?

4 0

3 years ago

An object ends up at a final position of x=-55.25 meters after a displacement of -189.34 meters after a displacement of -189.34

Travka [436]

The initial position of the object was found to be 134.09 m.

<u>Explanation:</u>

As displacement is the measure of difference between the final and initial points. In other words, we can say that displacement can be termed as the change in the position of the object irrespective of the path followed by the object to change the path. So

Displacement = Final position - Initial position.

As the final position is stated as -55.25 meters and the displacement is also stated as -189.34 meters. So the initial position will be

Initial position of the object = Final position-Displacement

Initial position = -55.25 m - (-189.34 m) = -55.25 m + 189.34 m = 134.09 m.

Thus, the initial position for the object having a displacement of -189.34 m is determined as 134.09 m.

4 0

3 years ago

Jet fighter planes are launched from aircraft carriers with the aid of their own engines and a catapult. If in the process of be

ANTONII [103]

Answer:

W=315 x 10⁵ J

Explanation:

Given that

F= 2.5 x 10⁵ N

d= 90 m

K.E.=5.4 x 10⁷ J

We know that work done by all force is equal to the change in kinetic energy

Lets take work done by catapult is W

W + F.d= K.E.

W= 5.4 x 10⁷ - 2.5 x 10⁵ x 90 J

W= (540 - 25 x 9) 10⁵ J

W=315 x 10⁵ J

5 0

3 years ago

A 2.5 kg tribble is placed in a bucket and whirled in a 1.4 m radius vertical circle at a constant tangential speed. If the forc

Over [174]

Given that,

Mass of a tribble, m = 2.5 kg

Radius, r = 1.4 m

The force on the tribble from the bucket does not exceed 10 times its weight.

To find,

The maximum tangential speed.

Solution,

The force acting on the tribble is equal to the centripetal force.

F = 10mg

The formula for the centripetal force is given by :

$F=\dfrac{mv^2}{r}$

v is maximum tangential speed

$v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{mgr}{m}} \\\\v=\sqrt{{10gr}} \\\\v=\sqrt{10\times 9.8\times 1.4} \\\\v=11.7\ m/s$

So, the maximum tangential speed is 11.7 m/s.

8 0

3 years ago

An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electro

valkas [14]

Answer:

Final velocity of electron, $v=6.45\times 10^5\ m/s$

Explanation:

It is given that,

Electric field, E = 1.55 N/C

Initial velocity at point A, $u=4.52\times 10^5\ m/s$

We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :

$v^2=u^2+2as$ ........(1)