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2 years ago

3. A block falls from a table 0.6 m high. It lands on a vertical spring which has a spring constant of k= 2400 N/m.

Physics

1 answer:

goldfiish [28.3K]2 years ago

8 0

<h2>SOLUTION :-</h2>

PLEASE CHECK THE ATTACHED FILE

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Assume light is traveling through an optical medium of n1 and is incident on a boundary with a different optical medium with n2.

Elza [17]

That n2 = 2*n1. That is, the index of refraction is twice as big in medium 2 since v=c/n

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3 years ago

In 5 meters, a person running at 0.8 m/s accelerates at 1.6 m/s2. How fast 16 points

frutty [35]

They were going at a velocity 4.07m/s

<u>Explanation:</u>

Distance s =5 m

initial velocity u= 0.8 m/s

Acceleration a =1.6m/s2

We have to calculate the velocity with which they were going afterwards i.e final velocity.

Use the equation of motion

$v^2=u^2+2as\\=0.8^2+2\times 1.6\times 5\\=16.64\\v=4.07 m/s$

They were going with a velocity 4.07 m/s afterwards.

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3 years ago

A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North. a) What is the resultant velocity of the motorb

Nostrana [21]

Explanation:

It is given that,

Velocity in East, $v_1=4\ m/s$

Velocity in North, $v_2=3\ m/s$

(a) The resultant velocity is given by :

$v=\sqrt{4^2+3^2}=5\ m/s$

(b) The width of the river is, d = 80 m

Let t is the time taken by the boat to travel shore to shore. So,

$t=\dfrac{d}{v}$

$t=\dfrac{80\ m}{5\ m/s}$

t = 16 seconds

(c) Let x is the distance covered by the boat to reach the opposite shore. So,

$x=v_2\times t$

$x=3\ m/s\times 16\ s$

x = 48 meters

Hence, this is the required solution.

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3 years ago

An airplane flying at 115 m/s due east makes a gradual turn following a circular path to fly south. The turn takes 15 seconds to

bulgar [2K]

To solve this exercise it is necessary to apply the concepts related to Centripetal and Perimeter acceleration of a circle.

The perimeter of a circle is defined by

$P = 2\pi r$

Where,

r= radius

While centripetal acceleration is defined by

$a=\frac{v^2}{r}$

Where,

v= velocity

r= radius

PART A)

The distance of a body can be defined based on the speed and the time traveled, that is

x = v*t

For our values the distance is equal to

x = 15*115=1725m

The plane when going to make the turn from east to south makes a quarter of the circumference that is

$\frac{P}{4} = \frac{2\pi r}{4}$

The same route you take is the distance traveled, that is

$x = \frac{P}{4}$

$x = \frac{2\pi r}{4}$

$1725 = \frac{2\pi r}{4}$

$r = 1098.17m$

PART B)

With the radius is possible calculate he centripetal acceleration,

$a = \frac{v^2}{r}$

$a = \frac{115^2}{1098.17}$

$a = 12.04m/s^2$

Therefore the radius of the curva that the plane follows in making the turn is 1098.17m with a centripetal acceleration of $12.04m/s^2$

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4 years ago

In a hydraulic lift, the radius of the small and large pistons are

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Answer:

the answer is 52.15cm×35.5

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