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3 years ago

A physicist drops astone off the top of cliff over looking a lake. She hears the splash 4 second after releasing the stone?

Physics

1 answer:

Angelina_Jolie [31]3 years ago

6 0

Answer:

Height of cliff(h) = 80 meter (Approx)

Explanation:

Given:

Time taken t = 4 sec

Gravitational acceleration = 10 m/s (Approx)

Find:

Height of cliff(h)

Computation:

h = ut + 1/2(g)(t)²

Height of cliff(h) = (0)(4) + 1/2(10)(4)²

Height of cliff(h) = 80 meter (Approx)

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The maximum energy a bone can absorb without breaking is surprisingly small. For a healthy human of mass 60 kg60 kg, experimenta

netineya [11]

Answer:

the maximum height a man can jump from and land rigidly upright on both feet without breaking his legs is 0.34 m

Explanation:

Mass of a healthy man = 60 kg

energy the bone can take without breaking = 200 J

If a healthy man jumps from a height 'h', he falls with an energy equal to the potential energy due to his initial height above the ground.

initial potential energy of the healthy man = mgh

where m = mass of the man

g = acceleration due to gravity = 9.81 m/s^2

h = the height above ground

==> PE = 60 x 9.81 x h = 588.6h

If we assume that all energy is absorbed in the leg bones in a rigid landing, then we can safely say that this calculated PE for a healthy man is equal to the energy his bone can absorb in the jump without breaking.

equating, we have

200 = 588.6h

the maximum height a man can jump from without breaking his legs = 200/588.6 = 0.34 m

When people jump from a height, the sudden deceleration to zero can impact a big force on the leg bones, shattering them. If the time spent in decelerating to zero is increased, the overall force on the leg bones is reduced greatly.

Bending the knees gradually on landing from a jump from a height, and then rolling increases the time spent decelerating, and reduces the impact force on the legs due to the landing. If you observe carefully you will see that this is what professional stunts men and acrobats do when they jump from a height.

5 0

3 years ago

You use 8x binoculars were used on a warbler (14cm long) in a tree 18cm away. What angle (in degrees) does the image of the warb

mafiozo [28]

Answer:

The angle it subtend on the retina is $\theta_z = 0.44586^o$

Explanation:

From the question we are told that

The length of the warbler is $L = 14cm = \frac{14}{100} = 0.14m$

The distance from the binoculars is $d = 18cm = \frac{18}{100} = 0.18m$

The magnification of the binoculars is $M =8$

Without the 8 X binoculars the angle made with the angular size of the object is mathematically represented as

$\theta = \frac{L}{d}$

$\theta = \frac{0.14}{0.18}$

$= 0.007778 rad$

Now magnification can be represented mathematically as

$M = \frac{\theta _z}{\theta}$

Where $\theta_z$ is the angle the image of the warbler subtend on your retina when the binoculars i.e the binoculars zoom.

$\theta_z = M * \theta$

=> $\theta_z =8 * 0.007778$

$= 0.0622222224$

Generally the conversion to degrees can be mathematically evaluated as

$\theta_z = 0.062222224 * (\frac{360 }{2 \pi rad} )$

$\theta_z = 0.44586^o$

7 0

3 years ago

A book is to be produced with pages of thickness 0.125mm.If the book is to be exactly 1cm thick what is the maximum number of pa

miskamm [114]

Answer:

1cm=10mm

10cm/0.125mm

80pages

5 0

3 years ago

_____ is a chemical property of matter.

german

Answer:

it is chemical reactivity (D)

Explanation:

it is because chemical properties are properties that can be measured and observed only when matter undergoes a change to become an entirely different kind of matter.Which they include Chemical reactivity,flammable,and the ability to rust.And Chemical reactivity is the ability of matter to react chemically with other substances.

8 0

3 years ago

A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in

dexar [7]

Answer:

K = 0.076 J

Explanation:

The height of the target, h = 0.860 m

The mass of the steel ball, m = 0.0120 kg

Distance moved, d = 1.50 m

We need to find the kinetic energy (in joules) of the target ball just after it is struck. Let t is the time taken by the ball to reach the ground.

$h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}}$