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3 years ago

14

If Galileo drops a cannon ball from the 60 meter high) Leaning Tower of Pisa, how fast will it be moving when it hits the ground

?

1 answer:

viva [34]3 years ago

5 0

Answer:

When the ball hits the ground, the velocity will be -34 m/s.

Explanation:

The height and velocity of the ball at any time can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the ball at time "t".

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity. (-9.8 m/s² considering the upward direction as positive).

v = velocity at time "t".

If we place the origin of the frame of reference on the ground, when the ball hits the ground its height will be 0. Then using the equation of height, we can calculate the time it takes the ball to reach the ground:

y = y0 + v0 · t + 1/2 · g · t²

0 = 60 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²

0 = 60 m - 4.9 m/s² · t²

-60 m / -4.9 m/s² = t²

t = 3.5 s

Now, with this time, we can calculate the velocity of the ball when it reaches the ground:

v = v0 + g · t

v = 0 m/s - 9.8 m/s² · 3.5 s

v = -34 m/s

When the ball hits the ground, the velocity will be -34 m/s.

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a particle is moving along a circular path having a radius of 4 in such that its position as a function of time is given by thet

ANTONII [103]

Answer:

Explanation:

Given

radius of circular path $r=4\ in.$

Position is given by

$\theta =\cos 2t---1$

Differentiate 1 to angular velocity we get

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega =-2\sin 2t----2$

Differentiate 2 to get angular acceleration

$\frac{\mathrm{d} \omega }{\mathrm{d} t}=-2^2\cos 2t ---3$

Net acceleration is the vector summation of tangential and centripetal force

$a_t=\alpha \times r$

$a_t=-4\cos 2t\times 4=-16\cos 2t$

$a_r=\omega ^2\cdot r$

$a_r=(-2\sin 2t)^2\cdot 4$

$a_r=16\sin^2(2t)$

$a_{net}=\sqrt{a_r^2+a_t^2}$

$a_{net}=\sqrt{(16\sin ^2(2t)+(-16\cos 2t)^2}$

$a_{net}=\sqrt{256\cos ^2(2t)+256\sin ^4(2t)}$

6 0

3 years ago

How much force is required to pull a spring 3.0 cm from

avanturin [10]

Answer:

I know that T= kx where T is the tension which equaka the force og gravity = mg = 1.37 * 10 = 13.7 x is the elongation of the spring so the length after dangling the object minus the original length.

I hope it helps

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4 years ago

The nonreflective coating on a camera lens with an index of refraction of 1.29 is designed to minimize the reflection of 634-nm

eimsori [14]

Answer:

minimum thickness of the coating = 122.868 nm

Explanation:

Given data

lens index of refraction = 1.29

wavelength = 634 nm

glass index of refraction = 1.53

to find out

minimum thickness of the coating

solution

we have given non reflective coating

so

we know that minimum thickness of the coating formula

minimum thickness of the coating = Wavelength / 4n

here n is coating index of refraction

so put here both value to get thickness

minimum thickness of the coating = Wavelength / 4n

minimum thickness of the coating = 634 / 4 ( 1.29 )

so minimum thickness of the coating = 122.868 nm

5 0

3 years ago

A 90 kg man stands in a very strong wind moving at 17 m/s at torso height. As you know, he will need to lean in to the wind, and

victus00 [196]

Answer:

a) $t=195.948N.m$

b) $\phi=13.6 \textdegree$

Explanation:

From the question we are told that:

Density $\rho=1.225kg/m^2$

Velocity of wind $v=14m/s$

Dimension of rectangle:50 cm wide and 90 cm

Drag coefficient $\mu=2.05$

a)

Generally the equation for Force is mathematically given by

$F=\frac{1}{2}\muA\rhov^2$

$F=\frac{1}{2}2.05(50*90*\frac{1}{10000})*1.225*17^2$

$F=163.29$

Therefore Torque

$t=F*r*sin\theta$

$t=163.29*1.2*sin90$

$t=195.948N.m$

b)

Generally the equation for torque due to weight is mathematically given by

$t=d*Mg*sin90$

Where

$d=sin \phi$

Therefore

$t=sin \phi*Mg*sin90$

$195.948=833sin \phi$

$\phi=sin^{-1}\frac{195.948}{833}$

$\phi=13.6 \textdegree$

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3 years ago

What is mightier than steel yet cowers from the sun?

JulijaS [17]

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4 years ago

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