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2 years ago

What would be the speed of an object just before hitting the ground if dropped 91.5 meters

Physics

1 answer:

Aleks04 [339]2 years ago

5 0

Answer:

about 42.35 m/s

Explanation:

Use the equation for accelerated motion (g), and with zero initial velocity that doesn't include time:

$v_f^2=v_i^2+2\,a\,\Delta x$

which for our case would reduce to:

$v_f^2=v_i^2+2\,a\,\Delta x\\v_f^2=0+2\,9.8\,(91.5)\\v_f^2= 1793.4\\v_f=\sqrt{1793.4} \\v_f \approx 42.35$

then the velocity just before hitting would be about 42.35 m/s

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Monochromatic light of wavelength 687 nm is incident on a narrow slit. On a screen 1.65 m away, the distance between the second

Sophie [7]

Answer:

a ) 1.267 radian

b ) 1.084 10⁻³ mm

Explanation:

Distance of screen D = 1.65 m

Width of slit d = ?

Wave length of light λ = 687 nm.

Distance of second minimum fro centre y = 2.09 cm

Angle of diffraction = y / D

= 2.09 /1.65

= 1.267. radian

Angle of diffraction of second minimum

= 2 λ / d

so 2 λ / d = 1.267

d = 2 λ / 1.267 = (2 x 687 ) /1.267 nm

=1084.45 nm = 1.084 x 10⁻³ mm.

3 0

2 years ago

it takes 90 j of work to stretch a spring 0.2 m from its equilibrium position. How muc work is needed to stretch it an additiona

Vinvika [58]

Work needed: 720 J

Explanation:

The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by

$E=\frac{1}{2}kx^2$

where

k is the spring constant

x is the stretching of the spring from the equilibrium position

In this problem, we have

E = 90 J (work done to stretch the spring)

x = 0.2 m (stretching)

Therefore, the spring constant is

$k=\frac{2E}{x^2}=\frac{2(90)}{(0.2)^2}=4500 N/m$

Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of

x = 0.2 + 0.4 = 0.6 m

Substituting,

$E'=\frac{1}{2}kx^2=\frac{1}{2}(4500)(0.6)^2=810 J$

Therefore, the additional work needed is

$\Delta E=E'-E=810-90=720 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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3 years ago

How are compounds and mixture alike ?

natita [175]

Compounds are atoms or molecules that are chemically bonded together to form a different molecule or compound. The process could be written as a chemical reaction. A mixture is when two elements or compounds are mixed together but do not react, ie no chemical reaction between the two occurs.

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3 years ago

A car with a mass of 2.0 * 10^3 kg is traveling at 15 m/s. what is the momentum of the car?

Allushta [10]

Hello,

Your answer to this problem is 400/3

Hope this helps!

3 0

2 years ago

Billy drops a water balloon from the roof of his house. Since the balloon began with an original velocity of zero how far above

VashaNatasha [74]

T=2,5s
a=g≈10m/s²

h=s=?

$s= \frac{1}{2} at^2 \Leftrightarrow h=\frac{1}{2}gt^2=\frac{1}{2}*10m/s^2 *(2,5s)^2=5m/s^2* \frac{25}{4} s^2=\boxed{31,25m}$

"Non nobis, Domine, non nobis, sed Nomini tuo da gloriam."

Regards M.Y.

7 0

3 years ago