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3 years ago

PLEASE HELP MEEE, Manager Michael Scott drops a watermelon from a 10 story

Physics

2 answers:

goldenfox [79]3 years ago

7 0

Answer:

V = square root (2 x 9.8 x 30)

V = square root ( 588 ) = 24.2 m/s

Explanation:

Lubov Fominskaja [6]3 years ago

6 0

V = square root (2 x 9.8 x 30)
V = square root ( 588 ) = 24.2 m/s

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A plucked guitar string produces a sound wave with a wavelength of 0.15 m and a velocity of 10.5 m/s, what is the frequency of t

Leno4ka [110]

We know,
V= f× wavelength
10.5= f×0.15
f=10.5/0.15
f= 70 Hz

8 0

3 years ago

How much protein would be consumed with a 3-ounce serving of

dalvyx [7]

Answer:

15kg

(i actually did this)

Explanation:

k12

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6 0

3 years ago

Two objects are dropped from rest from the same height. Object A falls through a distance Da and during a time t, and object B f

stiv31 [10]

Answer:

Da=(1/4)Db

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

When s = Da, t = t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times t^2\\\Rightarrow Da=\frac{1}{2}at^2$

When s = Db, t = 2t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times (2t)^2\\\Rightarrow Db=\frac{1}{2}a4t^2$

Dividing the two equations

$\frac{Da}{Db}=\frac{\frac{1}{2}at^2}{\frac{1}{2}a4t^2}=\frac{1}{4}\\\Rightarrow \frac{Da}{Db}=\frac{1}{4}\\\Rightarrow Da=\frac{1}{4}Db$

Hence, Da=(1/4)Db

3 0

4 years ago

calculate the pressure exerted by a mercury column of 76 cm high at its bottom.Given that density of mercury is 13600kg/m^3 and

Amanda [17]

Answer:

Explanation:I don't say you have to mark my ans as brainliest but if you think it has really helped you plz don't forget to thank me ...

5 0

3 years ago

How do you do this problem?

kvasek [131]

Explanation:

First, find the velocity of the projectile needed to reach a height h when fired straight up.

Given:

Δy = h

v = 0

a = -g

Find: v₀

v² = v₀² + 2aΔy

(0)² = v₀² + 2(-g)(h)

v₀ = √(2gh)

Now find the height reached if the projectile is launched at a 45° angle.

Given:

v₀ = √(2gh) sin 45° = √(2gh) / √2 = √(gh)

v = 0

a = -g

Find: Δy

v² = v₀² + 2aΔy

(0)² = √(gh)² + 2(-g)Δy

2gΔy = gh

Δy = h/2

5 0

3 years ago