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nlexa [21]
3 years ago
11

What compounds are you made of?

Physics
2 answers:
Vanyuwa [196]3 years ago
6 0
I am made of literally THOUSANDS of compounds ... too many to list here.
But the one compound that's most abundant, and actually comprises almost
80% of my entire beautiful body, is the compound DiHydrogen Oxide, with
the molecular formula H₂O .  This compound is commonly known as "water".
tamaranim1 [39]3 years ago
5 0
Thousands of compounds
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What is a stream’s discharge rate if it has a width of 10 meters, a depth of 2 meters, and a velocity of 2 meters per second?
julia-pushkina [17]

A stream’s discharge rate if it has a width of 10 meters, a depth of 2 meters, and a velocity of 2 meters per second will be 40 m^{3}/sec .

Discharge rate = velocity * area

                 = velocity * depth * width

                 = 2 * 2 * 10 = 40 m^{3}/sec

A stream’s discharge rate if it has a width of 10 meters, a depth of 2 meters, and a velocity of 2 meters per second will be 40 m^{3}/sec .

learn more about discharge rate

brainly.com/question/20709500?referrer=searchResults

#SPJ4

4 0
2 years ago
A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica
-BARSIC- [3]

Answer:

33.2 m

Explanation:

For the first object:

y₀ = 81.5 m

v₀ = 0 m/s

a = -9.8 m/s²

t₀ = 0 s

y = y₀ + v₀ t + ½ at²

y = 81.5 − 4.9t²

For the second object:

y₀ = 0 m

v₀ = 40.0 m/s

a = -9.8 m/s²

t₀ = 2.20 s

y = y₀ + v₀ t + ½ at²

y = 40(t−2.2) − 4.9(t−2.2)²

When they meet:

81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²

81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)

81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716

81.5 = 61.56t − 111.716

193.216 = 61.56t

t = 3.139

The position at that time is:

y = 81.5 − 4.9(3.139)²

y = 33.2

7 0
3 years ago
A Van de Graaff generator is one of the original particle accelerators and can be used to accelerate charged particles like prot
Rzqust [24]

Answer:

595391.482946 m/s

3.21875\times 10^{6}

Explanation:

E = Energy = 1.85 keV

I = Current = 5.15 mA

e = Charge of electron = 1.6\times 10^{-19}\ C

t = Time taken = 1 second

m = Mass of proton = 1.67\times 10^{-27}\ kg

Velocity of proton is given by

v=\sqrt{\dfrac{2E}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.85\times 10^3\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}}\\\Rightarrow v=595391.482946\ m/s

The speed of the proton is 595391.482946 m/s

Current is given by

I=\dfrac{\Delta Q}{t}\\\Rightarrow \Delta Q=It\\\Rightarrow \Delta Q=5.15\times 10^{-3}\times (1\ sec)\\\Rightarrow Q=5.15\times 10^{-3}\ C

Number of protons is

n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{5.15\times 10^{-3}}{1.6\times 10^{-19}}\\\Rightarrow n=3.21875\times 10^{6}\ protons

The number of protons is 3.21875\times 10^{6}

3 0
3 years ago
What’s a good hypothesis and and experiment for water?
Lelu [443]

Answer: You could do something like, "how does water react to being mixed with baking soda"...or something along those lines

Explanation:

6 0
3 years ago
A 15 kg bucket of water is suspended by a very light ropewrapped around a solid uniform cylinder 0.300 m in diamter withmass 12.
matrenka [14]

Answer:

Part a)

T = 42 N

Part b)

v_f = 11.8 m/s

Part c)

t = 1.7 s

Part d)

F = 159.7 N

Explanation:

Part a)

While bucket is falling downwards we have force equation of the bucket given as

mg - T = ma

for uniform cylinder we will have

TR = I\alpha

so we have

T = \frac{1}{2}MR^2(\frac{a}{R^2})

T = \frac{1}{2}Ma

now we have

mg = (\frac{M}{2} + m)a

a = \frac{mg}{(\frac{M}{2} + m)}

a = \frac{15 \times 9.81}{(6 + 15)}

a = 7 m/s^2

now we have

T = \frac{12 \times 7}{2}

T = 42 N

Part b)

speed of the bucket can be found using kinematics

so we have

v_f^2 - v_i^2 = 2 a d

v_f^2 - 0 = 2(7)(10)

v_f = 11.8 m/s

Part c)

now in order to find the time of fall we can use another equation

v_f - v_i = at

11.8 - 0 = 7 t

t = 1.7 s

Part d)

as we know that cylinder is at rest and not moving downwards

so here we can use force balance

F = T + Mg

F = 42 + (12 \times 9.81)

F = 159.7 N

5 0
3 years ago
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