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1 year ago

2H, +0,-> 2H,0 How many hydrogen atoms are on the left side of the equation?

Physics

1 answer:

marin [14]1 year ago

8 0

Answer:

Explanation:

The reaction between hydrogen and oxygen gives water (H2O)

2H,+O, -> 2H2O

Oxygen is a gas O, = O2

Balancing both sides

let , be x

2x = 4

$x = \frac{4}{2}$

x= 2

Therefore, 2H2 = 4 hydrogen

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We are designing a crude propulsion mechanism for a science fair demonstration. One of our team members stands on a skateboardth

Scrat [10]

Answer:

greater speed will be obtained for the elastic collision,

Explanation:

To answer this exercise we must find the speed that the sail acquires after each impact.

Let's start by hitting a ball of clay.

The system is formed by the candle and the clay balls, therefore the forces during the collision are internal and the moment is conserved.

initial instant. before the crash

p₀ = m v₀

where m is the mass of the ball and vo its initial velocity, we are assuming that the candle is at rest

final instant. After the crash

the mass of the candle is M

p_f = (m + M) v

the moment is preserved

p₀ = p_f

m v₀ = (m + M) v

v = $\frac{m}{m+M} \ v_o$

for when n balls have collided

v = $\frac{m}{n \ m + M}$ v₀

Now let's analyze the case of the bouncing ball (elastic)

initial instant

p₀ = m v₀

final moment

p_f = m v_{1f} + M v_{2f}

p₀ = p_f

m v₀ = m v_{1f} + M v_{2f}

m (v₀ - v_{1f}) = M v_{2f}

this case corresponds to an elastic collision whereby the kinetic energy is conserved

K₀ = K_f

½ m v₀² = ½ m v_{1f}² + ½ M v_{2f}²

v₁ = v_{1f} v₂ = v_{2f}

m (v₀² - v₁²) = M v₂²

let's use the identity

(a² - b²) = (a + b) (a-b)

we write our equations

m (v₀ - v₁) = M v₂ (1)

m (v₀ - v₁) (v₀ + v₁) = M v₂²

let's divide these equations

v₀ + v₁ = v₂

Let's look for the final speeds

we substitute in equation 1

m (v₀ - v₁) = M (v₀ + v₁)

v₀ (m -M) = (m + M) v₁

v₁ = $\frac{m-M}{m + M}$ v₀

we substitute in equation 1 to find v₂

$\frac{M}{m}$ v₂ = v₀ - $\frac{m-M}{m+M}$ v₀

v₂ = $\frac{m}{M} ( 1 - \frac{m-M}{m+M} ) \ v_o$

v₂ = $\frac{m}{M} ( \frac{2M}{m+M} ) \ \ v_o$

v₂ = $\frac{2m}{m +M} \ v_o$

Let's analyze the results for inelastic collision with each ball that collides with the sail, the total mass becomes larger so the speed increase is smaller and smaller.

In the case of elastic collision, the increase in speed is constant with each ball since the total mass remains invariant.

Consequently, greater speed will be obtained for the elastic collision, that is, the ball will bounce.

8 0

2 years ago

The gauge pressure at the bottom of a cylinder of liquid is 0.30atm. The liquid is poured into another cylinder with twice the r

likoan [24]

Answer:

$P_g' = 0.075 atm$

Explanation:

Gauge pressure at the bottom of the cylinder depends on the height of water in the cylinder

So here we can say that

$P_g = \rho g h$

now when liquid is filled to height "h" in base area "A" then gauge pressure of the liquid at the bottom is given as

$P_g = 0.30 atm$

now we put the whole liquid into another cylinder with twice radius of the first cylinder

So area becomes 4 times

now by volume conservation we can say that if area is increased by 4 times then height of liquid will decrease by 4 times

so we have

$h' = \frac{h}{4}$

so gauge pressure is given as

$P_g' = \frac{0.30}{4} = 0.075 atm$

5 0

3 years ago

A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the

Eva8 [605]

Answer:

The speed of the ball is 42.5 m/s

Explanation:

The initial kinetic energy of the ball is:

$K_1=\frac{1}{2} m v_0^2=\frac{1}{2}*0.140 kg*(35.0 m/s)^2$ = 85.75 J

The speed of the ball after leaving the bat is:

$K_2=K_1+W_{nc}\\ \frac{1}{2}mV^2= 85.75 J + 75 J\\ (\frac{1}{2}mV^2)2=( 160.75 J)2\\ mV^2= 321.5 J\\ V^2= \frac{321.5 J}{0.140kg} \\ V=\sqrt{\frac{321.5 J}{0.140kg}}$

V=47.92 m/s

Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:

$V_f^2-V^2=-2gh$

$V_f^2-(47.92 m/s)^2=-2*9.81m/s^2*25m$

$V_f^2=-2*9.81m/s^2*25m+(47.92 m/s)^2$

$V_f=\sqrt{-2*9.81m/s^2*25m+(47.92 m/s)^2}$

$V_f=42.5m/s$

3 0

2 years ago

Charge is flowing through a conductor at the rate of 420 C/min. If 742 J. of electrical energy are converted to heat in 30 s., w

Y_Kistochka [10]

Answer:

3.53 V

Explanation:

Electric charge: The is the rate of flow of electric charge along a conductor.

The S.I unit of electric charge is C.

Mathematically it is expressed as,

Q = It ............................ Equation 1

Where Q = electric charge, I = current, t = time.

I = Q/t.......................... Equation 2

From the question, charge flows through the conductor at the rate of 420 C/mim

Which means in 1 min, 420 C of charge flows through the conductor.

Hence,

Q = 420 C, t = 1 min = 60 seconds

Substitute into equation 2

I = 420/60

I =7 A

Also

P = VI......................... Equation 3

Where P = power, V = potential drop, I = current.

V = P/I................... Equation 4

Note: Power = Energy/time

From the question, P = 742/30 = 24.733 W. and I = 7 A.

Substitute these values into equation 4

V = 24.733/7

V = 3.53 V

Hence the potential drop across the conductor = 3.53 V

4 0

3 years ago

All music, _______ in the world, is from only ______ notes.

AnnZ [28]

Answer:

All music in the world, is form only two notes and those notes are described below in detailed explanation.

Explanation:

In the chromatic scale, there are basically seven central musical notes, designated A, B, C, D, E, F, and G. They individual express a separate pitch or frequency. For illustration, the "central" A note has a pitch of 450 Hz, and the "common" B note has a pitch of 495 Hz.

Varieties Of Musical Notes You Require To Understand

Semibreve (Whole Note)

Minim (Half Note)

Crotchet (Quarter Note)

5 0

2 years ago