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GREYUIT [131]
3 years ago
9

How does the law of conservation of mass apply to this reaction:Al+3HCl-->H2+AlCl3

Physics
1 answer:
Veronika [31]3 years ago
8 0
This reaction is unbalanced. If you balance it, it would be:


2Al + 6HCl --> 3H2 + 2AlCl3

That way there are two aluminums on each side, six hydrogens on each side, and six chlorines on each side.

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A 0.11 N m torque is applied to a fan that was initially at rest. The fan has moment of inertia of 0.034 kg m2. Determine the ki
Mars2501 [29]

Answer:

2.72*10-3 Joules

Explanation:

From Newton's second law of motion

F=ma

\tau=I*\alpha

given

\tau= 0.11Nm\\\\I=0.034kgm^2\\\\t= 8s\\\\\alpha=?

\alpha =0.11/0.034\\\\\alpha=3.23 rad/s^2

the angular velocity is

\omega = \alpha/t\\\\\omega =3.23/8\\\\\omega =0.4 rad/s

KE= 1/2*I* \omega^2

KE=1/2*0.034*0.4^2\\\\KE=1/2*0.034*0.16\\\\KE=0.00272\\\\KE=2.72*10^-3J

4 0
3 years ago
You are trying to overhear a most interesting conversation, but from your distance of 10.0 m , it sounds like only an average wh
Alexandra [31]

Answer:

r₂=0.1 m

Explanation:

Given that

r₁= 10 m  , β₁ = 20 dB

At r₂ ,β₂= 60 dB

As we know that intensity level of sound given as

\beta =10\ log\dfrac{I}{10^{-12}}

\beta _1=10\ log\dfrac{I_1}{10^{-12}}

20=10\ log\dfrac{I_1}{10^{-12}}

10² x 10⁻¹² = I₁

I₁=10⁻¹⁰ W/m²

\beta _2=10\ log\dfrac{I_2}{10^{-12}}

60=10\ log\dfrac{I_1}{10^{-12}}

10⁶ x  10⁻¹² = I₂

I₂ = 10⁻⁶ W/m²

I₁=10⁻¹⁰ W/m²

P = I A

P=Power ,I =Intensity  ,A=Area

\dfrac{I_1}{I_2}=\dfrac{r^2_2}{r^2_1}

\dfrac{10^{-10}}{10^{-6}}=\dfrac{r^2_2}{10^2}

r₂=0.1 m

4 0
3 years ago
A wave has a wavelength of 50 m and a speed of 200 m/s. What is the frequency of this wave in hertz?
Kobotan [32]

Answer:

freq= 200/50= 4 Hz

.............

3 0
3 years ago
What causes the pressure to increase as you go deeper below the crust?
Masteriza [31]

The deeper you go, the more rock must be supported so the more force is required and the pressure goes up.

8 0
3 years ago
Read 2 more answers
You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn f
Verdich [7]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. We will calculate the initial velocity of the object, and from it, we will calculate the final position. With the considerations made in the statement we will obtain the total height. Initial velocity of the acorn,

u = 0m/s

Also, it is given that the acorn takes 0.201s to pass the length of the meter stick.

s = ut+\frac{1}{2} at^2

Replacing,

1 = u(0.141)+ \frac{1}{2} (9.8)(0.141)^2

u =6.4013m/s

The height of the acorn above the meter stick can be calculated as,

v^2 = u^2 +2gh

h = \frac{v^2-u^2}{2g}

h = \frac{6.4013^2-0^2}{2(9.8)}

h = 2.0906m

Also the top of the meter stick is 1.87m above the ground hence the height of the acorn above the ground is

h = 2.0906+1.87

h = 3.9606m

4 0
3 years ago
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