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seraphim [82]
3 years ago
12

Identify the number of atoms on the reactant side of the chemical equation.

Physics
2 answers:
klio [65]3 years ago
5 0
Hi, I believe this should belongs to the chemistry section, but nevertheless here is the solution :

Number of Fe atoms: (2) + (2) =4
Number of O atoms. : (3)
Number of H atoms: (3 X 2)=6
Total atoms: 4+3+6=13 atoms
seropon [69]3 years ago
3 0

Answer:

The Correct Answer is: 2,3,and 6

Explanation:

There are  

2

 Fe (iron) atoms.

There are  

3

 O (oxygen) atoms.

There are  

6

 H (hydrogen) atoms.

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A solid sphere of radius 40.0 cm has a total positive charge of 16.2 μC uniformly distributed throughout its volume. Calculate t
Jobisdone [24]

Answer:

(a) E=0  :   0 cm from the center of the sphere

(b) E= 227.8*10³ N/C   :    10.0 cm from the center of the sphere

(c)E= 911.25*10³ N/C    :    40.0 cm from the center of the sphere

(d)E= 411.84 * 10³ N/C  :    59.5 cm from the center of the sphere

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

E=\frac{K*Q}{r^{2} } : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

E=k*\frac{Q}{a^{3} } *r :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q:  Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at  0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1

E= 227.8*10³ N/C

(c) Magnitude of the electric field at 40.0 cm from the center of the sphere

r=a, We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }

E= 911.25*10³ N/C

(d) Magnitude of the electric field at 59.5 cm from the center of the sphere  

r>a , We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }

E= 411.84 * 10³ N/C

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Explanation:

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A 0.877-kg ball has an elastic, head-on collision with a second ball that is initially at rest. The second ball moves off with h
Art [367]

Answer: the mass of the second ball is 2.631 kg

Explanation:

Given that;

m1 = 0.877 kg

Initial velocity = V0

Initial momentum = m1 × V0

final velocity of m1 is u1, final velocity of m2 is u2 = v0/2

now final momentum = m1 × u1 + m2 × u2

using momentum conservation;

m1×V0 = m1×u1 + m2×v0/2

m1×(v0 - u1) = m2×V0/2 ----- let this be equation 1

Now, for elastic collision;

m1×v0²/2 = m1×u1²/2 + m2×(v0/2)²/2

m1×(v0² - u1²) = m2×(v0/2)² --------- let this be equation 2

now; equation 2 / equation 1

: V0 + u1 = v0/2

2V0 + 2u1 = V0

2u1 = V0 - 2V0

u1 = -V0/2

now we insert in equ 1

 m1×3V0/2= m2×V0/2

m1 × 3 = m2

m2 = 0.877 × 3

m2 = 2.631 kg

Therefore, the mass of the second ball is 2.631 kg

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