Explanation:
The balanced chemical equation is given as:
2H₂O₂ → O₂ + 2H₂O
А There are two peroxide molecules on the reactant side.
The statement is correct because the left hand side is the reactant side. We can read the reaction as;
2 molecules of hydrogen peroxide decomposed to given 1 molecule of oxygen gas and 2 molecules of water .
D Mass must be conserved so there are an equal number of molecules in the reactants and products.
E Mass must be conserved so there are an equal number of atoms in the reactants and products.
Since the chemical equation is balanced, choices D and E are correct. The reaction will have equal number of molecules and atoms on both sides of the expression.
We use 1/o + 1/i = 1/f where o is the distance of the object, i as distance of the image and f is the focal length.
Substituting, <span>1/ 100 + 1 / i = - 1 /25 </span>
<span>i = - 20 cm </span>
<span>For the case of the problem,</span>
<span>o = (20 + 30) = 50 cm </span>
<span>f = 33.33. </span>Using 1<span> / i + 1 / o = 1/f , </span><span> </span><span>i = 100 cm </span>
<span>M = magnification = - i / o </span>
<span>m1 = -(-20)/100 = 20/100 = 0.2 </span>
<span>m2 = -100/50 = -2 </span>
<span>M = m1*m2 = -2 x 0.2 = -0.4.</span>
C because they are both going in a constant speed
Answer:
F = - 1,598 10⁻³ N
Explanation:
Electic strength is given by Coulomb's law
F = k q₁ q₂ / r²
Where k is the Coulomb constant that is worth 8.99 10⁸ N m²/C², q₁ and q₂ are the charges and r is the distance that separates the electric charges
In this case the charge of the two spheres is the same and of a different sign since when you remove the charge of a sphere that was initially neutral, it is left with that charge removed but of the opposite sign
q₁ = q₂ = 2.50 10¹³ electrons = 2.50 10¹³ 1.6 10⁻¹⁹
q₀ = 4.0 10⁻⁶ C
Let's calculate
F = - 8.99 10⁸ (4.0 10⁻⁶)² / 0.30²
F = - 1,598 10⁻³ N
Answer:
All the answers are solved and explained below.
Explanation:
Note: This questions lacks the initial and most necessary data to answer these following questions. I have found a related question. I will be considering that question to carry out the answers.
Question: A car with a mass of 1000 kg is at rest at a spotlight. when the light turns green, it is pushed by a net force of 2000 N for 10 s. (This was the information missing in this question).
Data Given:
m = 1000 kg
F = 2000N
t = 10s
Q1 Solution:
Acceleration = a = ?
F = ma
a = F/m
a = 2000/ 1000
a = 2 
Q2: Solution:
Change in velocity = Δv = ?
acceleration = change in velocity / time
a = Δv/t
Δv = axt
Δv = 2 x 10
Δv = 20 m/s
Q3: Solution:
Impulse = I = ?
Impulse = Force x time
I = 2000 x 10
I = 20000 Ns
Q4: Solution:
Change in Momentum = Δp = ?
Δp = mΔv
Δp = 1000 x 20
Δp = 20000 Kgm/s
Q5: Solution:
Final velocity of the car at the end of 10 seconds = vf = ?
Δp = m x Δv
Δp = m x (vf-vi)
Δp = 1000 x (vf - 0 )
20000 = 1000 x vf
vf = 20000/1000
vf = 20 m/s
Q6: Solution:
Change in momentum the car experiences as it continues at this velocity?
Δp = ?
Δp = mΔv
Δp = m x (0)
Δp = 0
Q7: Solution:
Impulse = Change in momentum
Impulse = Δp
Implulse = 0
Q8: Solution:
Change in momentum = Δp = mΔv
Δp = m(vf-vi)
Δp = 1000 x (0-20)
Δp = -20000 kgm/s
Q9: Solution:
Impulse = Δp
Impulse = -20000 Ns
Q10: Solution:
Impulse = ?
Impulse = F x t
F = impulse/t
F = -20000/4s
F = -5000 N
Q11: Solution:
F = ma
a = ?
a = F/m
a = -5000/1000
a = -5
