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Allushta [10]
3 years ago
14

...please help me :,)

Physics
1 answer:
natka813 [3]3 years ago
4 0

it doesnt show a picture or anything for me soorryy

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6
trapecia [35]

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.

5 0
3 years ago
An object is placed 100 cm in front of a diverging lens of focal length -25cm. A converging lens of focal length 33 1/3 cm is pl
bagirrra123 [75]
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>
8 0
4 years ago
Can someone pls help me with this?
alexira [117]
C because they are both going in a constant speed
6 0
3 years ago
Two very small spheres are initially neutral and separated by a distance of 0.30 m. Suppose that 2.50 1013 electrons are removed
quester [9]

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

4 0
4 years ago
1. What is the value of the acceleration that the car experiences? 2. What is the value of the change in velocity that the car e
tankabanditka [31]

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 m/s^{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 = -5m/s^{2}

6 0
3 years ago
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