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Lubov Fominskaja [6]
3 years ago
14

Which of the following equations is balanced correctly? A. 3 H2O → H2 + 3 O2 B. Cl2 + 2 KBr → KCl + Br2 C. 2 C2H2 + 5 O2 → 4 CO2

+ 2 H2O D. 2 C3H3 + O2 → 2 CO2 + H2O
Physics
2 answers:
posledela3 years ago
8 0
A. the carbons are unbalanced B. the hydrogens are unbalanced. D. the chlorines are unbalanced. That leaves C. to be correctly balanced.
slega [8]3 years ago
7 0
A. the carbons are unbalanced B. the hydrogens are unbalanced. D. the chlorines are unbalanced.
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An archer uses a bow to fire two similar arrows with the same string force. One
Sveta_85 [38]

Answer:

answers the correct one is C

Explanation:

For this exercise we must use the projectile launch ratios

the expressions for the initial velocities are

           sin θ =v_{oy} / vo

           cos θ = v₀ₓ / vo

           v_{oy} = v₀ sin θ

           v₀ₓ = v₀ cos θ

Range and flight time are requested

the expression for the scope is

          R = \frac{ v_{o}^2 \  sin  2 \theta}{g}

           

We calculate for each angle

θ = 45º

          R₄₅ = \frac{ v_{o}^2 \  sin  90}{g}

          R₄₅ = v₀² / g

 

θ = 60º

          R₆₀ = \frac{ v_{o}^2 \  sin  120}{g}

          sin  120 = sin 60

          R₆₀60 = sin  60  R₄₅

as the sine function has values ​​between 0 and 1, the range for this angle is less

Flight time is twice the time it takes to reach maximum altitude

           v_y = v_{oy} - gt

at the point of maximum height there is no vertical velocity vy = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

θ=45º angle

           t₄₅ = sin 45  v₀/g

           t₄₅ = 0.707 v₀/g

θ=60º angle

          t₆₀ = sin 60 v₀/ g

          t₆₀ = 0.86   v₀/g

         

 

the answer for this part is

           R₄₅ > R₆₀

            t₄₅ < t₆₀

when reviewing the different answers the correct one is C

6 0
2 years ago
If you are pushing a box toward your friend with a force of 20 N, and your friend is pushing the box toward you with a force of
yan [13]

Answer : the box will move backwards with a force 10 N

Explanation :

F1 = 20 N

F2 = 30 N

F = F2 - F1 = 30 - 20 = 10 N

7 0
2 years ago
Which change will always result in an increase in the gravitational force between two objects?
lora16 [44]

The gravitational force of two objects, by definition, is given by:

F = G * (\frac{m1m2}{d ^ 2})

Where,

G: gravitational constant

m1: mass of object number 1.

m2: mass of object number 2.

d: distance between both objects.

Therefore, according to the given equation, a change that always results in an increase in gravitational force is:

Increase in the mass of the objects and decrease in the distance between them.

Answer:

A change that will always result in an increase in the gravitational force between two objects is:

Increase in the mass of the objects and decrease in the distance between them.

8 0
3 years ago
Read 2 more answers
(a) You wish to determine the height of the smokestack of a local coal burning power plant. You convince a member of the mainten
kicyunya [14]

Answer:

a. 86.80 m

b. i. The mass of the bob

ii. The length of the pendulum

Explanation:

a. Determine the height of the smokestack.

Using T = 2π√(L/g) where T = period of pendulum = 18.7 s, L = length of pendulum = height of smokestack and g = acceleration due to gravity = 9.8 m/s².

So, making L subject of the formula, we have

T = 2π√(L/g)

T/2π = √(L/g)

squaring both sides, we have

(T/2π)² = L/g

L = (T/2π)²g

Substituting the values of the variables into the equation, we have

L = (T/2π)²g

L = (18.7 s/2π)²(9.8 m/s²)

L = (2.976 s)²(9.8 m/s²)

L = 8.857 s² × 9.8 m/s²

L = 86.796 m

L ≅ 86.80 m

b. What factors influence the period of a simple pendulum

The factors that influence the period of a simple pendulum are

i. The mass of the bob

ii. The length of the pendulum

5 0
3 years ago
The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5
KiRa [710]

Answer:

0.68 s

Explanation:

We are given that

Initial velocity of box=u=13m/s

Final velocity of box=v=11.5 m/s

Distance=d=8.5 m

We have to find the time taken by box to slow by this amount.

We know that

v^2-u^2=2as

Substitute the values

(11.5)^2-(13)^2=2a(8.5)

132.25-169=17a

-36.75=17a

a=\frac{-36.75}{17}=-2.2m/s^2

We know that

Acceleration=a=\frac{v-u}{t}

Substitute the values

-2.2=\frac{11.5-13}{t}

-2.2=\frac{-1.5}{t}

t=\frac{1.5}{2.2}=0.68 s

Hence, the time taken by box to slow by this amount=0.68 s

8 0
3 years ago
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