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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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Suppose the initial position of an object is zero, the starting velocity is 3 m/s and the final velocity was 10 m/s. The object
Nookie1986 [14]

Explanation:

We have,

The initial position of an object is zero.

The starting velocity is 3 m/s and the final velocity was 10 m/s.

The object moves with constant acceleration..

The area covered under the velocity-time graph gives displacement of the object. The correct option is "the area of the rectangle plus the area of the triangle under the line".

8 0
3 years ago
A cyclist rides at 6.20 m/s through a intersection. A stationary car begins to
Xelga [282]

Answer:

The width of the intersection is 20 meters

Explanation:

The speed with which the cyclist is riding, v₁ = 6.20 m/s

The rate at which the car starts to accelerate, a = 3.844 m/s²

The initial velocity of the car = The car is stationary at the start = 0 m/s

The time at which the cyclist and the car reach the other side of the intersection = The same time;

Let 't' represent the time at which the cyclist and the car both reach the other side of the intersection, we have;

The distance travelled by the cyclist = The distance traveled by the car

∴ v₁ × t = 1/2 × a × t²

Plugging in the values for 'v₁', and 'a' in the above equation, we get;

6.20 × t = 1/2 × 3.844 × t²

∴ 1.922·t² - 6.20·t = 0

∴ t·(1.922·t - 6.20) = 0

t = 0, or t = 6.20/1.922 = 100/31

The time at which the cyclist and the car both reach the other side of the intersection, t = 100/31 seconds

The with of the intersection, w = v₁ × t

∴ w = 6.20 × 100/31 = 100/5 = 20

The width of the intersection, w = 20 meters.

8 0
3 years ago
9. How much work is done when a 15kg box is lifted to a height of 2 meters?
torisob [31]

Answer: W = 294 J

Explanation: Solution:

Work is expressed as the product of force and the distance of the object.

W = Fd        where F = mg

W= Fd

   = mg d

   = 15 kg ( 9.8 m/s²) ( 2m )

   = 294 J

5 0
3 years ago
a football player kicks a ball with a mass of 0.42kg. The average acceleration of the football was 14.8 m/s2. How much force did
Zarrin [17]

Answer:

6.216 N

Explanation:

As for Newton's second law of motion

F=ma

where F= the acting force

           m=subjected mass

           a= the acceleration

applying F=ma to the football

F=m*a

 =0.42*14.8

 =6.216 N

6.216 N of a force is supplied to the ball

5 0
3 years ago
Material speed of light
Cloud [144]
The question is poor. Light doesn't refract on its way THROUGH anything. It refracts at the boundary BETWEEN two different media. The effect is greatest where the ratio of the speeds of light in the two media is greatest. On your list, that would be at the boundary between air or space and glass.
3 0
3 years ago
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