All categories

3 years ago

Why is it important to balance a chemical equation?

Physics

2 answers:

Thepotemich [5.8K]3 years ago

8 0

Hi there, the correct answer is D to obey the law conservation of mass, I took the quiz and this is the answer

Ostrovityanka [42]3 years ago

6 0

D I took the quiz and I'm pretty sure you will get it correct

You might be interested in

What is the power of a 12 V, 50 A light?

lora16 [44]

Answer:

42Amp

Dad's house is a message from

5 0

3 years ago

A diverging lens with a focal length of 14 cm is placed 12 cm to the right of a converging lens with a focal length of 21 cm. An

IRINA_888 [86]

Answer:

-0.233 m left of diverging lens and ( 0.12 - 0.233 ) = -113 m left of conversing

and

0.023 m right of diverging lens

Explanation:

given data

focal length f2 = 14 cm = -0.14 m

Separation s = 12 cm = 0.12 m

focal length f1 = 21 cm = 0.21 m

distance u1 = 38 cm

to find out

final image be located and Where will the image

solution

we find find image location i.e v2

so by lens formula v1 is

1/f = 1/u + 1/v ...............1

v1 = 1/(1/f1 - 1/u1)

v1 = 1/( 1/0.21 - 1/0.38)

v1 = 0.47 m

and

u2 = s - v1

u2 = 0.12 - 0.47

u2 = -0.35

so from equation 1

v2 = 1/(1/f2 - 1/u2)

v2 = 1/(-1/0.14 + 1/0.35)

v2 = -0.233 m

so -0.233 m left of diverging lens and ( 0.12 - 0.233 ) = -113 m left of conversing

and

for Separation s = 45 cm = 0.45 m

v1 = 1/(1/f1 - 1/u1)

v1 =0.47 m

and

u2 = s - v1

u2 = 0.45 - 0.47 =- 0.02 m

v2 = 1/(1/f2 - 1/u2)

v2 = 1/(-1/0.14 + 1/0.02)

v2 = 0.023

so here 0.023 m right of diverging lens

6 0

3 years ago

Two beams of coherent light travel different paths, arriving at point P. If the maximum destructive interference is to occur at

Nimfa-mama [501]

Answer:

The path difference between the two waves should be one-half of a wavelength

Explanation:

When two beams of coherent light travel different paths, arriving at point P. If the maximum destructive interference is to occur at point P , then the condition for it is that the path difference of two beams must be odd multiple of half wavelength. Symbolically

path difference = ( 2n+1 ) λ / 2

So path difference may be λ/2 , 3λ/ 2, 5λ/ 2 etc .

Hence right option is

The path difference between the two waves should be one-half of a wavelength.

5 0

3 years ago

PPPLLLLSSS HELP ME Identify the following as visible light, radio, ultraviolet, gamma, micro, infrared waves this is for the sec

Lorico [155]

I’m trying to get things expanded graph explanation sorry

6 0

3 years ago

A stone is thrown horizontally from 2.4m above the ground at 35m/s. The wall is 14m away and 1m high.At what height the stone wi

KIM [24]

The stone reaches the wall at a height of 1.62 m.

The stone lands at a point 24.5 m from the point of projection.

The stone is projected horizontally with a velocity u at a height h from the ground. The wall is located at a distance x from the point of projection. The stone takes a time t to reach the wall and in the same time the stone falls a vertical distance y.

The horizontal distance x is traveled with a constant velocity u.

$x=ut$

Calculate the time taken t.

$t=\frac{x}{u} \\ =\frac{14m}{35 m/s} \\ =0.40s$

The stone's initial vertical velocity is zero. It falls through a distance y in the time t under the action of acceleration due to gravity g.

$y=\frac{1}{2} gt^2\\ \frac{1}{2} (9.81m/s^2)(0.40s)^2\\ =0.784m$

The height h₁ of the stone above the ground when it reaches the wall is given by,

$h_1=h-y\\ =(2.4m)-(0.784m)\\ =1.616m=1.62m$

When the stone reaches the wall, its height from the ground is 1.62m.

The stone thus crosses over the wall, since the height of the wall is 1 m. It reaches the ground at a distance R from the point of projection. If the time taken by the stone to reach the ground is t₁, then,

$h=\frac{1}{2} gt_1^2$

Calculate the time taken by the stone to reach the ground.

$t_1=\sqrt{\frac{2h}{g} } \\=\sqrt{\frac{2(2.4m)}{9.81m/s^2} } \\ =0.699 s$

The horizontal distance traveled by the stone is given by,

$R=ut_1 \\ =(35m/s)(0.699s)\\ =24.5m$

The stone lands at point 24.5 m from the point of projection and 10.5 m from the wall.

3 0

3 years ago