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algol [13]
3 years ago
9

Which process represents a chemical change

Physics
2 answers:
omeli [17]3 years ago
8 0
A process that represents a chemical change is burning a piece of paper. I am not sure if this is a process but I hope this helps you!
Ganezh [65]3 years ago
4 0
Melting of ice could be one
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List atleast to conditions that could cause the pressure inside a sealed container of gas to increase
hjlf
There are exactly three ways that could happen:

1).  The container was heated, and the gas inside it got warmer.

2).  Some part inside the container moved somehow, and made
      the inside volume smaller, so the gas got scrunched into a
      smaller space.

3).  Somebody pumped some more gas into the container, so
      a greater amount of gas had to live in the same space.     
8 0
3 years ago
What is the ratio of the intensities of an earthquake P wave passing through the Earth and detected at two points 14 km and 49 k
Molodets [167]

Answer:

\dfrac{I_1}{I_2}=12.25

Explanation:

r_1 = 14 km

r_2 = 49 km

Intensity of a wave is inversely proportional to distance

I\propto \dfrac{1}{r^2}

So,

\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{49^2}{14^2}\\\Rightarrow \dfrac{I_1}{I_2}=12.25

The ratio of the intensities is \dfrac{I_1}{I_2}=12.25

6 0
3 years ago
A body of mass 2 kg at O has an initial velocity of 3m/s along OE and it is subjected to a force of 4N perpendicular to OE the d
timofeeve [1]
Perpendicular acceleration:
F = ma
a = 4 / 2 = 2 m/s²

Perpendicular distance:

s = ut + 1/2 at²
s = 0 x 4 + 1/2 x 2 x 4²
s = 16 m

Horizontal distance:
s = ut
= 3 x 4
= 12 m

Total distance = √(12² + 16²)
= 20 m.
3 0
3 years ago
A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta
jeyben [28]

Answer:

V_0

Explanation:

Given that, the range covered by the sphere, M, when released by the robot from the height, H, with the horizontal speed V_0 is D as shown in the figure.

The initial velocity in the vertical direction is 0.

Let g be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. V_0 will remain constant throughout the projectile motion.

So, if the time of flight is t, then

D=V_0t\; \cdots (i)

Now, from the equation of motion

s=ut+\frac 1 2 at^2\;\cdots (ii)

Where s is the displacement is the direction of force, u is the initial velocity, a is the constant acceleration and t is time.

Here, s= -H, u=0, and a=-g (negative sign is for taking the sigh convention positive in +y direction as shown in the figure.)

So, from equation (ii),

-H=0\times t +\frac 1 2 (-g)t^2

\Rightarrow H=\frac 1 2 gt^2

\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)

Similarly, for the launched height 2H, the new time of flight, t', is

t'=\sqrt {\frac {4H}{g}}

From equation (iii), we have

\Rightarrow t'=\sqrt 2 t\;\cdots (iv)

Now, the spheres may be launched at speed V_0 or 2V_0.

Let, the distance covered in the x-direction be D_1 for V_0 and D_2 for 2V_0, we have

D_1=V_0t'

D_1=V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_1=\sqrt 2 D [from equation (i)]

\Rightarrow D_1=1.41 D (approximately)

This is in the 3 points range as given in the figure.

Similarly, D_2=2V_0t'

D_2=2V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_2=2\sqrt 2 D [from equation (i)]

\Rightarrow D_2=2.82 D (approximately)

This is out of range, so there is no point for 2V_0.

Hence, students must choose the speed V_0 to launch the sphere to get the maximum number of points.

7 0
3 years ago
Rays of light strike a bumpy road and are reflected.
ser-zykov [4K]

Answer:

last option is the correct one

6 0
3 years ago
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