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Sati [7]
3 years ago
8

How does the law of conservation of mass apply to chemical changes?

Physics
2 answers:
Paha777 [63]3 years ago
8 0
Matter cannot be created or destroyed, all of the atoms that are present in the reactants must show up as products
Zielflug [23.3K]3 years ago
5 0

Answer:

The Law of Conservation of Mass states that matter cannot be created nor destroyed. It can only be transformed from one state or substance to another. This means that this law applies to both physical and chemical changes. In physical changes, the substance before and after the phase change still has the same mass.

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All ball is thrown up with a vertical velocity of 54 m/s and a horizontal velocity of 39 m/s. Calculate how many seconds it will
VikaD [51]

5.5 s

Explanation:

The time it takes for the ball to reach its maximum height can be calculated using

v_y = v_{0y} - gt \Rightarrow t = \dfrac{v_{0y}}{g}

since v_y = 0 at the top of its trajectory. Plugging in the numbers,

t = \dfrac{(54\:\text{m/s})}{(9.8\:\text{m/s}^2)} = 5.5\:\text{s}

8 0
3 years ago
Which has greater kinetic energy, a car traveling at 30 km/hr or a car of half the mass traveling at 60 km/hr?
Mumz [18]
Half mass car because it's traveling faster

5 0
3 years ago
Read 2 more answers
A scientist is designing a device that will mimic Earth's atmosphere by
mafiozo [28]

Answer:

1. Ultraviolet light (UV)

2. X-rays

3. Gamma-rays

Explanation:

Though there are different types of energy or electromagnetic waves with varying wavelengths, including the likes of Gamma X-rays, ultraviolet light, visible light, infrared radiation, and microwave radiation.

What is more certain is that the atmosphere blocked the high-energy waves from getting to the earth surface or biosphere such as Ultraviolet light (UV), X-rays and Gamma-rays

8 0
3 years ago
A position vector with magnitude 10 m points to the right and up. its x-component is 6.0 m. part a what is the value of its y-co
lidiya [134]

The position vector can be transcribed as:

A<span> = 6 i + y j                           </span>

i <span>points in the x-direction and j points in the y-direction.</span>

The magnitude of the vector is its dot product with itself:

<span>|A|2 = A·A</span>

<span>102  = (6 i + y j)•(6 i+ y j)            Note that i•j = 0, and  i•i  = j•j = 1 </span>

<span>100  = 36 + y2       </span>

<span>64    = y2</span>

<span>get the square root of 64 = 8</span>

<span>The vertical component of the vector is 8 cm.</span>

3 0
3 years ago
Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w
Tatiana [17]

Answer:

The average velocity is

266\frac{m}{s},274\frac{m}{s} and 117\frac{m}{s} respectively.

Explanation:

Let's start writing the vertical position equation :

L(t)=2t^{3}+t^{2}-5t+1

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

V_{avg}=\frac{Displacement}{Time} = Δx / Δt = \frac{x2-x1}{t2-t1}

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

t2-t1=8s-5s=3s

For the position variation we use the vertical position equation :

x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m

x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

\frac{798m}{3s}=266\frac{m}{s}

For the second time interval :

t1 = 4 s → t2 = 9 s

x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m

x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

\frac{1370m}{5s}=274\frac{m}{s}

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m

x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

\frac{702m}{6s}=117\frac{m}{s}

5 0
3 years ago
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