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svet-max [94.6K]
3 years ago
12

Which statement is true about the difference between calcium carbonate and calcium oxide?

Physics
1 answer:
Kay [80]3 years ago
7 0
Kalsium Karbonat is true


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A race car accelerates from 0 m/s to 30.0 m/s with a displacement of
worty [1.4K]

Answer:

4. 10.0 m/s²

Explanation:

I) if initial velocity is 'v₀', the final velocity is 'v', the accelaration is 'a', the distance is 'L' and elapsed time if 't', then:

1. \ a=\frac{v-v_0}{t};

2. \ L=\frac{at^2}{2}.

II) using these two equations after substitution v₀=0; v=30 and L=45:

\left \{{{45 =\frac{at^2}{2}} \atop {a=\frac{30-0}{t} }} \right.

\left \{ {{at^2=90} \atop {at=30}} \right. \  \ \left \{ {{a=10} \atop {t=3}} \right. => \ a=10\frac{m}{s^2}

6 0
3 years ago
A kite is 100m above the ground. if there are 200m of string out what is the angle between the string and the horizontal? (assum
jeka57 [31]

Answer :  The angle between the string and the horizontal is 30 degrees

Explanation:  Imagine this a a triangle where the length of the string (200m) is the hypotenuse and the height of the kite is the opposite side (100m) .

Let the angle between the string and the horizontal be theta.

Now  sin (Theta) = opposite side/hypotenuse

                             =  100/200 = 1/2

Therefore Theta = Sin ⁻¹ ( 1/2 )

Theta = 30 degrees

4 0
3 years ago
The center of the Hubble space telescope is 6940 km from Earth’s center. If the gravitational force between Earth and the telesc
Law Incorporation [45]
The gravitational force between two objects is given by:
F=G \frac{m_1 m_2}{r^2}
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is the separation between the two objects

The distance of the telescope from the Earth's center is r=6940 km=6.94 \cdot 10^6 m, the gravitational force is F=9.21 \cdot 10^4 N and the mass of the Earth is m_1=5.98 \cdot 10^{24} kg, therefore we can rearrange the previous equation to find m2, the mass of the telescope:
m_2 =  \frac{Fr^2}{Gm_1}= \frac{(9.21 \cdot 10^4 N)(6.94\cdot 10^6)^2}{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})} =11121 kg
6 0
3 years ago
Read 2 more answers
In what regions of the electromagnetic spectrum is the atmosphere transparent enough to allow observations from the ground?
MissTica

Answer:

Visible Light and Radio waves

Explanation:

The earth's atmosphere is transparent to a few windows in the electromagnetic spectrum. it is completely transparent to allow observation from the ground in visible light rang 380 to 740 nano meters. Also in the range of radio wave as communication are done from space to ground in the form of radio waves.

it is Partially transparent to Microwave and infrared range.

3 0
3 years ago
8. Chemical weathering occurs more rapidly in what kind of climates?
Lady bird [3.3K]

I think it's b..................

5 0
3 years ago
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