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3 years ago

What is the only force acting on an object orbiting earth

Chemistry

2 answers:

ludmilkaskok [199]3 years ago

3 0

Gravity because the earth's gravity is what's causing it to orbit it

adell [148]3 years ago

3 0

Answer:

The correct answer is gravity,

Explanation:

Once the object leaves the atmosphere, it is placed in a certain position and begins to orbit the earth. At this point, the possibility of the object remaining in orbit due to gravity will be determined.

Gravity is the force of attraction that exerts all mass on another, and the more mass it has, the greater it will be.

That is why satellites are continuously attracted to the earth.

Have a nice day!

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For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow

lora16 [44]

Answer:

Rate constant of the reaction is $3.3\times 10^{-3} M^{-2} s^{-1}$ .

Explanation:

A + B + C → D + E

Let the balanced reaction be ;

aA + bB + cC → dD + eE

Expression of rate law of the reaction will be written as:

$R=k[A]^a[B]^b[C]^c$

Rate(R) of the reaction in trail 1 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.30 M$

$R=9.0\times 10^{-5} M/s$

$9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c$ ...[1]

Rate(R) of the reaction in trail 2 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.90 M$

$R=2.7\times 10^{-4} M/s$

$2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c$ ...[2]

Rate(R) of the reaction in trail 3 ,when :

$[A]=0.60 M,[B]=0.30 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c$ ...[3]

Rate(R) of the reaction in trail 4 ,when :

$[A]=0.60 M,[B]=0.60 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c$ ...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

$R=k[A]^2[B]^0[C]^1$

Rate constant of the reaction = k

$9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1$

$k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}$

$k=3.3\times 10^{-3} M^{-2} s^{-1}$

7 0

3 years ago

When might you receive a blood transfusion Help me pls

umka21 [38]

Explanation:

in the case of blood loss, you need blood from someone with your blood type or with universal donor type

5 0

3 years ago

The molar solubility of ag2s is 1.26 × 10-16 m in pure water. calculate the ksp for ag2s.

MAXImum [283]

Answer is: Ksp for silver sulfide is 8.00·10⁻⁴⁸.
Reaction of dissociation: Ag₂S(s) → 2Ag⁺(aq) + S²⁻(aq).
s(Ag₂S) = s(S²⁻) = 1.26·10⁻¹⁶ M.
s(Ag⁺) = 2s(Ag₂S) = 2.52·10⁻¹⁶ M; equilibrium concentration of silver cations.
Ksp = s(Ag⁺)² · s(S²⁻).
Ksp = (2.52·10⁻¹⁶ M)² · 1.26·10⁻¹⁶ M.
Ksp = 6.35·10⁻³² M² · 1.26·10⁻¹⁶ M.
Ksp = 8.00·10⁻⁴⁸ M³.

6 0

3 years ago

The octet rule states that atoms will gain, lose, or share valence electrons to get a full outer shell (orbital). Explain in you

bezimeni [28]

Answer:

In covalent bonding, the octet rule is important because sharing electrons gives both atoms a full valence shell. As a result, each atom can consider the shared electrons to be part of its own valence shell.

np :)

6 0

3 years ago

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Leto [7]

I would pick the first option in the third option

4 0

3 years ago