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Lerok [7]
3 years ago
11

How would the attractive force between two spheres change if the mass of one sphere was doubled?

Physics
1 answer:
Dominik [7]3 years ago
8 0
The force of attraction between two objects can be illustrated using Newton's Law of Universal Gravitation.
The relation between the different parameters is shown in the attached image.

Now, from the relation, we can deduce that the force between the two objects is directly proportional to the masses of the two objects.

This means that, if the mass of one object is doubled, then the force between the two objects will also be doubled.

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When y⁷ is multiplied by y⁹ the answer is?
Allushta [10]

Answer:

y^16

Explanation:

who need to add the exponents only

7 + 9 = 16

therefore, the answer is y^16

5 0
2 years ago
A jogger accelerates at a constant rate as she travels 5.0
VladimirAG [237]

Answer:

2.0s

Explanation:

...

3 0
3 years ago
Problema en la cual aplicaste velocidades, impulso, conservación del movimiento y de la energía.
Kazeer [188]
Huh huh what? ¿Can’t you translate?
6 0
3 years ago
Q:-What is speed????​
Margarita [4]

Explanation:

In everyday use and in kinematics, the speed of an object is the magnitude of the rate of change of its position with time or the magnitude of the change of its position per unit of time; it is thus a scalar quantity.

SI unit: m/s, m s−1

s=d/t

5 0
3 years ago
A first order reaction, A -> products, has a rate reaction of .00250 Ms-1 when [A] = . 484 M. (a) What is the rate constant,
tamaranim1 [39]

Answer: a)  The rate constant, k, for this reaction is 0.00516s^{-1}

b) No t_{\frac{1}{2}} does not depend on concentration.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

A\rightarrow products

Given: Order with respect to A = 1

Thus rate law is:

a) Rate=k[A]^1

k= rate constant

0.00250=k[0.484]^1

k=0.00516s^{-1}

The rate constant, k, for this reaction is 0.00516s^{-1}

b) Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  

t = age of sample

a = let initial amount of the reactant  

a - x = amount left after decay process  

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{2.303}{k}\log\frac{100}{50}

t_{\frac{1}{2}}=\frac{0.69}{k}

Thus t_{\frac{1}{2}} does not depend on concentration.

8 0
3 years ago
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