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2 years ago

8

Which of the following never cause a change in the motion of an object? A. Net forces B. Unbalanced forces O C. Balanced forces

O D. Contact forces

1 answer:

Karolina [17]2 years ago

5 0

Answer:

balanced force never changes its motion

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Two electrons with a charge of magnitude 1.6×10-19 C in an atom are separated by 1.5×10-10 m, the typical size of an atom. What

vesna_86 [32]

Answer:

$1.02\cdot 10^{-8} N$ , repulsive

Explanation:

The magnitude of the electric force between two charged particles is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges of the two particles

r is the separation between the two charges

The force is:

- repulsive if the two charges have same sign

- Attractive if the two charges have opposite signs

In this problem, we have two electrons, so:

$q_1=q_2=1.6\cdot 10^{-19}C$ is the magnitude of the two electrons

$r=1.5\cdot 10^{-10} m$ is their separation

Substituting into the formula, we find the electric force between them:

$F=(8.99\cdot 10^9)\frac{(1.6\cdot 10^{-19})^2}{(1.5\cdot 10^{-10})^2}=1.02\cdot 10^{-8} N$

And the force is repulsive, since the two electrons have same sign charge.

4 0

3 years ago

Please help and check all that apply and I will mark brainliest if it’s correct

Yuri [45]

A syncline is visable

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3 years ago

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sergejj [24]

Organ, integument, dermis

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3 years ago

The atomic number of a nucleus decreases during which nuclear reactions?

Goryan [66]

A related type of beta decay actually decreases the atomic number of the nucleus when a proton becomes a neutron. Due to charge conservation, this type of beta decay involves the release of a charged particle called a “positron” that looks and acts like an electron but has a positive charge.

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3 years ago

If two point masses 1kg & 4kg are seperated by a distance of 2m. Magnitude of gravitational force exerted by 1kg on 4kg is ?

Aliun [14]

Answer:

F = G Newtons

Explanation:

Given:

Mass of 1st body = $1\:kg$
Mass of 2nd body = $4\:kg$

To Find:

Magnitude of gravitational force

Solution:

Here, we have a formula

$F=\dfrac{G.M_{1}.M_{2}}{r^{2}}$

<u>Substituting the values</u>

$\implies\:\:F = \dfrac{G(1)(4)}{2^{2}}$

$\implies\:\:F = \dfrac{4G}{4}$

$\implies\:\:F = \dfrac{\cancel{4}G}{\cancel{4}}$

$\implies\:\:\red{F = G}$

Know More:

The applied formula for the above solution is

${\boxed{F_{G}=\dfrac{G.M_{1}.M_{2}}{r^{2}}}}$

where,

F $_{G}$ = Gravitational force
G = Gravitational constant
M $_{1}$ = mass of 1st body
M $_{2}$ = mass of 2nd body
r = distance between two bodies

6 0

3 years ago