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MA_775_DIABLO [31]

3 years ago

5

IF ANYONE WOULD HELP ME AS FAST AS POSSIBLE CAUSE I HAVE TO SUBMIT THIS AT 11:59 and I wanna finish early to study for my quiz s

o if anyone could help me fast I would give you 20 Points

2 answers:

harina [27]3 years ago

8 0

I think the answers are box #1 and #3. It is number one for a fact. However, it depends where the start is at #3.

8_murik_8 [283]3 years ago

6 0

A and D are true.

B and C aren't.

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PLS HELP 30 POINTS IF U HELP

USPshnik [31]

Answer:

its D so yeah.........

Explanation:

tell me if it's wrong

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2 years ago

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PLZ HELP!!!!!!! My favorite playground equipment is the swings. In 1-3 sentences explain a pattern of motion that u predict u'll

In-s [12.5K]

My favorite playground equipment is the swings.

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3 years ago

While playing Quidditch you throw the quaffle straight down to the person below you at

oee [108]

Answer:

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3 years ago

A 2kg Book is sitting on a table.a 10n Force is pulling to the right.a 3n Force is pulling to the left what is the net force act

salantis [7]

The net force = sum of all forces acting on the body

If we take left side as -ve and right side as +ve,
then,

The net force here would be equal to,
10N + (- 3N)
= 7N.

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3 years ago

A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.

iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

$F = \frac{kq_1q_2}{d^2}$

Here

k = Coulomb's Constant

$q_{1,2} =$ Charge of each object

d = Distance

Our values are given as,

$q_1 = 1 \mu C$

$q_2 = 6 \mu C$

d = 1 m

$k = 9*10^9 Nm^2/C^2$

a) The electric force on charge $q_2$ is

$F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}$

$F_{12} = 54 mN$

Force is positive i.e. repulsive

b) As the force exerted on $q_2$ will be equal to that act on $q_1$ ,

$F_{21} = F_{12}$

$F_{21} = 54 mN$

Force is positive i.e. repulsive

c) If $q_2 = -6 \mu C$ , a negative sign will be introduced into the expression above i.e.

$F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}$

$F_{12} = F_{21} = -54 mN$

Force is negative i.e. attractive

6 0

3 years ago