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3 years ago

Given: Q1 = 3.0 × 10-5 C

Physics

2 answers:

Igoryamba3 years ago

8 0

F=KQ1Q2/R2
F=9 x 10^9 x 3 x 10^-5 x 4 x 10^-5 / 9
F=12 x 10^-1
F=1.2 N

Mariulka [41]3 years ago

4 0

$F=k\dfrac{Q_1Q_2}{r^2}= \\ \\ =9\cdot 10^9\cdot \dfrac{3\cdot 10^{-5}\cdot 4\cdot 10^{-5}}{9}=...$

You just do the calculations.

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A coin placed on the cover of a book just begins to move when the cover makes an angle of 38 degrees with the horizontal. What i

RSB [31]

Answer:

m g sin theta = force of object along incline due to gravity

N μ = frictional of incline on object where N is the normal force

N = m g cos theta force perpendicular to incline

m g sin theta = N μ = μ m g cos theta

μ = tan theta = tan 38 = .78

6 0

2 years ago

Help please!!!

TiliK225 [7]

Answer:

Cu 8.92

Explanation:

The formula for density is mass/volume. If you were to divide 62.44 by 7, you would get 8.92. Since copper is the only metal in this table that has a density of 8.92, that is the answer.

7 0

4 years ago

Stan is driving north on his scooter at 8m/s, accelerates 11m/s (North) in 4s, drives a constant velocity for the next 15s, and

kow [346]

A) Acceleration: $a_1 = 0.75 m/s^2, a_2 = 0, a_3 = -1.57 m/s^2$

B) The total displacement is 209.5 m north

C) The average velocity is 8.06 m/s north

Explanation:

Acceleration is defined as:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to v

Here we have:

- In the first segment,

u = 8 m/s north

v = 11 m/s north

t = 4 s

So the acceleration is

$a_1 = \frac{11-8}{4}=0.75 m/s^2$ (north)

- In the second segment, Stan drives at a constant velocity: so the final velocity is equal to the initial velocity,

u = v

Therefore, the acceleration is zero: $a_2 = 0$

- In the third segment,

u = 11 m/s (north)

v = 0 (he comes to a stop)

t = 7 s

So the acceleration is

$a=\frac{0-11}{7}=-1.57 m/s^2$

And the negative sign means the acceleration is south, opposite to the direction of motion.

In a uniformly accelerated motion, the displacement can be calculated as:

$s=ut+\frac{1}{2}at^2$

where

u is the initial velocity

a is the acceleration

t is the time

- For the first segment, we have

$u = 0\\a = 0.75 m/s^2\\t=4 s$

So the displacement is

$s_1 = 0+\frac{1}{2}(0.75)(4)^2=6 m$

- For the second segment, we have

$u = 11 m/s\\a = 0\\t=15 s$

So the displacement is

$s_2 = (11)(15)+0=165 m$

- For the third segment, we have

$u = 11\\a = -1.57 m/s^2\\t=7 s$

So the displacement is

$s_3 = (11)(7)+\frac{1}{2}(-1.57)(7)^2=38.5 m$

So the total displacement is:

s = 6 m + 165 m + 38.5 m = 209.5 m

In the north direction (positive direction)

The average velocity is given by:

$v=\frac{d}{t}$

where

d is the total displacement

t is the total time

Here we have:

d = 209.5 m

t = 26 s

Therefore, the average velocity is

$v=\frac{209.5}{26}=8.06 m/s$ (north)

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

7 0

4 years ago

The force you have to overcome to start an object moving is _____

MAVERICK [17]

The forece you have to overcome to start an object moving is called Inertia.

3 0

4 years ago

A trap-jaw ant snaps its mandibles shut at very high speed, a good trait for catching small prey. But an ant can also slam its m

irakobra [83]

Answer:

v = 0.509 m/s

Explanation:

Given that,

Mass of the ant, $m = 12\ mg = 1.2\times 10^{-5}\ kg$

Force exerted, $F = 47\ mN = 0.047\ N$

Time, $t=0.13\ ms=0.00013\ s$

To find,

Speed with which it leave the ground.

Solution,

The change in momentum is equal to the product of force and time as :

$F\times t=m(v-u)$

Since, u = 0

$F\times t=mv$

$v=\dfrac{Ft}{m}$

$v=\dfrac{0.047\ N\times 0.00013\ s}{1.2\times 10^{-5}\ kg}$

v = 0.509 m/s

So, the speed with which it will leave the ground is 0.509 m/s.

3 0

3 years ago