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3 years ago

The most basic unit of charge is the _____.

2 answers:

8 0

I believe that the answer to the question provided above is the most basic unit of charge is the <span>ampere (A).</span>
Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.

postnew [5]3 years ago

3 0

Answer:

Coulomb

Explanation:

The basic unit of charge is Coulomb.

The other units of charge are stat Coulomb, esu, etc.

1 Coulomb = 3 x 10^9 stat Coulomb

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Celina has a water sample that’s contaminated with salt and microorganisms. Which method should she use to purify the water?

Dmitry_Shevchenko [17]

Celina should perform the following method to get pure water:

Boil the water in a pan, as the water boils it changes into gas (water vapours) these vapours should be cooled down through the process of condensation which changes gas into liquid form, Through this process germs would be killed due to high temperature and the salt will remain in the pan as salt has higher boiling point than water,

7 0

3 years ago

Read 2 more answers

The amount of matter in an object is its _____.

Andrew [12]

The amount of matter in an object ismass....anything that occupies spaca and has weight is called matter.....

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3 years ago

2. At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a n

AleksandrR [38]

Answer:

Explanation:

Let the reference origin be location of ship B in the beginning. We can then create the equation of motion for ship A and ship B in term of time t (hour):

A = 12 - 12t

B = 9t

Since the 2 ship motions are perpendicular with each other, we can calculate the distance between 2 ships in term of t

$d = \sqrt{A^2 + B^2} = \sqrt{(12 - 12t)^2 + (9t)^2}$

For the ships to sight each other, distance must be 5 or smaller

$d \leq 5$

$\sqrt{(12 - 12t)^2 + (9t)^2} \leq 5$

$(12 - 12t)^2 + (9t)^2 \leq 25$

$144t^2 - 288t + 144 + 81t^2 - 25 \leq 0$

$225t^2 - 288t + 119 \leq 0$

$(15t)^2 - (2*15*9.6)t + 9.6^2 + 26.84 \leq 0$

$(15t^2 - 9.6)^2 + 26.84 \leq 0$

Since $(15t^2 - 9.6)^2 \geq 0$ then

$(15t^2 - 9.6)^2 + 26.84 > 0$

So our equation has no solution, the answer is no, the 2 ships never sight each other.

8 0

3 years ago

A particle moves along the x axis. It is initially at the position 0.180 m, moving with velocity 0.060 m/s and acceleration -0.3

shusha [124]

Answer:

$x_2 = -2.3356$

$v = -1.384 \ m/s$

Explanation:

From the question we are told that

The initial position of the particle is $x_1 = 0.180 \ m$

The initial velocity of the particle is $u = 0.060 \ m/s$

The acceleration is $a = -0.380 \ m/s^2$

The time duration is $t = 3.80 \ s$

Generally from kinematic equation

$v = u + at$

=> $v = 0.060 + (-0.380 * 3.80)$

=> $v = -1.384 \ m/s$

Generally from kinematic equation

$v^2 = u^2 + 2as$

Here s is the distance covered by the particle, so

$(-1.384)^2 = (0.060)^2 + 2(-0.380)* s$

=> $s = -2.5156 \ m$

Generally the final position of the particle is

$x_2 = x_1 + s$

=> $x_2 = 0.180 + (-2.5156)$

=> $x_2 = -2.3356$

6 0

3 years ago

A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The

lina2011 [118]

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s

Explanation:

1) We can find the speed of the object by conservation of energy:

$E_{i} = E_{f}$

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s$

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s$

3) When the object is at the equilibrium position, the speed of the object is:

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s$

I hope it helps you!

8 0

3 years ago