What is the minimum number of points necessary to determine a straight line?

Can at transform plate boundaries, two plate move toward each other true or false

nekit [7.7K]

the answer to the question is false

7 0

2 years ago

The velocity as a function of time of a moving particle is given by v = α+ βt2 , where α and β are constants and t is time in s.

marin [14]

The acceleration of the particle at 3s is [tex]a = 6 \beta [/tex]

$v = \alpha + \beta {t}^{2}$

<h3>How to calculate acceleration </h3>

$a = \frac{dv}{dt} \\ v = \alpha + \beta {t}^{2} \\ \frac{dv}{dt} = 0 + 2 \beta t \\$

if Time is given as 3s

therefore, Acceleration is

$a = 2 \beta t \\ a = 2 \beta 3 \\ a = 2 \times \beta \times 3 \\ a = 6 \beta$

Acceleration is

$a = 6 \beta$

Read more about velocity:

brainly.com/question/19365526

6 0

2 years ago

A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive

Ivahew [28]

Answer:

a

The x- and y-components of the total force exerted is

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

b

The magnitude of the force is

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction of the force is

$\theta =327.43 ^o$ Clockwise from x-axis

Explanation:

From the question we are told that

The magnitude of the first charge is $q_1 = +5.00nC = 5.00*10^{-9}C$

The magnitude of the second charge is $q_2 = -2.00nC = -2.00*10^{-9}C$

The position of the second charge from the first one is $d_{12} = 4.00i \ cm = \frac{4.00i}{100} = 4.00i *10^{-2} m$

The magnitude of the third charge is $q_3 = +6.00nC = 6.00*10^{-9}C$

The position of the third charge from the first one is $\= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} = (4i + 3j) *10^{-2}m$

$|d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m$

$|d_{31}| =5 *10^{-2} m$

The position of the third charge from the second one is

$\= d_{32} = 3j cm = 3j *10^{-2}m$

$|d_{32}| =(\sqrt{ 3^2}) *10^{-2} m$

$|d_{32}| =3 *10^{-2} m$

The force acting on the third charge due to the first and second charge is mathematically represented as

$F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}$

Substituting values

$F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2} \\ \ + \ \ \ \ \ \ \ \ \ \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2}$

$F_{31 +32} = 2.16 *10^{-5} (4i + 3j) - 12*10^{-5} j$

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of $F_{31 +32}$ is mathematically evaluated as

$|F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}$

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction is obtained as

$tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}$

$\theta = tan ^{-1} [-0.63889]$

$\theta = - 32.57 ^o$

$\theta = 360 - 32.57$

$\theta =327.43 ^o$

5 0

3 years ago

Pam, wearing a rocket pack, stands on frictionless ice. She has a mass of 49 kg. The rocket supplies a constant force for 22.0 m

ss7ja [257]

Answer:

Magnitude of the force is 4350N

Explanation:

As the woman accelerates at a distance of 22 m to go from rest to 62.5 m / s, we can use the kinematics to find the acceleration

v² = v₀² + 2 a x

v₀ = 0

a = v² / 2x

a = 62.5²/(2 × 22)

a = 88.78m/s²

the time you need to get this speed

v = v₀ + a t

t = v / a

t = 62.5 / 88.78

t = 0.704s

Let's caculate the magnitude of the force

F = ma

= 49 × 88.78

= 4350.22

≅ 4350N

Magnitude of the force is 4350N

t = 1,025 s

a = 55.43 m / s²

4 0

2 years ago

What is it called when something dissolves into something else

Evgen [1.6K]

Answer:

The correct answer is dissolving and the end product is a solution.

Explanation:

Solubility is an property of a substance to get dissolved into another. When a substance, termed as solute, dissolves in another substance, termed as solvent, we get a solution.

Larger molecules of the solute breaks because of weak covalent bonds between their molecules and pairs up with the smaller molecules of the solvent and thus dissolves in it. Not all elements can serve solutes or solvents as some have strong forces that bind them together. For example sugar is a solute in water but oil is not soluble in water.

7 0

3 years ago