Physics, 14.09.2019 02:20 lalalalal5

You come across a mysterious pond of an unknown liquid during your normal work day (lucky you). being curious, you submerge an upright large stick into the liquid. you note that 1.5 m of the stick is submerged. although you are holding the stick out 21 cm in front of you, the bottom of the stick appears 31 cm away. what is the mysterious liquid’s index of refraction?

Answers: 1

Show answers

Answers

Answer from: Quest

what are the options.

explanation:

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

Answer from: Quest

What does the start look like

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

Other questions on the subject: Physics

Physics, 21.06.2019 20:40, denjayjr681

If an object with an initial temperature of 300 k increases its temperature by 1°c every minute, by how many degrees fahrenheit will its temperature have increased in 10 minutes? (a) 6°f (b) 10°f (c) 18°f (d) 30°f

Answers: 3

continue

Physics, 21.06.2019 21:50, mckinley2006

What is the length x of the side of the triangle below? (hint: use the cosine function)

Answers: 1

continue

Physics, 21.06.2019 23:30, aprilwilson3p8711m

Truck traveling west on the highway experiences an acceleration to the east. what could be the cause of this acceleration

Answers: 1

continue

Physics, 22.06.2019 08:40, Hazeleyes13

An isolated conducting spherical shell carries a positive charge. part a which statement (or statements) about the electric field and the electric potential inside and outside the spherical shell is correct? which statement (or statements) about the electric field and the electric potential inside and outside the spherical shell is correct? electric potential inside the shell is constant and outside the shell is changing as 1/r2 both the electric potential and the electric field does change with r inside and outside the spherical shell electric potential inside and outside the shell is constant, but not zero electric potential inside the shell is constant and outside the shell is equal to zero electric field inside and outside the shell is constant (does not change with the position r), but is not equal to zero electric field inside and outside the shell is changing as 1/r (where r is the distance from the center of the sphere) electric field inside is equal to zero and outside the shell is constant, but not zero electric potential inside the shell is constant and outside the shell is changing as 1/r electric field inside and outside the shell is changing as 1/r2 electric field inside is equal to zero and outside the shell is changing as 1/r2 electric field inside and outside the shell is zero electric field inside is constant and outside the shell is changing as 1/r

Answers: 3

continue

You know the right answer?

You come across a mysterious pond of an unknown liquid during your normal work day (lucky you). bein...