3 years ago

What is the equation of the line that is perpendicular to y= -4x +5 and passes through the point (4,-3)?

Physics

1 answer:

Ostrovityanka [42]3 years ago

5 0

Y=4x-19

Perpendicular because it has an opposite reciprocal slope

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Constants Part A wo charged dust particles exert a force of 7.5x102 N n each other What will be the force if they are moved so t

alexandr1967 [171]

Answer:

New force, $F'=48\times 10^3\ N$

Explanation:

It is given that,

Force acting between two charged particles, $F=7.5\times 10^2\ N$

We need to find the force if they are moved so they are only one-eighth as far apart.

The force between two charged particles separated at a distance of r is given by :

$F=k\dfrac{q_1q_2}{r^2}$ ............(1)

If the charges are one-eighth as far apart then, r' =(1/8)r and new force is given by :

$F'=k\dfrac{q_1q_2}{(\dfrac{r}{8})^2}$ ..........(2)

Dividing equation (1) and (2) :

$\dfrac{F}{F'}=\dfrac{1}{64}$

$F'=7.5\times 10^2\ N\times 64$

F' = 48000 N

$F'=48\times 10^3\ N$

Hence, this is the required solution.

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Hydrogen H

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Froghopper insects have a typical mass of around 11.3 mg and can jump to a height of 58.8 cm. The takeoff velocity is achieved a

allochka39001 [22]

Answer:

2874.33 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{v^2-0^2}{2\times h}\\\Rightarrow v^2=2ah\ m/s$

Now H-h = 0.588 - 0.002 = 0.586 m

The final velocity will be the initial velocity

$v^2-u^2=2as\\\Rightarrow 0^2-u^2=2gs\\\Rightarrow -2ah=2\times g(H-h)\\\Rightarrow -2a0.002=2\times g0.586\\\Rightarrow a=-\frac{0.586\times -9.81}{0.002}\\\Rightarrow a=2874.33\ m/s^2$

Acceleration of the frog is 2874.33 m/s²

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