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2 years ago

Do cat worship a cat god such as beerus?

Physics

2 answers:

Inessa05 [86]2 years ago

5 0

No they do not there not humans

nika2105 [10]2 years ago

4 0

Ummm No i dont think cats worship anything but people worship cats doe

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A colony of troglodytes has been in a lengthy feud with

Troyanec [42]

Answer:

h = 181.73 m

Explanation:

given,

distance between the neighbors = 42 m

speed of the rock rolling = 6.9 m/s

vertical velocity of ball = 0 m/s

height of the cliff = ?

time taken by the ball to travel 42 m

d = s x t

s is the horizontal speed of the ball equal to 6.9 m/s

$t =\dfrac{d}{s}$

$t =\dfrac{42}{6.9}$

t = 6.09 s

same time will be taken by the ball to travel vertical distance

Using equation of motion

$h = u t + \dfrac{1}{2}gt^2$

$h = 0+ \dfrac{1}{2}gt^2$

$h = \dfrac{1}{2}\times 9.8 \times 6.09^2$

h = 181.73 m

7 0

2 years ago

In a television picture tube, electrons strike the screen after being accelerated from rest through a potential difference of 27

Vera_Pavlovna [14]

Answer:

9.74 x 10^7 m/s

Explanation:

V = 27000 V

energy of electrons = e x V

K = 1.6 x 10^-19 x 27000 = 43200 x 10^-19 J

Energy = 1/2 m v^2

43200 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2

v^2 = 9.495 x 10^15

v = 9.74 x 10^7 m/s

4 0

3 years ago

What is the reaction force to a foot pushing down on the floor?

Helga [31]

Answer:

C. The floor pushing back against the foot

Explanation:

5 0

2 years ago

Read 2 more answers

Calculate the potential difference across a 25- resistor if a 0.3-A current is flowing through it.

alekssr [168]

Voltage = (current) x (resistance)

Voltage = (0.3 A) x (25 ohms)

Voltage = (0.3 x 25) (A-Ω)

<em>Voltage = 7.5 volts</em>

8 0

3 years ago

Read 2 more answers

A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find

sukhopar [10]

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge $q_{1}=9\times10^{-6}\ C$ at origin

Second charge $q_{2}=6\times10^{-6}\ C$

Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

$E=\dfrac{kq}{r^2}$

$0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}$

$\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}$

$\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1$

$\dfrac{1}{d}=1.82$

$d=\dfrac{1}{1.82}$

$d=0.55\ m$

Hence, The coordinates of the point is (0,0.55).

3 0

2 years ago