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3 years ago

Is it inhumane to keep a cat in one room?

1 answer:

4 0

So there are a couple of variables to that question. if the room has everything a cat needs then no. if it had food, water, a bed, stuff to keep it occupied, a litter box, sunlight, and some other form of life to keep it company then no it wouldn't be.
But if you locked it in a dark room without any of that stuff yes the cat would die

You might be interested in

An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much

cupoosta [38]

Answer:

time will elapse before it return to its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × $10^{7}$ m/s

uniform electric field E = 1.18 × $10^{4}$ N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

put here all value

9.11 × $10^{-31}$ kg ×a = 1.18 × $10^{4}$ × 1.602 × $10^{-19}$

a = 20.75 × $10^{14}$ m/s²

so acceleration is 20.75 × $10^{14}$ m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × $10^{7}$ = 0 + 20.75 × $10^{14}$ (t)

t = 11.80 × $10^{-9}$ s

so time will elapse before it return to its staring point is

time = 2t

time = 2 ×11.80 × $10^{-9}$

time is 23.6 × $10^{-9}$ s

time will elapse before it return to its staring point is 23.6 ns

7 0

3 years ago

Two ladders of uniform density and equal mass m are propped up against

-BARSIC- [3]

Answer:

i need some more coins LMAI

7 0

3 years ago

Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150

Phoenix [80]

Answer:

The temperature of the core raises by $2.8^{o}C$ every second.

Explanation:

Since the average specific heat of the reactor core is 0.3349 kJ/kgC

It means that we require 0.3349 kJ of heat to raise the temperature of 1 kg of core material by 1 degree Celsius

Thus reactor core whose mass is $1.60\times 10^{5}kg$ will require

$0.3349\times 1.60\times 10^{5}kJ\\\\=0.53584\times 10^{5}kJ$

energy to raise it's temperature by 1 degree Celsius in 1 second

Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by

$\frac{150\times 10^{6}}{0.53584\times 10^{8}}=2.8^{o}C/s$

5 0

3 years ago

A car is running at the speed of 45km/hr see a child 25 meter ahead and suddenly apllies a brakes. If the retradation of the car

kenny6666 [7]

Answer:

The car stops in 7.78s and does not spare the child.

Explanation:

In order to know if the car stops before the distance to the child, you take into account the following equation:

$x=x_o+v_ot-\frac{1}{2}at^2$ (1)

vo: initial speed of the car = 45km/h

a: deceleration of the car = 2 m/s^2

t: time

xo: initial distance to the child = 25m

x: final distance to the child = 0m

It is necessary that the solution of the equation (1) for time t are real.

You first convert the initial speed to m/s, then replace the values of the parameters and solve the quadratic polynomial for t:

$45\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=12.5\frac{m}{s}$

$0=25+12.5t-2t^2\\\\2t^2-12.5t-25=0\\\\t_{1,2}=\frac{-(-12.5)\pm \sqrt{(-12.5)^2-4(2)(-25)}}{2(2)}\\\\t_{1,2}=\frac{12.25\pm 18.87}{4}\\\\t_1=7.78s\\\\t_2=-1.65s$

You take the first value t1 because it has physical meaning.

The solution for t is real, then, the car stops in 7.78s and does not spare the child.

4 0

3 years ago

I need help!! i’ll give 20 points

larisa86 [58]

Answer:

780 m to travel north

Explanation:

6 m over = 750

53 degree so it will take about 2 min to reach the destination

8 0

3 years ago