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2 years ago

Help my cat is constipated what do I do

Physics

1 answer:

storchak [24]2 years ago

7 0

Answer:

Increase Water Consumption

Try a New Diet

Help Your Cat Maintain a Healthy Weight

Increase Exercise and Enrichment

Minimize Stress and Anxiety

Add More Litter Boxes

Try Fiber or Probiotics

Ask Your Vet About Over-the-Counter Laxatives

Explanation:

<h2><em>Hope this helps ;)</em></h2>

You might be interested in

The force of gravity between two objects depends on ___________.

murzikaleks [220]

The answer is D. the mass of the two objects and the distance between them

4 0

2 years ago

What force must be used to do 224 joules of work on an object over a distance of 32 meters

Tems11 [23]

Considering that Work, W, is:
W=F·d
You have: 224=F·32
So that F=7 N

3 0

2 years ago

The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at posit

SpyIntel [72]

Answer:

Magnitude of the Frictional force = (mv₀²)/2x₁

Explanation:

For the frictional force to stop the box, it has to produce the deceleration of the box; thereby being the opposing force to the box's motion.

According to Newton's first law of motion

Frictional force = (mass of the box) × (deceleration experienced by the box)

Let the mass of the box be m

Then,

Frictional force = ma

Then we can obtain the deceleration using the equations of motion

v² = u² + 2ax

u = Initial velocity = v₀ m/s

v = Final velocity = 0 m/s (since the box comes to rest at the end)

x = horizontal distance covered = (x₁ - x₀) = x₁ (since x₀ = 0)

a = ?

v² = u² + 2ax

0 = (v₀)² + 2ax₁

2ax₁ = - v₀²

a = - (v₀²)/(2x₁) (minus sign, because it's a deceleration)

Magnitude of the Frictional force = ma = (mv₀²)/2x₁

4 0

2 years ago

A car accelerates from rest at 1.0 m/s2 for 20.0 second along a straight road. It then moves at a constant speed for half an hou

Whitepunk [10]

Total distance = 36500 m

The average velocity = 19.73 m/s

<h3>Further explanation</h3>

Given

vo=initial velocity=0(from rest)

a=acceleration= 1 m/s²

t₁ = 20 s

t₂ = 0.5 hr = 1800 s

t₃= 30 s

Required

Total distance

Solution

State 1 : acceleration

$\tt d=vo.t+\dfrac{1}{2}at^2\\\\d=\dfrac{1}{2}\times 1\times 20^2\rightarrow vo=0\\\\d=200~m$

$\tt vt=vo+at\\\\vt=at\rightarrow vo=0\\\\vt=1\times 20\\\\vt=20~m/s$

State 2 : constant speed

$\tt d=v\times t\\\\d=20\times 1800\\\\d=36000~m$

State 3 : deceleration

$\tt vt=vo+at\rightarrow vt=0(stop)\\\\vo=-at\\\\20=-a.30~s\\\\a=-\dfrac{2}{3}m/s^2(negative=deceleration)$

$\tt d=vot+\dfrac{1}{2}at^2\\\\d=20.30-\dfrac{1}{2}.\dfrac{2}{3}.30^2\\\\d=300~m$

Total distance : state 1+ state 2+state 3

$\tt 200 + 36000 + 300=36500~m$

the average velocity = total distance : total time

$\tt avg~velocity=\dfrac{36500}{20~s+1800~s+30~s}=19.73~m/s$

4 0

2 years ago

A square piece of tin has 12 inches on a side. An open box is formed by cutting out equal square pieces at the corners and bendi

Citrus2011 [14]

Answer:

Explanation:

Given a square Piece whose side is 12 inches

Now square pieces are cut from each corner to make it a open box

Suppose x is the length of square piece at each corner

then

base square has a length of $12-2x$

Dimension of new box is $(12-2x)\times (12-2x)\times x$

Volume $V=(12-2x)\times (12-2x)\times x$

$V=\left ( 12-2x\right )^2\cdot x$

For maximum volume differentiate with respect to x we get

$\Rightarrow\frac{\mathrm{d} V}{\mathrm{d} x}=2\times \left ( 12-2x\right )\times \left ( -2\right )\cdot x+\left ( 12-2x\right )^2=0$

we get x=6 and 4 but at x=6 volume becomes zero therefore x=4 is valid

$V=\left ( 12-2\cdot 4\right )^2\cdot 4$

$V=4^3$

$V=64\ in.^3$

6 0

2 years ago