1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
konstantin123 [22]
2 years ago
14

An 85 kg dog is sitting on a couch what is the weight of the dog

Physics
2 answers:
White raven [17]2 years ago
8 0
If the couch is anywhere on Earth, then the dog's weight is 833 Newtons.
That's about 187 pounds ... Easy there big boy.  Nice doggy ...

Actually, it would be exactly the same if he were sitting on a chair or sleeping on the floor.
sergejj [24]2 years ago
4 0
The dog is 85kg or 187lbs (pounds)
You might be interested in
An electron enters the gap between the plates of a capacitor at the center of the gap traveling parallel to theplates at 2.0 x 1
Svetlanka [38]

Answer:

How far will the electron travel beforehitting a plate is 248.125mm

Explanation:

Applying Gauss' law:

Electric Field E = Charge density/epsilon nought

Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12

Therefore E = 1.0 x 10^-6/8.85× 10^-12

E= 1.13×10^5N/C

Force on electron F=qE

Where q=charge of electron=1.6×10^-19C

Therefore F=1.6×10^-19×1.13×10^5

F=1.808×10^-14N

Acceleration on electron a = Force/Mass

Where Mass of electron = 9.10938356 × 10^-31

Therefore a= 1.808×10^-14 /9.11 × 10-31

a= 1.985×10^16m/s^2

Time spent between plate = Distance/Speed

From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2

Therefore Time = 0.01/2×10^6

Time =5×10^-9s

How far the electron would travel S =ut+ at^2/2 where u=0

S= 1.985×10^16×(5×10^-9)^2/2

S=24.8125×10^-2m

S=248.125mm

4 0
3 years ago
What is the BEST example of a purebred
ki77a [65]

no se   nada pvodebrhtre

8 0
3 years ago
A cylinder of mass 250 kg and radius 2.60 m is rotating at 4.00 rad/s on a frictionless.
aleksandrvk [35]

Answer:

The angular momentum of a cylinder, when it is rotating with constant angular velocity is Lini =Iωi

. When two cylinders are added to the rotating cylinder, which are identical in their dimensions, the moment of inertia of the entire system increases (since mass increases). The final moment of inertia will be 3I

Since friction exist, all the cylinders start rotating with same angular velocity, the new angular velocity can be calculated using conservation of angular momentum

Thus, Iωi =3Iωf ⟹ωf =ωi/3 = 0.33ωi

8 0
3 years ago
An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _
Viefleur [7K]
Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
where 
f=5 cm is the focal length
p=10 cm is the distance of the object from the mirror
q is the distance of the image from the mirror.

Rearranging the equation, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}
so, the distance of the image from the mirror is q=10 cm.

1b) The image height is given by the magnification equation:
\frac{h_i}{h_o}=- \frac{p}{q}
where h_i is the heigth of the image and h_o=1 cm is the height of the object. By rearranging the equation and using p and q, we find
h_i=-h_o  \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
Rearranging it, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}
so, the distance of the image from the mirror is
q= \frac{10}{4}cm= 2.5 cm

3) As before, we find the distance of the image from the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
Rearranging it, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}
so, the distance of the image from the mirror is
q= \frac{10}{4}cm= 2.5 cm

And now we can use the magnification equation to find the image height:
\frac{h_i}{h_o}=- \frac{p}{q}
Rearranging it, we find
h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm
and the negative sign means the image is inverted.
5 0
3 years ago
What do you get when you subtract the force of air resistance from the force<br> of gravity?
atroni [7]

Answer:

Net force

Explanation:

3 0
3 years ago
Read 2 more answers
Other questions:
  • Which property do metalloids share with nonmetals
    8·2 answers
  • A gas is __________ and assumes __________ of its container whereas a liquid is __________ and assumes __________ of its contain
    10·1 answer
  • You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 1515 lb and was trav
    6·2 answers
  • There are 3.78 liters in one gallon, therefore 3.78L = 1 Gallon. If I need 58 half-gallon containers for my boat, determine the
    5·1 answer
  • you are running at a rate of 8 meters per second west. You ran a total distance of 400 meters. how long did it take to complete
    10·1 answer
  • Hey! Can someone help with this question? Thx :)
    10·1 answer
  • n April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min , 43.6 s . Suppose "Pre" was at the 7.57 km mark at a
    6·1 answer
  • Show that any three linear operators A, B, and Ĉ satisfy the following (Ja- cobi) identity (10 pt) [[A, B] Ĉ] + [[B,C), A] + [[C
    12·1 answer
  • A ball is launched vertically upward from the edge of a cliff. The ball reaches its maximum height 1.6 seconds after launch. Bar
    10·1 answer
  • 3. Specify the wrong sentences.
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!