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Maslowich
1 year ago
9

Consider the surface with parametric equations .

Physics
1 answer:
balandron [24]1 year ago
5 0

There are two ways to express the tangent plane equation. Given that it is the sum of the two distinct tangent vectors, its parametric equation is r=r0+sru+trv.

<h3>What does the parametric surface in calculus mean?</h3>

A surface in Euclidean space is considered to be parametric if it can be described by a parametric equation with two variables. Parametric representation is another reasonably general technique for describing a surface in addition to implicit representation.

On the other hand, tangent planes to a surface are planes that are "parallel" to the surface at the point where they barely touch it. Remember that this provides us with a point on the The tangent plane equation can be written in two different ways. The following point emerges from the surface and tangent plane meeting at (x0,y0):

learn more about Tangent planes refer

brainly.com/question/17748591

#SPJ4

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The energy associated with the random motion of molecules or atoms within a substance is
Yuliya22 [10]

Answer:

C

technically B too but youre teachers not that smart so there you go

6 0
3 years ago
Add these two velocity vectors to find the magnitude of their resultant vector.
hammer [34]

The  magnitude of their resultant vector is 4.6 meters/seconds

Since we are to add the  velocity vectors in order to  find the magnitude of their resultant vector.

Hence:

Resultant vector magnitude=5.8 meters/seconds + (1.2 meters/seconds)

Resultant vector magnitude=5.8 meters/seconds-1.2 meters/seconds

Resultant vector magnitude 4.6 meters/seconds

Inconclusion The  magnitude of their resultant vector is 4.6 meters/seconds

Learn more here:

brainly.com/question/11134601

6 0
3 years ago
In this experiment, you will use a track, a toy car, and some washers to explore Newton’s first two laws of motion. You will mak
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How can we experimentally verify newton's laws?

7 0
3 years ago
Read 2 more answers
Assume that the driver begins to brake the car when the distance to the wall is d=107m, and take the car's mass as m-1400kg, its
Evgen [1.6K]

Answer:

Explanation:

a ) Let let the frictional force needed be F

Work done by frictional force = kinetic energy of car

F x 107 = 1/2 x 1400 x 35²

F = 8014 N

b )

maximum possible static friction

= μ mg

where μ is coefficient of static friction

= .5 x 1400 x 9.8

= 6860 N

c )

work done by friction for μ = .4

= .4 x 1400 x 9.8 x 107

= 587216 J

Initial Kinetic energy

= .5 x 1400 x 35 x 35

= 857500 J

Kinetic energy at the at of collision

= 857500 - 587216

= 270284 J

So , if v be the velocity at the time of collision

1/2 mv² = 270284

v = 19.65 m /s

d ) centripetal force required

= mv₀² / d which will be provided by frictional force

= (1400 x 35 x 35) / 107

= 16028 N

Maximum frictional force possible

= μmg

= .5 x 1400 x 9.8

= 6860 N

So this is not possible.

4 0
3 years ago
To lift a load of 100 kg a distance of 1 m an effort of 25 kg must be applied over an inclined plane of length 4 m. What must be
andreyandreev [35.5K]
Work = force * distance.
We must produce twice as much energy as we are lifting the weight twice as high.
But we are not increasing the force so we must increase the length of the ramp ( distance ) instead.
The new length will be twice as great as the previous length.
So 8 metres is required. 


25 kg * 8 m = work = 100 kg * 2 m
7 0
3 years ago
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