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Xelga [282]
3 years ago
6

What is the largest component of M1?

Physics
1 answer:
mariarad [96]3 years ago
8 0

The largest component of M1 is- Currency.

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How many significant figures are there in the measurement (4.07×10^17m)
Alex17521 [72]

Answer:

2...................

6 0
3 years ago
Find a unit vector in the direction in which f increases most rapidly at P and give the rate of chance of f in that direction; f
Burka [1]

Answer:

Check attachment for complete question

Question

Find a unit vector in the direction in which

f increases most rapidly at P and give the rate of change of f

in that direction; Find a unit vector in the direction in which f

decreases most rapidly at P and give the rate of change of f in

that direction.

f (x, y, z) = x²z e^y + xz²; P(1, ln 2, 2).

Explanation:

The function, z = f(x, y,z), increases most rapidly at (a, b,c) in the

direction of the gradient and decreases

most rapidly in the opposite direction

Given that

F=x²ze^y+xz² at P(1, In2, 2)

1. F increases most rapidly in the positive direction of ∇f

∇f= df/dx i + df/dy j +df/dz k

∇f=(2xze^y+z²)i + (x²ze^y) j + (x²e^y + 2xz)k

At the point P(1, In2, 2)

Then,

∇f= (2×1×2×e^In2+2²)i +(1²×2×e^In2)j +(1²e^In2+2×1×2)

∇f=12i + 4j + 6k

Then, unit vector

V= ∇f/|∇f|

Then, |∇f|= √ 12²+4²+6²

|∇f|= 14

Then,

Unit vector

V=(12i+4j+6k)/14

V=6/7 i + 2/7 j + 3/7 k

This is the increasing unit vector

The rate of change of f at point P is.

|∇f|= √ 12²+4²+6²

|∇f|= 14

2. F increases most rapidly in the positive direction of -∇f

∇f=- (df/dx i + df/dy j +df/dz k)

∇f=-(2xze^y+z²)i - (x²ze^y) j - (x²e^y + 2xz)k

At the point P(1, In2, 2)

Then,

∇f= -(2×1×2×e^In2+2²)i -(1²×2×e^In2)j -(1²e^In2+2×1×2)

∇f=-12i -4j - 6k

Then, unit vector

V= -∇f/|∇f|

Then, |∇f|= √ 12²+4²+6²

|∇f|= 14

Then,

Unit vector

V=-(12i+4j+6k)/14

V= - 6/7 i - 2/7 j - 3/7 k

This is the increasing unit vector

The rate of change of f at point P is.

|∇f|= √ 12²+4²+6²

|∇f|= 14

6 0
4 years ago
Read 2 more answers
Contrast the image formed by a convex lens when an object is located more than twice the focal length from the lens with the ima
matrenka [14]

Answer:

\begin{array}{lll}&\underline{Object \ at \ more \ than \ 2\times Focus} & \underline{Object   \ at \ less \ than \  Focus}\\\\1.  \ Location \ of \ image &Same \ side \ as \ object&Opposite \ side \ of \ lens\\\\2. \ Orientation \ of \ image &Inverted&Upright\\\\3. \ Type \ of \ image&Real&Virtual\\\\4. \ Size\ of \ image&Smaller \ than \ object& Larger \ than \ object\end{array}

Explanation:

The location, orientation, size, and type of image formed by a convex lens are related to the position of the image location in front of the lens

Object >2·F = The image formed by a convex lens when the object is located more than twice the focal length from the lens

Object < F = The image formed by the convex lens when the object is located between the lens and the focal length

\begin{array}{lll}&\underline{Object \ > \ 2\cdot F} & \underline{Object  < F}\\\\1.  \ Location \ of \ image &Same \ side \ as \ object&Opposite \ side \ of \ lens\\\\2. \ Orientation \ of \ image &Inverted&Upright\\\\3. \ Type \ of \ image&Real&Virtual\\\\4. \ Size\ of \ image&Smaller \ than \ object& Larger \ than \ object\end{array}

8 0
3 years ago
3. A bicycle has a momentum of 25.00 kg* m/s and a velocity of 2.5 m/s . What is the bicycle's
Anna35 [415]

Answer:

10 kg

Explanation:

The question is most likely asking for the mass of the bicycle.

Momentum is the product of an object's mass and velocity. Mathematically:

p = m * v

Where p = momentum

m = mass

v = velocity

Hence, mass is:

m = p / v

From the question:

p = 25 kgm/s

v = 2.5 m/s

Mass is:

m = 25 / 2.5 = 10 kg

The mass of the bicycle is 10 kg.

In case the question requires the Kinetic energy of the bicycle, it can be gotten by using the formula

K. E = ½ * p * v

K. E. = ½ * 25 * 2.5 = 31.25 J

5 0
4 years ago
An object with a mass of 2 kg has a force of 4 N acting on it. What is the acceleration of the object?
STatiana [176]
Answer: 2 m/s^2

Explain

£F=ma
net force = mass x acceleration
4N is a force acting on it
4N=(2kg) a
2kg is the mass
you divide the 2 over
and A= 2 m/s^2

8 0
3 years ago
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