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kykrilka [37]
2 years ago
15

In real number multiplication, if uv1=uv2 and u does not equal zero we can cancel out u and conclude that v1=v2. Does the same r

ule hold for the dot product?
Physics
1 answer:
Leya [2.2K]2 years ago
6 0

The conclusion v₁=v₂ can not be made. Let u=(1,1,1),v₁=(2,3,-5),v₂=(4,-1,-3). The dot product of u.v₁=uv₂ is satisfied,but v₁≠v₂.The correct answer is c.

<h3>What is dot product?</h3>

The dot product, also known as the scalar product, is an algebraic operation that yields a single integer from two equal-length sequences of numbers.

The dot product of two vectors' Cartesian coordinates is commonly used in Euclidean geometry.

Because there may be more than the 1 vectors satisfying the expression.

\rm uv_1  = 1 \times 2 +1 \times 3 +1 \times(-5)=0

\rm uv_2= 4-1-3 =0

The v₁ is not found equal to the v₂.

The conclusion v₁=v₂ can not be made. Let u=(1,1,1),v₁=(2,3,-5),v₂=(4,-1,-3). The dot product of u.v₁=uv₂ is satisfied,but v₁≠v₂.The correct answer is c.

Hence, the correct answer is c.

To learn more about the dot product, refer to the link;

brainly.com/question/26550859

#SPJ4

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velocity of car=5.855 m/s

Explanation:

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A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a
stira [4]

Answer:

Explanation:

Given

acceleration is given by

a=-g-c_Dv^2

where \ddot{y}=a

\dot{y}=v

Also acceleration is given by

a=v\frac{\mathrm{d} v}{\mathrm{d} s}

ds=\frac{v}{a}dv

\int ds=\int \frac{v}{-g-0.001v^2}dv

\Rightarrow Let -g-0.001v^2=t

-0.001\times 2vdv=dt

vdv=-\frac{dt}{0.002}

at\ v_0=50\ m/s,\ t=-g-0.001(50)^2

t=-g-2.5

at v=0,\ t=-g

\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}

\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}

s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}

s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})

s=113.608\ m

when air drag is neglected maximum height reached is

h=\frac{v_0^2}{2g}

h=\frac{50^2}{2\times 9.8}

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3 0
3 years ago
Does the mass of a parachute affect terminal velocity?
vovangra [49]

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A submarine is stranded on the bottom of the ocean with its hatch 21.0 m below the surface. calculate the force (in n) needed to
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Answer:

28,400 N

Explanation:

Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:

p_{top} = p_{atm} + \rho g h=1.013\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(21.0 m)=3.071 \cdot 10^5 Pa

On the lower part of the hatch, there is a pressure equal to

p_{bot}=p_{atm}=1.013\cdot 10^5 Pa

So, the net pressure acting on the hatch is

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which acts from above.

The area of the hatch is given by:

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So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:

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8 0
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The period of wave-a is 1/2 second.

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