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Alina [70]
3 years ago
13

Use vector notation to describe the points that lie in the configuration. (Let s and t be elements of the Reals). the plane span

ned by v1=(2,7,0) and v2=(0,2,7). I(s,t)=
Physics
1 answer:
aniked [119]3 years ago
5 0

Answer:

The answer is l(s, t) = (2s,7s + 2t,7t)

Explanation:

A general vector notation that can describe the points which lie in the configuration of the plane spanned by v1 and v2, is evaluated, thus:

V = v1 + v2

Where s and t are the elements of v1(2,7,0) and v2(0,2,7), respectively, and projected in x, y and z coordinates, we have:

V = sv1 + sv2, implying that:

V = (2sx + 7sy + 0sz) + (0tx + 2ty + 7tz)

Or, V= (2sx + 7sy) + (2ty + 7 tz)

Therefore, V = (2s,7s + 2t,7t)

and l(s, t) = (2s,7s + 2t,7t).

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olga55 [171]

Answer:

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Explanation:

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3 years ago
g as measured from the earth, a spacecraft is moving at speed .80c toward a second spacecraft moving at speed .60c back toward t
cupoosta [38]

Answer:

the speed of the first spacecraft as viewed from the second spacecraft is 0.95c

Explanation:

Given that;

speed of the first spacecraft from earth v_a = 0.80c

speed of the second spacecraft from earth v_b = -0.60 c

Using the formula for relative motion in relativistic mechanics

u' = ( v_a - v_b ) / ( 1 - (v_bv_a / c²) )

we substitute

u' = ( 0.80c - ( -0.60c)  ) / ( 1 - ( ( 0.80c × -0.60c) / c² ) )

u' = ( 0.80c + 0.60c ) /  ( 1 - ( -0.48c² / c² ) )

u' = 1.4c /  ( 1 - ( -0.48 ) )

u' = 1.4c /  ( 1 + 0.48 )

u' = 1.4c / 1.48

u' = 0.9459c ≈ 0.95c  { two decimal places }

Therefore, the speed of the first spacecraft as viewed from the second spacecraft is 0.95c

7 0
3 years ago
Pleaseee someone help me TEST TOMORROW
zubka84 [21]
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3 years ago
A large box of mass M is pulled across a horizontal, frictionless surface by a horizontal rope with tension T. A small box of ma
Nesterboy [21]

Answer:

T = g μ_s ( M+m )

78.4 N

Explanation:

When both of them move with the same acceleration , small box will not slip over the bigger one. When we apply force on the lower box, it starts moving with respect to lower box. So a frictional force arises on the lower box which helps it too to go ahead . The maximum value that this force can attain is mg μ_s . As a reaction of this force, another force acts on the lower box in opposite direction .

Net force on the lower box

= T - mg μ_s = M a    ( a is the acceleration created by net force in M )

Considering force on the upper box

mg μ_s = ma

a = g μ_s

Put this value of a in the equation above

T - m gμ_s = M g μ_s

T = mg μ_s + M g μ_s

=  g μ_s ( M+m )

2 )

Largest tension required

T = 9.8 x  .50 x ( 10+6 )

= 78.4 N

5 0
4 years ago
Which term describes the energy of position?
scoray [572]

Answer:

B

Explanation:

3 0
4 years ago
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