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3 years ago

15

What light is NOT in the visible light spectrum

1 answer:

Evgesh-ka [11]3 years ago

8 0

If you mean with the visible eye, then its radiation. we cant see radiation.
get a little more detail about the question.

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You have decided to study the effect of loud noise on plant growth. You put one plant in a quiet room and the other plant in you

Aleksandr [31]

B. control

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3 years ago

Lightning can sometimes occur on hot and humid summer evenings when there are no thunderstorms.

mart [117]

Answer:

(B) False

Explanation:

No, it is not possible to have thunder without lightning. Thunder is a direct result of lightning.

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4 years ago

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What is/are the energy transformation(s) that take place when using a wind turbine to generate usable energy?

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Answer:

the answer is C

Explanation:

C) friction - mechanical - electrical

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3 years ago

An object is moving north at a constant speed. A force of 5 N begins to push the object east at the same moment that a force of

Svetllana [295]

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7 0

3 years ago

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A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.03 s la

Nookie1986 [14]

Answer:

$h=53.09m$ (2)

$v_{min}>5.05m/s$

$v_{max}$

Explanation:

<u>a)Kinematics equation for the first ball:</u>

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h$ initial position is the building height

$v_{o}=8.9m/s$

The ball reaches the ground, y=0, at t=t1:

$0=h+v_{o}t_{1}-1/2*g*t_{1}^{2}$

$h=1/2*g*t_{1}^{2}-v_{o}t_{1}$ (1)

Kinematics equation for the second ball:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h$ initial position is the building height

$v_{o}=0$ the ball is dropped

The ball reaches the ground, y=0, at t=t2:

$0=h-1/2*g*t_{2}^{2}$

$h=1/2*g*t_{2}^{2}$ (2)

the second ball is dropped a time of 1.03s later than the first ball:

t2=t1-1.03 (3)

We solve the equations (1) (2) (3):

$1/2*g*t_{1}^{2}-v_{o}t_{1}=1/2*g*t_{2}^{2}=1/2*g*(t_{1}-1.03)^{2}$

$g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)$

$g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)$

$-2v_{o}t_{1}=g*(-2.06*t_{1}+1.06)$

$2.06*gt_{1}-2v_{o}t_{1}=g*1.06$

$t_{1}=g*1.06/(2.06*g-2v_{o})$

vo=8.9m/s

$t_{1}=9.81*1.06/(2.06*9.81-2*8.9)=4.32s$

t2=t1-1.03 (3)

t2=3.29sg

$h=1/2*g*t_{2}^{2}=1/2*9.81*3.29^{2}=53.09m$ (2)

b) $t_{1}=g*1.06/(2.06*g-2v_{o})$

t1 must : t1>1.03 and t1>0

limit case: t1>1.03:

$1.03>9.81*1.06/(2.06*g-2v_{o})$

$1.03*(2.06*9.81-2v_{o})$

$20.8-2.06v_{o}$

$(20.8-10.4)/2.06$

$v_{min}>5.05m/s$

limit case: t1>0:

$g*1.06/(2.06*g-2v_{o})>0$

$2.06*g-2v_{o}>0$

$v_{o}$

$v_{max}$

8 0

4 years ago