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3 years ago

Most scientific questions are based on

2 answers:

4 0

Most scientific questions are based on observations. Scientists observe something repeatedly, and they do the experiments over and over again, so as to get to the bottom of some of their questions.

Oksana_A [137]3 years ago

4 0

<span>opinions, hypotheses, experimental data, observations</span>

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What is (a) the x component and (b) the y component of the net electric field at the square's center

Sav [38]

Answer:

What is (a) the x component and (b) the y component of the net electric field at the square's center

8 0

3 years ago

Answer the question with step.

Maru [420]

Answer:

f1/f2 =W1/W2 = 1/3

.0 f2 = 3f1

As ,

1/F= 1/f1 +1/f2

...1/40 = 1/f1 - 1/3f1

f1=> 80/3 cm

... f2 = 2f1 = 3 x 80/3 = 80 cm

7 0

3 years ago

Do YOU think aliens are real? HELP PLS

nikdorinn [45]

Answer:Yes

Explanation:yeah i think it could be possible because the universe is so far and wide its probably infinite and if we did i personally believe that we would be absolutely be annihilated by them because we would be so far behind technology wise we would have no use for the\

<h2>YOU FELL FOR IT FOOL THUNDER CROSS SPLIT ATTACK</h2><h2></h2>

6 0

3 years ago

An object is located in air, 25 cm from the vertex of the concave surface of a block of glass (as viewed from the air side of th

Marizza181 [45]

Your question is missing one part as "magnification"

i have completed the missing part below

Answer:

a. $d_{i}=-0.0566cm$

b. $M=441.69$

Explanation:

For this type of numerical we will use the following formulas

$\frac{n1}{d_{o} }+\frac{n_{2} }{d_{i} }=\frac{n_{2}-n_{1} }{R}$ ......... Eq1

where,

$n_{1}$ $=$ refractive index of the medium surrounding refracting surface/object

i.e. $n_{1}$ = $n_{air}$

$n_{2}$ $=$ refractive index of the refracting surface/object

i.e. $n_{2}=n_{glass}$

$d_{0}=$ distance of object from the vertex of the refracting surface

$d_{i}=$ distance of image from the vertex of the refracting surface

$R=$ radius of curvature of the refracting surface

$M=\frac{d_{0} }{d_{i} }$ ........... Eq2

where,

$M=$ magnification

Convention:

$R>0\for\ the\ convex\ refractive\ surface\ of\ curvature\\\ R$

Given:

$n_{1}$ = $n_{air}$ $=1.0$

$n_{2}=n_{glass}$ $=1.5$

$d_{0}=$ $25cm$

$R=-11$ $cm$ because refraction surface is concave

Required:

a. $d_{i}=$ $?$

b. $M=?$

Solution:

a. putting values in eq1, we get

$\frac{1.0}{25}+\frac{1.5}{d_{i} }=\frac{1.5-1.0}{-11}$

$\frac{1.5}{d_{i} }=-0.045-0.040$

$d_{i}=(-0.085)(\frac{1}{1.5} )$

$d_{i}=-0.0566$ cm

b. $M=\frac{25}{-0.05665}$

$M=441.69$

5 0

4 years ago

Ball 1, with a mass of 20kg is moving to the right at 20 m/s. At what velocity should ball 2, with a mass of 40kg, move so that

Ket [755]

This is a conservation of momentum question. Initial momentum of the system is the momentum of ball 1 plus the momentum of ball 2. The final momentum of the system should be 0 since the balls stand still after the collision.

mv + mv = 0

mv = -mv

(20)(20) = -(40)Vi

400 = -40Vi

Vi = -10

So ball 2 should travel at 10m/s to the left

4 0

3 years ago