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kondor19780726 [428]
3 years ago
5

I need the answer to (c): Describe how the student accurately traces the path of the light emerging from the block through side

CD.

Physics
1 answer:
Artyom0805 [142]3 years ago
3 0

Explanation:

i guess using a ruler.so first on the opposite side use your eye to alighn the ruler to the light from Ray box then draw a line or you can use pins and place them alighned to the Ray box an draw a line connecting the pins

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A bullet is fired at an angle θ above the horizontal with an initial velocity of 800 m/s from the top of an 80 m high tower. Wha
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Answer:

pahingi po ng pic pls para masagutang kopo iyan

6 0
3 years ago
An 88 kg worker stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale reads 900 N. Fi
Aleks [24]

Answer:

The magnitude is "3.8 m/s²", in the upward direction.

Explanation:

The given values are:

Mass,

m = 88 kg

Scale reads,

T = 900 N

As we know,

⇒  N=mg

On substituting the given values, we get

⇒      =88\times 9.8

⇒      =862.4 \ N

Now,

⇒  T=mg-ma

On substituting the given values in the above equation, we get

⇒  900=862.4-9.8 a

On subtracting "862.4" from both sides, we get

⇒  900-862.4=862.4-9.8 a-862.4

⇒              37.6=-9.8a

⇒                   a=-\frac{37.6}{9.8}

⇒                   a=3.8 \ m/s^2 (upward direction)

8 0
3 years ago
The force between a pair of charges is 900 newtons. The distance between the charges is 0.01 meters. If one of the charges is 2e
Maksim231197 [3]

Answer:

\fbox{strength  \: of \:  the \:  other \:  charge  =  - 0.0196 Ke \: Coulomb}

Explanation:

Given:

Force between pair of charges= 900 newtons

The distance between the charges = 0.01 meters

Strength of Charge first q1 = 2e-10 Coulomb

To find:

Strength of Charge second q2 = ____ Coulomb?

Solution:

We know that,

Force between two charges separate by distance r is given by the equation,

|F| =  K_e \frac{q1 \cdot \: q2}{ {r}^{2} }  \\ 900 =K_e  \frac{(2e - 10)\cdot \: q2}{ {0.01}^{2} } \\ 900 \times  {10}^{ - 4}  =  K_e {(2e - 10)\cdot \: q2} \\ q2 =   \frac{9 \times  {10}^{ - 2} }{(2e - 10) K_e}  \\  \\  \fbox{We \:  know \:  that \:  e = 2.71 } \\  substituting \: the \: value \: \\ q2 =  \frac{9 \times  {10}^{ - 2} }{(2 \times 2.71 - 10)K_e}  \\ q2 =  \frac{0.09}{ - 4.58 K_e}  \\ q2 =  \frac{-0.0196}{K_e}\: coulomb

\fbox{strength  \: of \:  the \:  other \:  charge  =  - 0.0196 Ke \: Coulomb}

<em><u>Thanks for joining brainly community!</u></em>

3 0
2 years ago
What are some ways that scientists would collect data and make observations to help them learn more about the severity of the li
nika2105 [10]
Experiments, tests, and trials
6 0
4 years ago
Three equal 1.55-μC point charges are placed at the corners of an equilateral triangle whose sides are 0.500 m long. What is the
kati45 [8]

Answer:

0.12959085 J

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

q = Charge = 1.55 μC

d = Distance between charge = 0.5 m

Electric potential energy is given by

U=k\dfrac{q^2}{d}

In this system with three charges which are equidistant from each other

U=k\dfrac{q^2}{d}+k\dfrac{q^2}{d}+k\dfrac{q^2}{d}

\\\Rightarrow U=k\dfrac{3q^2}{d}\\\Rightarrow U=8.99\times 10^9\times \dfrac{3\times (1.55\times 10^{-6})^2}{0.5}\\\Rightarrow U=0.12959085\ J

The potential energy of the system is 0.12959085 J

6 0
3 years ago
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