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Brums [2.3K]
3 years ago
11

__________energy might also be released during a chemical reaction

Physics
1 answer:
pentagon [3]3 years ago
7 0
Kinetic energy i think
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Just need the answer
marin [14]

Answer:

1.  1, 2, 4 all show some form of refraction as the bending of a light ray when passing from one media to another.

Explanation:

Number 4 is the most accurate as it also shows some light being reflected and the bending of the refracted light ray in the correct direction for going from a medium of low refractive index (air) into a higher refractive index material (crown glass)

8 0
3 years ago
Which scenario is an application of Newtons Second Law of Motion?
svlad2 [7]
D would be the answer because The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
4 0
3 years ago
A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.
alex41 [277]

The magnitude of the unknown height of the projectile is determined as 16.1 m.

<h3>Magnitude of the height</h3>

The magnitude of the height of the projectile is calculated as follows;

H = u²sin²θ/2g

H = (36.6² x (sin 29)²)/(2 x 9.8)

H = 16.1 m

Thus, the magnitude of the unknown height of the projectile is determined as 16.1 m.

Learn more about height here: brainly.com/question/1739912

#SPJ1

3 0
2 years ago
A 1.0-μm-diameter oil droplet (density 900 kg/m3) is negatively charged with the addition of 39 extra electrons. It is released
GREYUIT [131]

Answer:

6.75\mu C/m^2

Explanation:

We are given that

Diameter,d=1\mu m=1\time 10^{-6} m

1\mu m=10^{-6} m

Radius,r=\frac{d}{2}=\frac{1}{2}\times 10^{-6}=0.5 \times 10^{-6} m

Density,\rho=900kg/m^3

Total number of electrons,n=39

Charge on electron =1.6\times 10^{-19} C

Total charge=q=ne=39\times 1.6\times 10^{-19}=62.4\times 10^{-19} C

Distance,s=2mm=2\times 10^{-3} m

Mass =density\times volume=900\times \frac{4}{3}\pi r^3=900\times \frac{4}{3}\pi(0.5\times 10^{-6})^3=4.7\times 10^{-16} kg

Initial velocity,u=0

Final speed,v=4.5 m/s

v^2-u^2=2as

(4.5)^2-0=2a(2\times 10^{-3})

20.25=4a\times 10^{-3}

a=\frac{20.25}{4\times 10^{-3}}=5062.5m/s^2

Force,F=ma

qE=ma

q(\frac{\sigma}{2\epsilon_0})=ma

\sigma=\frac{2\epsilon_0ma}{q}=\frac{2\times 8.85\times 10^{-12}\times 4.7\times 10^{-16}\times 5062.5}{62.4\times 10^{-19}}

\epsilon_0=8.85\times 10^{-12}

\sigma=6.75\times 10^{-6}C/m^2=6.75\mu C/m^2

6 0
3 years ago
Refer to the attached image!!!​
dimaraw [331]

The time of motion of the track star is determined as 0.837 s.

<h3>Time of motion of the track star</h3>

The time of motion of the track star is calculated as follows;

T = (2u sinθ)/g

where;

  • T is time of motion
  • g is acceleration due to gravity
  • θ is angle of projection

T = (2 x 12 x sin20)/9.8

T = 0.837 s

Learn more about time of motion here: brainly.com/question/2364404

#SPJ1

6 0
2 years ago
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