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EleoNora [17]
3 years ago
6

Which of the following is an example of a physical change?

Physics
1 answer:
Stels [109]3 years ago
7 0

Answer:

1 question is- I believe 4. Melting Question 2 is Density

Explanation:

THOUGHT ABOUT IT!!!

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An electric field around two charged objects is shown.
mamaluj [8]

Answer: D

<u></u>

X: positive

Y: negative

Explanation:

It's either A or D and I chose A and got it wrong.

3 0
3 years ago
If I have a picosecond laser how would that be expressed in terms of zeptoseconds? In terms of petaseconds?
kolbaska11 [484]
Picosecond = 10 ^ -12 seconds.
Zeptosecond = 10^ -18 seconsds
Petaseonds  =  10^15  seconds

To express Picoseconds into any of other two, you have to divide  10^-12 by the power index of the one in question

1Picosecond :   10^-12  /  10^-18 =    10^ (-12- 18) =  10^ (-12+18)= 10^6  zeptoseconds

1Picosecond :   10^-12  /  10^15 =    10^ (-12-15) =  10^-27  Petaseconds.

1Picosecond = 10^6  zeptoseconds

1Picosecond = 10^-27  Petaseconds




3 0
3 years ago
When an 8 V battery is connected to a resistor, a 2 A current flows in the resistor. What is the resistor's value?
Vinvika [58]

Answer:

B

Explanation:

V=IR    I= curren V=volt R=resistor

8=2.R   8/2=R R=4

5 0
3 years ago
An electron is initially moving at 1.4 x 107 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N
algol13

Answer:

K.E = 15.57 x 10⁻¹⁷ J

Explanation:

First, we find the acceleration of the electron by using the formula of electric field:

E = F/q

F = Eq

but, from Newton's 2nd Law:

F = ma

Comparing both equations, we get:

ma = Eq

a = Eq/m

where,

E = electric field intensity = 120 N/C

q = charge of electron = 1.6 x 10⁻¹⁹ C

m = Mass of electron = 9.1 x 10⁻³¹ kg

Therefore,

a = (120 N/C)(1.6 x 10⁻¹⁹ C)/(9.1 x 10⁻³¹ kg)

a = 2.11 x 10¹³ m/s²

Now, we need to find the final velocity of the electron. Using 3rd equation of motion:

2as = Vf² - Vi²

where,

Vf = Final Velocity = ?

Vi = Initial Velocity = 1.4 x 10⁷ m/s

s = distance = 3.5 m

Therefore,

(2)(2.11 x 10¹³ m/s²)(3.5 m) = Vf² - (1.4 x 10⁷)²

Vf = √(1.477 x 10¹⁴ m²/s² + 1.96 x 10¹⁴ m²/s²)

Vf = 1.85 x 10⁷ m/s

Now, we find the kinetic energy of electron at the end of the motion:

K.E = (0.5)(m)(Vf)²

K.E = (0.5)(9.1 x 10⁻³¹ kg)(1.85 x 10⁷ m/s)²

<u>K.E = 15.57 x 10⁻¹⁷ J</u>

4 0
3 years ago
What mediums are best for light waves?
ycow [4]
The strength of the electric and magnetic fields there is no physical "distance" of oscillation here. nothing is actually moving up and down if you draw light as a sinusoidal wave, the up and down motion is the strength of the EM fields cheers
4 0
3 years ago
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