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Oksi-84 [34.3K]
3 years ago
11

Question 3 (3 points)

Physics
2 answers:
topjm [15]3 years ago
8 0

Answer:

1. element symbol = Kr

2. atomic mass = 83.80

3. Atomic number = 36

Tomtit [17]3 years ago
3 0
I think it is but 1. Element symbol
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Help with this. <br><br><br><br><br><br> ....
Dovator [93]

Answer:

Explanation:

The x component is the adjacent side making up the given angle (39.4)

The vector is the hypotenuse.

The definition of the cos (x) is adjacent / hypotenuse.

cos(39.4) = adjacent / 47.3      Multiply both sides by 47.3

47.3 * cos(39.4) = adjacent      Cos(39.4) = 0.7727

adjacent = 36.55

4 0
3 years ago
What color is reflected off of most plant leaves?
serg [7]
Green is reflected off of most plant leaves.
3 0
3 years ago
Read 2 more answers
Define mechanics and describe its three major divisions
Andrew [12]

<u>Mechanics</u> is the branch of physics which deals with the study of motion of material objects.

<u><em>Divisions</em></u>

There are three major division of mechanics

Statics

Kinematics

Dynamics.

4 0
3 years ago
The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
s2008m [1.1K]

Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

6 0
3 years ago
A cubical Gaussian surface surrounds two positive charges, each has a charge q 1 1 = + 3.90 × 10 − 12 3.90×10−12 C, and three ne
Masteriza [31]

Answer:

The electric flux is zero because charge is zero.

Explanation:

Given that,

Positive charge q_{1}=3.90\times10^{-12}\ C

Negative charge q_{2}=-2.60\times10^{-12}\ C

We need to calculate the total charged

Using formula of charge

Q_{enc}=2q_{1}+3q_{2}

Put the value into the formula

Q_{enc}=2\times3.90\times10^{-12}+3\times(-2.60\times10^{-12})

Q_{enc}=0

We need to calculate the electric flux

Using formula of electric flux

\phi=\dfrac{Q_{enc}}{\epsilon_{0}}

Put the value into the formula

\phi=\dfrac{0}{8.85\times10^{-12}}

Hence, The electric flux is zero because charge is zero.

7 0
3 years ago
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