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Katarina [22]
3 years ago
9

Which of these is transparent?

Physics
2 answers:
forsale [732]3 years ago
8 0
A. a glass jar. obvious.
jolli1 [7]3 years ago
3 0
The answer is definitely A. A transparent material is a material that allows light to pass through it and can be seen through. A glass jar can be seen through. Some people might say B, which is a mirror, but a mirror cannot be seen through, it only shows reflection. Hope i helped. Have a nice day and Happy New Year.
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Young's Modulus refers to changes in the a Volume b- Length c- Body layers
Novosadov [1.4K]

Explanation:

Young' modulus is the ratio of normal stress to the longitudinal strain. Mathematically, it is given by :

Normal stress is given by force per unit area. Longitudinal strain is the change in length per unit original length.

The mathematical definition of Young's modulus is given by :

Y=\dfrac{F/A}{\Delta L/L}..........(1)

Where

\Delta L is the change in length

F is the force

A is the area of cross section

So, the Young's modulus refers to the change in length of the object. Hence, the correct option is (b) "length".

8 0
3 years ago
Two small nonconducting spheres have a total charge of 93.0 μC . Part A
Travka [436]

Answer:

charge on each

Q1 = 2.06 ×10^{-5}  C

Q2 = 7.23 × 10^{-5} C

when force were attractive

Q1 = 1.07 × 10^{-4} C

Q2 = -1.39 × 10^{-5} C

Explanation:

given data

total charge = 93.0 μC

apart distance r  = 1.14 m

force exerted  F = 10.3 N

to find out

What is the charge on each and What if the force were attractive

solution

we know that force is repulsive mean both sphere have same charge

so total charge on two non conducting sphere is

Q1 + Q2 = 93.0 μC  = 93 ×10^{-6} C

and

According to Coulomb's law force between two sphere is

Force F = \frac{K Q1 Q2 }{r^2}      .........1

Q1Q2 = \frac{F*r^2}{k}

here F is force and r is apart distance and k is 9 × 10^{9} N-m²/C² put all value we get

Q1Q2 = \frac{ 10.3*1.14^2}{9*10^9}

Q1Q2 = 1.49 ×  10^{-9} C²

and

we have  Q2 = 93 ×10^{-6} C - Q1

put here value

Q1²  - 93 ×10^{-6}  Q1 + 1.49 ×  10^{-9} = 0

solve we get

Q1 = 2.06 ×10^{-5}  C

and

Q1Q2 = 1.49 ×  10^{-9}

2.06 ×10^{-5}  Q2 = 1.49 ×  10^{-9}

Q2 = 7.23 × 10^{-5} C

and

if force is attractive we get here

Q1Q2 = - 1.49 ×  10^{-9} C²

then

Q1²  - 93 ×10^{-6}  Q1 - 1.49 ×  10^{-9} = 0

we get here

Q1 = 1.07 × 10^{-4} C

and

Q1Q2 = - 1.49 ×  10^{-9}

2.06 ×10^{-5}  Q2 = - 1.49 ×  10^{-9}

Q2 = -1.39 × 10^{-5} C

5 0
3 years ago
Please help me with my Physical Science! 50 POINTS
DaniilM [7]

Answer:

1, When Jane brakes, the brakes slow the car wheels turning and the road surface exerts a backwards force on the tires, causing the car to decelerate. The pocket book tends to continue on in a straight line (Newton's first law). If she brakes hard enough that the friction between the book and the car seat is insufficient to decelerate the book as fast as the car is decelerating, the book will slide off the seat, and gravity pulls it to the floor

2.

When the diver uses his / her force to depress the springboard, the springboard pushes him back with equal force

3.Newton's Second Law (F=ma)

4. 5 N

5. 19.5 N

65kg * 0.3 m/s^2

6.0.2 N/s

10kg divided by 2N

7.-Walking then pushing the moving forward

-Dribbling

-Basketball is pushed but bounces back

Explanation:

6 0
4 years ago
Read 2 more answers
WILL GIVE BRAINLIEST!!!!!!!!!!!! 50 PTS!!!!!!!!!!!!!!! 10 SCIENCE QUESTIONS!!!!
Jet001 [13]
1.b 2.c 3.a 4.b 5.d 6.a 7.a 9.idk 10.c hope thes helps have a nice day and God bless
8 0
3 years ago
calculate the spring constant if a weight of 250N is added to a spring which increases in length by 20cm
ZanzabumX [31]
Since, F = k . ∆x

Therefore, k = F / ∆x = 250 / 0.2 = 1250 N/m

(ps: convert 20 cm into 0.2 m)
8 0
3 years ago
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