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icang [17]
3 years ago
12

Powers of 10

Physics
1 answer:
Viktor [21]3 years ago
7 0

Answer:

es inglés plis españolejjdkdjdd

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1. Write down the readings on the side of Figs. 6 (a), (b) and (c) respectively. What is the least count of instrument scale for
alexandr402 [8]

Explanation:

This should be right. if any doubt post a comment.

3 0
2 years ago
How do you find the atomic number of a atom?
Vlada [557]

Answer:  To find the atomic mass of an atom of an element, add up the mass of protons and neutrons.

Explanation: Find the atomic mass of an isotope of carbon that has 7 neutrons. You can see from the periodic table that carbon has an atomic number of 6, which is its number of protons.

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3 years ago
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9. The three types of stress that act on Earth's rocks are compression, tension, and A. strain. B. shear. C. tephra. D. shale.
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2 years ago
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Two conducting parallel plates 5.0 × 10−3 meter apart are charged with a 12-volt potential
photoshop1234 [79]

Answer: 2.4×10^-3 v/m

Explanation: distance between plates of capacitor (d) =5.0×10^-3m

Potential difference between plates (v) = 12v

Force on electronic charge (f) = 3.8×10^-16 N

Strength of electric field (E) =?

The formulae that relates potential difference, eoectiic field strength and distance between plates is given as

v = Ed

By substituting the parameters, we have that

12 = E × 5.0×10^-3

E = 12/ 5.0 × 10^-3

E = 2.4×10^-3 v/m

7 0
3 years ago
A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s
Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g=4.67\times 10^{-3} kg

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1,m_1=1177g=1.177kg

Mass of block 2,m_2=1626 g=1.626 kg

Velocity of block 1,v_1=0.681m/s

(a)

Let velocity of the second block  after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

mv=m_1v_1+(m+m_2)v_2

4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2

1.66719=0.801537+1.63067v_2

1.66719-0.801537=1.63067v_2

0.865653=1.63067v_2

v_2=\frac{0.865653}{1.63067}

v_2=0.531m/s

Hence, the  velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

K_i=\frac{1}{2}mv^2

k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)

k_i=297.59 J

Final kinetic energy after collision

K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2

K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2

K_f=0.5028 J

Now, he ratio of the total kinetic energy after the collision to that before the collision

=\frac{k_f}{k_i}=\frac{0.5028}{297.59}

=0.00169

5 0
3 years ago
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