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Vlad1618 [11]
3 years ago
13

Can anyone help me with 11 and 12 please

Physics
1 answer:
dexar [7]3 years ago
7 0
The 1st one is basically B because it will stay in motion with the same speed and in the same direction unless acted on by an unbalanced force and the 2cd one is A because most of is transformed into thermos energy. hope this helps!
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NEEDD SOME HELP ASAP HELP MEHHH<br> 50 pOINTS
Vadim26 [7]
An electron shell can hold 2(n^2) electrons (technically) where n is the shell number, i.e. shell 1 can hold 2, shell 2 can hold 8, 3 holds 18 and so on.
The atomic number of Nitrogen is 7, i.e. it has 7 electrons (to match its 7 protons, assuming it isn't an ion).

With the atomic number, you simply start from shell 1 and work out. So we put 2 electrons in shell 1, leaving us with 5 left. Shell 2 can hold 6 so we can fit all 5 in.

In other words, you should have 2 electron shells on the atom, shell 1 with 2 e- and shell 2 with 5 e-.
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2 years ago
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How much work is accomplished when a force of 300N pushes a box across the floor for a distance of 100 meters?
Nesterboy [21]

So the correct ans is B.

hope it helps u.

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3 years ago
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A resistor, capacitor, and switch are all connected in series to an ideal battery of constant terminal voltage. Initially, the s
erma4kov [3.2K]

Answer:

The voltage across the resistor is zero, and the voltage across the capacitor is equal to the terminal voltage of the battery.

Explanation:

This is because when a capacitor is charged no current or voltage flows through it so it will have a voltage equal to the terminal voltage of the battery

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3 years ago
A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?
Tcecarenko [31]
The weight of the meterstick is:
W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance 
d_1 = 0.50 m - 0.40 m=0.10 m
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
(mg) d_2 = 0.20 Nm
from which we find the value of d2:
d_2 =  \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
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3 years ago
A 0.500 kg rock is whirled in a vertical circle of a radius 0.60 m . the velocity of the rock at the bottom of the swing is 4.0
OLEGan [10]

Explanation:

Centripetal acceleration is:

a = v² / r

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a = 26.6 m/s²

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