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Maslowich
3 years ago
5

Can any wan help pls

Physics
2 answers:
Nimfa-mama [501]3 years ago
4 0

Answer:

b

Explanation:

because of bonds and the bowl is used for heating purpose

Goryan [66]3 years ago
4 0

Answer:

I think b not sure but I hope its right and is that the nwea test??

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A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread non uniformly throu
Aloiza [94]
In other words a infinitesimal segment dV caries the charge 
<span>dQ = ρ dV </span>

<span>Let dV be a spherical shell between between r and (r + dr): </span>
<span>dV = (4π/3)·( (r + dr)² - r³ ) </span>
<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
<span>= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) </span>
<span>drop higher order terms </span>
<span>= 4·π·r²·dr </span>

<span>To get total charge integrate over the whole volume of your object, i.e. </span>
<span>from ri to ra: </span>
<span>Q = ∫ dQ = ∫ ρ dV </span>
<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
<span>= ∫ri→ra { 4·π·b·r } dr </span>
<span>= 2·π·b·( ra² - ri² ) </span>

<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
<span>= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) </span>
<span>= 3.77×10⁻⁸C </span>
<span>= 37.7nC</span>
6 0
4 years ago
True or false: seismographs from three locations are studied to calculate the epicenter of am earthquake
astraxan [27]
True........don't blame me if i'm wrong.<span />
7 0
3 years ago
A stone is dropped from the upper observation deck of a tower, 750 m above the ground. (assume g = 9.8 m/s2.) (a) find the dista
Gekata [30.6K]
(a) The stone moves by uniform accelerated motion, with constant acceleration g=9.81 m/s^2 directed downwards, and its initial vertical position at time t=0 is 750 m. So, the vertical position (in meters) at any time t can be written as
y(t)= y_0 -  \frac{1}{2}gt^2= 750 - 4.9 t^2

(b) The time the stone takes to reach the ground is the time at which the vertical position of the stone becomes zero: y(t)=0. So, we can write
750-4.9 t^2 = 0
from which we find the time t after which the stone reaches the ground:
t= \sqrt{\frac{750 m}{4.9 m/s^2 }}= 12.37 s

(c) The velocity of the stone at time t can be written as
v(t) = -gt
because it is an accelerated motion with initial speed zero. Substituting t=12.37 s, we find the final velocity of the stone:
v(12.37 s)=-(9.81 m/s^2)(12.37 s)=-121.3 m/s

(d) if the stone has an initial velocity of v_0 = 6 m/s, then its law of motion would be
y(t)=y_0 - v_0t -  \frac{1}{2}gt^2
and we can find the time it needs to reach the ground by requiring again y(t)=0:
0=750 - 6t - 4.9 t^2
which has two solutions: one is negative so we neglect it, while the second one is t=11.78 s, so this is the time after which the stone reaches the ground.

5 0
3 years ago
What best decribes the science method
fiasKO [112]
Making prediction , hypothesis and guessing to see if it’s right , and leading
3 0
3 years ago
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Which statement best describes the density of the outer planets?
gavmur [86]

Answer:

Saturn is mainly composed of the lightest two gases known, hydrogen and helium. It is the only planet in our solar system whose density is less than water.

Explanation:

5 0
4 years ago
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