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svp [43]
3 years ago
15

Jenna dives 20 meters into the ocean. how much pressure does she feel?

Physics
1 answer:
dexar [7]3 years ago
7 0
3 bar. 1 bar normal air pressure and 2 bar for being 20 m underwater
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How is the scientific use of the term digital different from the common use
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We commonly know refer to something 'digital' has to something electronic that can be visibly seen such as a watch, clock, camera, screen, etc.  It really refers to stored energy or electricity that's not natural.  But the word 'digital' in science refers to the depiction of data<span> or </span>information<span> in </span>figures<span> (such as in a </span>table<span>) in contrast to as a </span>chart<span>, </span>graph<span>, </span>drawing<span>, or other pictorial </span>form.<span>

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A student has the following equipment - copper wire, nail made of iron,a battery, paperclips.
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The first step would be to create an electromagnet. You can create an electromagnet by winding a copper wire around the nail, the connect both ends to the battery. A current would start flowing around the nail through the wire, creating an electromagnet with its own magnetic field. Next, bringing the electromagnet to the mixture of copper and iron would slowly attract the pieces of iron (as copper is weakly magnetic). Do this slowly and the iron pieces would all slowly be separated from the copper pieces.
4 0
3 years ago
The charge density of a uniformly charged disk 0.420 m in diameter is 2.92 ✕ 10−2 C/m2. What is the magnitude of the electric fi
iragen [17]

Answer:

E = {(Charge Density/2e0)*(1 - [z/(sqrt(z^2 - R^2))]}

R is radius = Diameter/2 = 0.210m.

At z = 0.2m,

Put z = 0.2m, and charge density = 2.92 x 10^-2C/m2, and constant value e0 in the equation,

E can be calculated at distance 0.2m away from the centre of the disk.

Put z = 0.3m and all other values in the equation,

E can be calculated at distance 0.3m away from the centre of the disk

3 0
3 years ago
Balance the equation​
mote1985 [20]
Is this a chemistry question ?
7 0
2 years ago
A 0.877-kg ball has an elastic, head-on collision with a second ball that is initially at rest. The second ball moves off with h
Art [367]

Answer: the mass of the second ball is 2.631 kg

Explanation:

Given that;

m1 = 0.877 kg

Initial velocity = V0

Initial momentum = m1 × V0

final velocity of m1 is u1, final velocity of m2 is u2 = v0/2

now final momentum = m1 × u1 + m2 × u2

using momentum conservation;

m1×V0 = m1×u1 + m2×v0/2

m1×(v0 - u1) = m2×V0/2 ----- let this be equation 1

Now, for elastic collision;

m1×v0²/2 = m1×u1²/2 + m2×(v0/2)²/2

m1×(v0² - u1²) = m2×(v0/2)² --------- let this be equation 2

now; equation 2 / equation 1

: V0 + u1 = v0/2

2V0 + 2u1 = V0

2u1 = V0 - 2V0

u1 = -V0/2

now we insert in equ 1

 m1×3V0/2= m2×V0/2

m1 × 3 = m2

m2 = 0.877 × 3

m2 = 2.631 kg

Therefore, the mass of the second ball is 2.631 kg

7 0
3 years ago
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