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Kruka [31]
4 years ago
15

What is the approximate average ocean depth along the continental shelves bordering north america?

Physics
1 answer:
Flura [38]4 years ago
7 0
Less than 150 M aprox.
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CAN SOMEONE PLEASE HELP ME WITH MY PHYSICS QUESTIONS? I NEED CORRECT ANSWERS ONLY!
BARSIC [14]
<h2>Right answer: acceleration due to gravity is always the same </h2><h2 />

According to the experiments done and currently verified, in vacuum (this means there is not air or any fluid), all objects in free fall experience the same acceleration, which is <u>the acceleration of gravity</u>.  

Now, in this case we are on Earth, so the gravity value is 9.8\frac{m}{s^{2}}  

Note the objects experience the acceleration of gravity regardless of their mass.

Nevertheless, on Earth we have air, hence <u>air resistance</u>, so the afirmation <em>"Free fall is a situation in which the only force acting upon an object is gravity" </em>is not completely  true on Earth, unless the following condition is fulfiled:

If the air resistance is <u>too small</u> that we can approximate it to <u>zero</u> in the calculations, then in free fall the objects will accelerate downwards at 9.8\frac{m}{s^{2}} and hit the ground at approximately the same time.  


5 0
4 years ago
A circular loop of wire 78 mm in radius carries a current of 114 A. Find the (a) magnetic field strength and (b) energy density
Finger [1]

Explanation:

It is given that,

Radius of loop, r = 78 mm = 0.078 m

Current, I = 114 A

(a) Magnetic field strength of the circular loop is given by :

B=\dfrac{\mu_oI}{2R}

B=\dfrac{4\pi \times 10^{-7}\times 114}{2\times 0.078}

B = 0.000918 T

or

B=9.18\times 10^{-4}\ T

(b) Energy density at the center of the loop is given by :

U=\dfrac{B^2}{2\mu_o}

U=\dfrac{(0.000918)^2}{2\times 4\pi \times 10^{-7}}

U=0.335\ J/m^3

Hence, this is the required solution.

7 0
3 years ago
Someone help me with letter d. and e. I’m supposed to solve for y.
Bumek [7]

Answer:

t=\frac{a}{\sqrt{1-\frac{y^2}{c^2}}}\\\frac{y^2}{c^2}=1-\frac{a^2}{t^2}\\y^2=\frac{c^2t^2-c^2a^2}{t^2}\\y=\frac{c}{t}\sqrt{t^2-a^2}

t=\frac{a}{\sqrt{1-\frac{v^2}{y^2}}}\\\frac{v^2}{y^2}=1-\frac{a^2}{t^2}\\\\\frac{v^2}{y^2}=\frac{t^2-a^2}{t^2}\\y^2=\frac{v^2t^2}{t^2-a^2}\\y=\frac{vt}{\sqrt{t^2-a^2}}

7 0
3 years ago
if three resistors are connected in series each have the same value of 5 ohms what is the total resistance ​
Sphinxa [80]
15 you get 3 Multiply it by 5 and get 15
7 0
3 years ago
Two people push on a boulder to try to move it. The mass of the boulder is 825 kg. One person pushes north with a force of 64 N.
stira [4]

Answer:

0.09 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity.

The S.I unit of acceleration is m/s².

From the question, expression for acceleration is given as

F' = ma  

Using Pythagoras Theory,

√(F₁²+F₂²) = ma................... Equation 1

Where F₁ = Force of the First person on the boulder, F₂ = Force of the Second person on the boulder, F' = resultant force acting on the boulder, m = mass of the boulder, a = acceleration of the boulder.

make a the subject of the equation

a = √(F₁²+F₂²) /m................ Equation 2

Given: m = 825 kg, F₁ = 64 N, F₂ = 38 N,

Substitute into equation 2

a = [√(64²+38²)]/825

a = {√(5540)}/825

a = 74.43/825

a = 0.09 m/s²

4 0
3 years ago
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