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Westkost [7]
3 years ago
6

A scuba diver measures an increase in pressure of around 10^5 Pa upon descending by 10 m, what is the change in force per square

centimetre on the diver's body?
Physics
1 answer:
IceJOKER [234]3 years ago
7 0

Answer:

Explanation:

Given that, .

Pressure around scuba is

P = 10^5 Pa

1 Pa = 1 N/m²

Then

P = 10^5 N/m²

Descending height

h = 10m

Change in force per unit square centimetre

We know that,

Pressure = Force / Area

Then,

Force / Area is the required question we are finding

Then,

Force / Area = 10^5 N / m²

So, let convert the m² to cm²

100cm = 1m

(100cm)² = (1m)²

10⁴cm² = 1m²

Then,

Force / Area = 10^5 N/m² × 1m² / 10⁴cm²

Force / Area = 10 N/cm²

So, the force per unit square centimeters is 10.

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The magnetic field over a certain range is given by B~ = Bx ˆı + By ˆ, where Bx = 2 T and By = 4 T. An electron moves into the
krek1111 [17]

Answer:

Explanation:

The force exerted in a magnetic field is given as

F = q (v × B)

Where

F is the force entered

q is the charge

v is the velocity

B is the magnetic field

Given that,

The magnetic field is

B = 2•i + 4•j. T

The velocity of the electron is

v = 2•i + 6•j + 8•k. m/s

Also, the charge of an electron is

q = -1.602 × 10^-19 C.

Then note that,

V×B is the cross product of the speed and the magnetic field

Then,

F = q (V×B)

F = -1.602 × 10^-19( 2•i + 4•j +8•k × 2•i + 4•j)

Note

i×i=j×j×k×k=0

i×j=k. j×i=-k

j×k=i. k×j=-i

k×i=j. i×k=-j

F = -1.602 × 10^-19[(2•i + 4•j +8•k) × (2•i + 4•j)]

F = -1.602 × 10^-19 [2×2•(i×i) + 2×4•(i×j) + 4×2•(j×i) + 4×4•(j×j) + 8×2•(k×i) + 8×4•(k×j)]

F = -1.602 × 10^-19[4•0 + 8•k + 8•-k + 16•0 + 16•j + 32•-i]

F = -1.602 × 10^-19(0 + 8•k - 8•k + 0 + 16•j - 32•i)

F = -1.602 × 10^-19(16•j - 32•i)

F = -1.602 × 10^-19 × ( -32•i + 16•j)

F = 5.126 × 10^-18 •i - 2.563 × 10^-18 •j

Then, the x component of the force is

Fx = 5.126 × 10^-18 N

Also, the y component of the force is

Fy = -2.563 × 10^-18 N

8 0
3 years ago
How many amps are there in the current of a circuit if the voltage source is 140v and the resistance is 2 ohms?
VLD [36.1K]
Voltage (V)= Current (I) * Resistance (R)
I=V/R=140/2=70A

Hope this helps!
7 0
3 years ago
Why is the weight of the ruler not include in the calculation of moment?​
Feliz [49]

Answer:

Since the distance between the pivot and its weight is zero, the moment of its weight about the pivot (= weight × 0) is zero. Hence, the weight of the ruler can be ignored.

Hope this helps:)

3 0
3 years ago
If 2.0×10^−4 C of charge passes a point in 2.0×10^−6 s , what is the rate of current flow?1.0×10^−10 A1.0×10^2 A4.0×10^−1 A4.0×1
Ipatiy [6.2K]

This problem is about the rate of the current. It's important to know that refers to the quotient between the electric charge and the time, that's the current rate.

I=\frac{Q}{t}

Where Q = 2.0×10^−4 C and t = 2.0×10^−6 s. Let's use these values to find I.

\begin{gathered} I=\frac{2.0\times10^{-4}C}{2.0\times10^{-6}\sec } \\ I=1.0\times10^{-4-(-6)}A \\ I=1.0\times10^{-4+6}A \\ I=1.0\times10^2A \end{gathered}

<em>As you can observe above, the division of the powers was solved by just subtracting their exponents.</em>

<em />

<h2>Therefore, the rate of the current flow is 1.0×10^2 A.</h2>
3 0
1 year ago
Two circular holes, one larger than the other, are cut in the side of a large water tank whose top is open to the atmosphere. Ho
lidiya [134]

Answer:\frac{r_1}{r_2}=1.565

Explanation:

Given

two holes are made with different sizes

Hole 1 is large in size and hole 2 is small

If the volume flow rate of water is same for both the hole then small hole must be below the large hole because for same flow rate, velocity of water is large while cross-sectional area is small so it compensate to give same flow for both the holes.

Now for radius apply Bernoulli's theorem at hole 1 and 2

P_1+\rho gh_1=P_{atm}+\frac{1}{2}\rho v_1^2

P_2+\rho g6h_2=P_{atm}+\frac{1}{2}\rho v_2^2

if hole 1 is h distance below water surface then h_2=6h

and P_1=P_2=P_{atm}

Also v_1=\sqrt{2gh}

v_2=\sqrt{2g(6h)}

and Q=A_1v_1=A_2v_2

A=\pi r^2

thus \dfrac{r_1}{r_2}=\sqrt{\dfrac{v_2}{v_1}}

\dfrac{r_1}{r_2}=\sqrt{\dfrac{\sqrt{6h}}{\sqrt{h}}}

\frac{r_1}{r_2}=1.565

5 0
3 years ago
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