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3 years ago

What replaces a cold current that sinks to the ocean floor?

Physics

1 answer:

FrozenT [24]3 years ago

5 0

Well simple the warm water then replaces the cold current that sinks to the ocean floor.

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Can ya pls answer dis rq!

lana66690 [7]

Answer:

For vegetarians beans

Explanation:

Because it's plant protein and very proteinious

5 0

2 years ago

Which of the following represents a joule?

Andru [333]

Answer:

d. N*m

Explanation:

Workdone which is measured in Joules is calculated using the formula;

Workdone = force * distance

Where;

Force is measured in Newton

Distance is measured in meters.

Therefore, workdone also has the measurement of Newton-meters.

5 0

3 years ago

What causes dust particles to move with brownian motion answers?

valina [46]

When atoms that are shaking with kinetic energy collide with the dust particles, it causes the dust particles to move with the Brownian motion.

Hope this helps ;)

3 0

3 years ago

Two current-carrying wires are parallel to each other. The current in one is increased by a factor of 2 , the current in the oth

ladessa [460]

Answer:

The new force $F_N$ will be $\frac{6}{13}$ times the old force F. The change then will be $\Delta F=-\frac{7}{13}F$

Explanation:

The force between two current-carrying parallel wires is calculated with the formula:

$F=\frac{\mu_0I_1I_2\Delta L}{2\pi r}$

where r is the distance between them, $\Delta L$ a portion of length of the wires we consider, $I_1$ and $I_2$ their current intensity and $\mu_0=4\pi\times10^{-7}N/A^2$ the vacuum permeability.

If the current in one wire is increased by a factor of 2, the current in the other is increased by a factor of 3 , and the distance between the wires is decreased by a factor of 13, then we would have a new situation N where (considering the previous variables as an initial situation):

$I_{N1}=2I_1\\I_{N2}=3I_2\\r_N=13r\\$

And the force then will be:

$F_N=\frac{\mu_0I_{N1}I_{N2}\Delta L}{2\pi r_N}=\frac{\mu_02I_13I_2\Delta L}{2\pi 13r}=\frac{6(\mu_0I_1I_2\Delta L)}{13(2\pi r)}=\frac{6}{13}F$

So the change will be:

$\Delta F=F_N-F=\frac{6}{13} F-F=(\frac{6}{13} -1)F=-\frac{7}{13}F$

5 0

3 years ago

A small block is placed at height h on a frictionless 30 degree ramp. Upon being released the block slides down the ramp and the

Kamila [148]

After leaving the plane, the block will have an unknown speed (S),

which can be broken into x,y components.

The x,y kinematics are: x – 1

x0 - 0 V - ? V0 - Scos(-30)

a – 0

t - t

y - 0

y0 – 1

V - ?

V0 - Ssin(-30)

a - -9.8

t – t

We then use x=x0+v0t+.5at^2

in the x case: 1=0+Scos(-30)+.5(0)t^2

Solving for t gives t=1/ Scos(-30)

in the y case,

with t-substitution:

0=1+Ssin(-30)*1/Scos(-30)+.5(-9.8)(1/Scos(-30))^-2

In the middle velocity term, S cancels out. Multiplying all known numbers as well as squaring the third term gives:

0=1-.5774-6.5333/S^2

Solving for S = S = 3.9319 m/s

Now with a mark on final ramp speed, we can now make a 3rd kinematics equation. The acceleration will be altered from gravity:

Slide force = 9.8*sin(30) = 4.9 m/s^2.

x - ?

x0 – 0

V - 3.9319

V0 – 0

a - 4.9

t - ?

So the equation we use is V2 = V02+2a(x-x0). 3.93192=0+2*4.9*(x-0)

Solving for x gives x=1.5775 m up the ramp.

So we now look for the y component of the ramp length:

1.5775*sin(30) = .78875 m 'high' on the ramp.

8 0

3 years ago