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Dennis_Churaev [7]
3 years ago
13

Downwelling is the process that moves cold, dense water from the ocean surface to the seafloor near the polar regions. how can d

ownwelling affect the ocean water around the poles?
Physics
1 answer:
mash [69]3 years ago
5 0

Explanation:

Downwelling is the process where cold and heavy dense water moves down into the ocean floor and warm light dense water rises to the surface. As a result of downwelling, the water high dense water which rises to the water surface brings the oxygen rich water to the surface for the marine animals to breathe properly. Also when the ocean surface water becomes little warmer it becomes a little comfortable for the marine animals to survive in this severely cold climatic conditions at polar reasons.

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A 20-cm long solenoid consists of 100 turns of a coil of radius r = 3.0 cm. A current of Io in the coiled wire produces a magnet
Romashka-Z-Leto [24]

Answer:

vi) Double the current in the wire, and double the number of turns in the 20-cm long solenoid

Explanation:

The magnetic field inside the solenoid and the current flowing in the coil of solenoid are related to each other by the following equation

B₀=μ₀nI₀

Where,

B₀ is the magnetic field in the middle of solenoid

n is the number of turns in the coil of solenoid

I₀ is the current flowing in the coil of solenoid

In the above equation, as μ₀ is a constant so the magnetic field will be directly proportional to the number of turns multiplied by the current. So, changing the radius of the coil or length of the coil will have no effect on the magnetic field.

As we have to increase the magnetic field by 4 times, we need to double the current as well as the number of turns as mentioned in the option vi.

3 0
3 years ago
Read 2 more answers
Water is discharged from a pipeline at a velocity v (in ft/sec) given by v = 1306p(1/2), where p is the pressure (in psi). If th
shepuryov [24]

Answer:

a=38.5 ft/sec^{2}

Explanation:

Note that acceleration is the rate change of velocity i.e

acceleration=\frac{change in velocity}{change in time}\\a=\frac{dv}{dt} \\.

Since the velocity is giving as a variable dependent on the pressure, we have to differentiate implicitly both side with respect to time,i.e

\frac{dv}{dt}=1306*(1/2)p^{-1/2}\frac{dp}{dt} \\

if we substitute value for the pressure as giving in the question and also since the rate change of pressure is 0.354psi/sec, we have

a=653*0.1667*0.354\\

a=\frac{dv}{dt}=653(36)^{-1/2}*0.354\\  a=38.5 ft/sec^{2}

3 0
3 years ago
A 0.140-kg baseball is dropped from rest. It has a speed of 1.20 m/s just before it hits the ground, and it rebounds with an upw
irga5000 [103]

Answer:

22 N upward

Explanation:

From the question,

Applying newton's second law of motion

F = m(v-u)/t....................... Equation 1

Where F = Average force exerted by the ground on the ball, m = mass of the baseball, v = final velocity, u = initial velocity, t = time of contact

Note: Let upward be negative and downward be positive

Given: m = 0.14 kg, v = -1.00 m/s, u = 1.2 m/s, t = 0.014 s

Substitute into equation 1

F = 0.14(-1-1.2)/0.014

F = 0.14(-2.2)/0.014

F = 10(-2.2)

F = -22 N

Note the negative sign shows that the force act upward

6 0
3 years ago
Convert 1,265,341 mm to m.
Thepotemich [5.8K]
The answer in Meters is going to to 1265.341
7 0
3 years ago
Read 2 more answers
One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side
anyanavicka [17]

Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

L1^2=L2^2=L^2=2^2+(H/2)^2  Replacing this value in the previous equation:

\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34  Solving for H:

H=\sqrt{52}

We can now, calculate the angle between L1 and the 2m segment:

\alpha = atan(\frac{H/2}{2})=60.98°

If we make a sum of forces in the midpoint of the rope we get:

-2*T*cos(\alpha ) + F = 0  where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

T=Fc=\frac{F}{2*cos(\alpha) } = 89.67N

7 0
3 years ago
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