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Firdavs [7]
4 years ago
7

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

Physics
1 answer:
FrozenT [24]4 years ago
5 0
Well simple the warm water then replaces the cold current that sinks to the ocean floor.
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May you please help me! I need answers fast!!
nataly862011 [7]
1. sliding 2. static 4. rolling
5 0
3 years ago
Using 6400 km as the radius of Earth, calculate how high above Earth’s surface you would have to be in order to weigh 1/16th of
FrozenT [24]
Gravity obeys the inverse square law.  At 6400 km above the center of the Earth (Earth's surface) you weigh x.  Twice that reduces your weight to 1/4th.  Four times that height reduces your weight to 1/16th.  4 times 6400 km is 25,600 km.  But that is above the center of the earth, and the question requests the height above the surface, so we deduct 6400 km to arrive at our final answer:  19,200 km.

Incidentally, it doesn't exactly work the opposite way.  At the center of the Earth the mass would be equally distributed around you, and you would therefore be weightless.
6 0
4 years ago
URGENT! 50pts! Help with rotational speed questions please?
EastWind [94]
1). If you want the wheel to turn as one solid piece, then all parts of it must have the same RATE of rotation. If one part was going 2 RPM and another part was going 3 RPM, then there's no way both parts could stay hooked together. Their SPEED depends on their distance from the center, but they all have to make the same RPM. 2). The CENTRIPETAL acceleration of anything that's rotating is. m-v-squared/ R. The object's speed depends on its radius, and it's acceleration varies directly with the square of the speed. So if you move in to half the radius, the acceleration becomes 1/4 of the original value.
7 0
3 years ago
Even when the head is held erect, as in the figure below, its center of mass is not directly over the principal point of support
alexandr1967 [171]

We are asked to determine the force required by the neck muscle in order to keep the head in equilibrium. To do that we will add the torques produced by the muscle force and the weight of the head. We will use torque in the clockwise direction to be negative, therefore, we have:

\Sigma T=r_{M\perp}(F_M)-r_{W\perp}(W)

Since we want to determine the forces when the system is at equilibrium this means that the total sum of torque is zero:

r_{M\perp}(F_M)-r_{W\perp}(W)=0

Now, we solve for the force of the muscle. First, we add the torque of the weight to both sides:

r_{M\perp}(F_M)=r_{W\perp}(W)

Now, we divide by the distance of the muscle:

(F_M)=\frac{r_{W\perp}(W)}{r_{M\perp}}

Now, we substitute the values:

F_M=\frac{(2.4cm)(50N)}{5.1cm}

Now, we solve the operations:

F_M=23.53N

Therefore, the force exerted by the muscles is 23.53 Newtons.

Part B. To determine the force on the pivot we will add the forces we add the vertical forces:

\Sigma F_v=F_j-F_M-W

Since there is no vertical movement the sum of vertical forces is zero:

F_j-F_M-W=0

Now, we add the force of the muscle and the weight to both sides to solve for the force on the pivot:

F_j=F_M+W

Now, we plug in the values:

F_j=23.53N+50N

Solving the operations:

F_j=73.53N

Therefore, the force is 73.53 Newtons.

8 0
1 year ago
Particles can enter the air by
8090 [49]

Answer:

the answer is diffusion

7 0
3 years ago
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