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3 years ago

What type of climate would be near cold ocean currents? PLEASE HELP QUICK!!

Physics

1 answer:

olya-2409 [2.1K]3 years ago

5 0

Answer:

hot humid with lots of rain.

Explanation:

ocean currents act as conveyer belts of warm and cold water sending heat to the polar regions and helping the tropical areas cool off, thus influencing both weather and climate. the tropics are particularly rainy because heat absorption , and thus ocean evaporation, is highest.

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In a nuclear reactor, the control rods are: A.made of graphite B.used to keep the reactor cool C.used to absorb free neutrons D.

algol13

I think its E. the control rods are used to control <span>the fission rate of uranium and plutonium. Hope this helps!! </span>

4 0

3 years ago

Read 2 more answers

suppose you are walking home after school. The distance from school to your home is five kilometers. on foot you can get home in

agasfer [191]

If it takes ten minutes to travel 5 km, then it'll take 60 minutes to travel thirty, making your biking speed 30 kilometers per hour. 25 minutes divided by ten is 2.5. multiply 5 by 2.5 to get 12.5, your average walking speed.

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3 years ago

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child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a s

Strike441 [17]

Answer:

θ = 13.7º

Explanation:

According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:

$\Delta K + \Delta U = W_{nc} (1)$

ΔK, is the change in kinetic energy, as follows:

$\Delta K = \frac{1}{2}* m* (v_{f} ^{2} - v_{0} ^{2}) (2)$

ΔU, is the change in the gravitational potential energy.
If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:

$\Delta U = 0 - m*g*h = -m*g*d* sin\theta (3)$

Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

$W_{nc} = F_{f} * d * cos 180\º \\\\ = 0.2*m*g*d* cos 180\º = -0.2*m*g*d (4)$

Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

$\frac{1}{2}* (v_{f} ^{2} - v_{0} ^{2}) = g*d* sin\theta -0.2*g*d$

Replacing by the givens and the knowns, we can solve for sin θ, as follows: $\frac{1}{2}*( (4.30 m/s) ^{2} - (0.75 m/s)^{2}) = 9.8 m/s2*25.5m* sin\theta -0.2*9.8m/s2*25.5m\\ \\ 8.56 (m/s)2 = 250(m/s)2* sin \theta -50 (m/s)2\\ \\ sin \theta = \frac{58.6 (m/s)2}{250 (m/s)2} = 0.236$ ⇒ θ = sin⁻¹ (0.236) = 13.7º

8 0

2 years ago

To have momentum, an object must be ______. A. heavy B. massive C. moving D. accelerating

Amiraneli [1.4K]

Momentum of the object can be calculated by multiplying the mass of the object and the velocity of the moving object. In this case, the starting situation should be the object should be moving, else there is no velocity and thus momentum is equal to zero .Answer is C

3 0

3 years ago

Read 2 more answers

Three children are riding on the edge of a merry‑go‑round that has a mass of 10^5 and a radius of 1.80 m . The merry‑go‑round is

zimovet [89]

Answer: 22.01 rpm.

Explanation:

If we assume no external torques are present, the angular momentum must be conserved.

L1 = L2

I1 . ω1 = I2 ω2 (1)

I1 = MR2 /2 + m1. R2 + m2. R2 + m3. R2

I2 = MR2 /2 + m1. R2 + m3. R2

(as the 28.0 kg child moves to the center, he has no part anymore in the rotational inertia)

As I units are SI units, it is advisable to convert the angular velocity to rad/sec, as follows:

22.0 rev/min. (2π/rev) . (1min/60 sec) = 2.3 rad/sec

Solving for ω2 in (1):

ω2 = 2.31 rad/sec = 2.31 (1 rev/2 π). (60sec/1min) = 22.01 rpm

3 0

2 years ago