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Marrrta [24]
3 years ago
15

Why are ocean currents important?

Physics
1 answer:
Lelu [443]3 years ago
4 0
For the considerably longer periods– decades to millennia – which are relevant for climate change, the slightly larger heat capacity of the deep ocean<span> is </span>important. Ocean currents<span> and mixing by winds and waves can transport and redistribute heat to deeper </span>ocean<span> layers.</span>
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In anticipation of a long 10 upgrade, a bus driver accelerates at a constant rate of 5 ft/s2 while still on a level section of a
fgiga [73]

Answer:

S = 20903.4 ft

Explanation:

First we will determine the acceleration of the bus while it is moving upward

The equilibrium  equation would be

F - W sin\theta = ma'\\ma - W sin\theta = ma'\\m = \frac{W}{g} \\\frac{W}{g}*a - W sin\theta = \frac{W}{g} * a'\\\frac{a}{g} - sin\theta = \frac{a'}{g}\\\frac{x}{y} \frac{5}{32.2} - sin 10 = \frac{a'}{32.2} \\\a' = -0.59

Let the displacement be S_0

As per newton's third law of motion

v^2 -u^2 = 2as\\u = 80 \frac{mi}{h} =   117.33\frac{ft}{s}\\v = 50 \frac{mi}{h} =   73.33\frac{ft}{s}\\73.33 ^ 2 - 117.33^2 =  2 * (-0.59) * (S-S_0)\\\\S_0 = 0\\S = 20903.4

3 0
3 years ago
A body undergoes SHM of amplitude 3cm and frequency 20Hz. What is the
Harrizon [31]

Answer:

0 m/s

Explanation:

Acceleration is a maximum at the lowest and highest positions.  The velocity is 0 at these positions.

7 0
4 years ago
Agent Bond is standing on a bridge, 17 m above the road below, and his pursuers are getting too close for comfort. He spots a fl
faust18 [17]

Answer:

2 poles

Explanation:

We are told Agent Bond is standing on a bridge, 17 m above the road below.

Now, the bed of the truck is 1.5 m above the road.

It means the distance hewill need to fall = 17 - 1.5 = 15.5 m

acceleration due to gravity = 9.8 m/s²

Initial velocity = 0 m/s

Thus, to find the time, we will use the equation;

s = ut + ½at²

Plugging in the relevant values gives;

15.5 = 0 + ½9.8t²

Multiply through by 2 to give;

15.5 × 2 = 9.8t²

31 = 9.8t²

t² = 31/9.8

t = √(31/9.8)

t = 1.78 sec

We are told that he spots a flatbed truck approaching at 28 m/s and that the telephone poles are 22m apart.

Thus; The truck will pass the poles at a rate of; Velocity/distance between poles = 28/22 = 1.273 poles per second

Since, we got the time to be 1.78 seconds, then we can find the number of poles by multiplying it with the rate of poles per seconds.

Thus;

Number of poles = 1.78 second × 1.273 poles/ seconds = 2.27 poles

This is approximately 2 poles

6 0
3 years ago
A car is traveling at a velocity of 20 m/s. the driver applies the brakes, and the car slows with an acceleration of a = -4.0 me
sasho [114]
   
\displaystyle\\&#10;V = a\times t\\\\&#10;t= \frac{V}{a} =   \frac{20}{4} = 5~s\\\\&#10;D =  \frac{a\times t^2}{2} =  \frac{4\times 5^2}{2} = 2 \times 25 = 50~m



3 0
3 years ago
What is the velocity of sound in air and vacuum​
guapka [62]

Answer:

approximately 2.99 × 10⁸ m/s

8 0
3 years ago
Read 2 more answers
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