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VMariaS [17]
3 years ago
13

One of the predicted problems due to global warming is that ice in the polar ice caps will melt and raise sea level everywhere i

n the world. Where is this problem the most worrisome?
Physics
1 answer:
babunello [35]3 years ago
8 0

Answer: The global warming is affecting even the most cold places on earth, where we can find the polar ice caps and the glaciers. These places belong to the arctic glacial ocean.

Explanation: In the pass of time, due the human action on the enviroment, the desappear of the polar ice cover are increasing day after day. We, people, are loosing the prime earth air condicioner sistem, which helps to regulate the climate sistem since the entire history of humanity.

In the year of 2018, researches estimated that the total volume of Arctic sea ice in late summer declined, by two estimates, by 75 percent in half a century.

We must know that the decrease of the arctic sea ice have deep effects on the climactic sistem of earth and, considering the polar ice caps desappearence, the sea temperature will increase a lot, due the reflection of ultraviolet rays on the blue ocean, (not anymore on the white ice), transforming what should be the earth air condicioner sistem to a strong heater.

You might be interested in
Select the correct answer.
Natali [406]

its b hoped i helped

8 0
3 years ago
Read 2 more answers
An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f
konstantin123 [22]
A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force  F=m \frac{v^2}{r}, directed toward the center of the circle (so, horizontally)
- the weight of the plane: W=mg, downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is r= \frac{260 m}{2} = 130 m, so the centripetal force acting on the plane is
F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N
On the vertical axis, we have two forces: the weight
W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to a=12 m/s^2, we can find the magnitude of this other force F by using Newton's second law:
F+mg=ma
F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
R= \sqrt{(F_c^2+(W+F)^2}=
= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2}  =2.23 \cdot 10^6 N

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
\tan \theta =  \frac{-(W+F)}{F_c}= -0.5
And so, the angle is
\theta = \arctan (-0.5)=-26.8 ^{\circ}
 
7 0
3 years ago
Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi
tigry1 [53]

Answer:

a

The  radial acceleration is  a_c  = 0.9574 m/s^2

b

The horizontal Tension is  T_x  = 0.3294 i  \ N

The vertical Tension is  T_y  =3.3712 j   \ N

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

   The length of the string is  L =  10.7 \ cm  =  0.107 \ m

     The mass of the bob is  m = 0.344 \  kg

     The angle made  by the string is  \theta  =  5.58^o

The centripetal force acting on the bob is mathematically represented as

         F  =  \frac{mv^2}{r}

Now From the diagram we see that this force is equivalent to

     F  =  Tsin \theta where T is the tension on the rope  and v is the linear velocity  

     So

          Tsin \theta  =   \frac{mv^2}{r}

Now the downward normal force acting on the bob is  mathematically represented as

          Tcos \theta = mg

So

       \frac{Tsin \ttheta }{Tcos \theta }  =  \frac{\frac{mv^2}{r} }{mg}

=>    tan \theta  =  \frac{v^2}{rg}

=>   g tan \theta  = \frac{v^2}{r}

The centripetal acceleration which the same as the radial acceleration  of the bob is mathematically represented as

      a_c  =  \frac{v^2}{r}

=>  a_c  = gtan \theta

substituting values

     a_c  =  9.8  *  tan (5.58)

     a_c  = 0.9574 m/s^2

The horizontal component is mathematically represented as

     T_x  = Tsin \theta = ma_c

substituting value

   T_x  = 0.344 *  0.9574

    T_x  = 0.3294 \ N

The vertical component of  tension is  

    T_y  =  T \ cos \theta  = mg

substituting value

     T_ y  =  0.344 * 9.8

      T_ y  = 3.2712 \ N

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is  

         

       T  = T_x i  + T_y  j

substituting value  

      T  = [(0.3294) i  + (3.3712)j ] \  N

         

3 0
3 years ago
Why do temperatures on the moons surface vary ncyb more than temperatures on earth
ollegr [7]

Answer:

The moon has no atmosphere

Explanation:

The temperatures on the surface of the Moon vary much more than those on Earth because the moon has no atmosphere (third answer in the list), and therefore there are no molecules that could retain residual heat and make the change from day to night a softer transition.

5 0
3 years ago
A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?
stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, m_1 = 300\ g = 0.3\ kg

Initial speed of the cart, u_1=1.2\ m/s

Mass of the larger cart, m_2 = 2\ kg

Initial speed of the larger cart, u_2=0

After the collision,

Final speed of the smaller cart, v_1=-0.81\ m/s (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let v_2 is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

m_1u_1+m_2u_2-m_1v_1=m_2v_2

v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}

v_2=0.301\ m/s

So, the speed of the large cart after collision is 0.301 m/s.

4 0
3 years ago
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