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3 years ago

15

Air at the poles tends to flow close to the surface toward the equator. What can you conclude about the characteristics of this

polar air?

1 answer:

LenaWriter [7]3 years ago

5 0

Answer:

That the polar air has has more pressure than the air at the equator.

Explanation:

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A boy throws a ball up into the air with a speed of 8.2 m/s. The ball has a mass of 0.3 kg. How much gravitational potential ene

diamong [38]

We can use the law of conservation of energy to solve the problem.

The total mechanical energy of the system at any moment of the motion is:
$E=U+K = mgh + \frac{1}{2}mv^2$
where U is the potential energy and K the kinetic energy.

At the beginning of the motion, the ball starts from the ground so its altitude is h=0 and therefore its potential energy U is zero. So, the mechanical energy is just kinetic energy:
$E_i = K_i = \frac{1}{2}mv^2 = \frac{1}{2}(0.3 kg)(8.2 m/s)^2=10.09 J$

When the ball reaches the maximum altitude of its flight, it starts to go down again, so its speed at that moment is zero: v=0. So, its kinetic energy at the top is zero. So the total mechanical energy is just potential energy:
$E_f = U_f$
But the mechanical energy must be conserved, Ef=Ei, so we have
$U_f = K_i$
and so, the potential energy at the top of the flight is
$U_f = K_i = 10.09 J$

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2 years ago

Read 2 more answers

John flies directly east for 20km, then turns to the north and flies for another 10 km before dodging a flock of geese. what’s t

KIM [24]

The distance is 30 km and the displacement is 22.4 km North East

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3 years ago

An ice cream truck is going 25m/s to the East. It accelerates to 45m/s in the same direction over 5s. What is its acceleration?

Naya [18.7K]

Hello!

We can use the kinematic equation:
$a = \frac{v_f - v_i}{t}$

a = acceleration (m/s²)

vf = final velocity (45 m/s)
vi = initial velocity (25 m/s)

t = time (5 sec)

Plug in the givens:
$a = \frac{45-25}{5} = \frac{20}{5} = \boxed{4 m/s^2}$

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2 years ago

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º. What is the coefficient of

Alex_Xolod [135]

Given :

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.

To Find :

The coefficient of static friction between the box and the plane.

Solution :

Vertical component of force :

$mg\ sin\ \theta = 120\times 10 \times sin\ 47^\circ{}=877.62 \ N$

Horizontal component of force(Normal reaction) :

$mg\ cos\ \theta = 120\times 10 \times cos\ 47^\circ{}=818.40 \ N$

Since, box is on the verge of slipping :

$mg\ sin\ \theta= \mu(mg \ cos\ \theta)\\\\\mu = tan \ \theta\\\\\mu = tan\ 47^o\\\\\mu = 1.07$

Therefore, the coefficient of static friction between the box and the plane is 1.07.

Hence, this is the required solution.

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3 years ago

What is the source of the radioactive nuclei present in spent fuel rods?

Debora [2.8K]

The answer is <span>nuclear fission. T</span>he source of the radioactive nuclei present in spent fuel rods is nuclear fission. Nuclear fission<span> is the </span>process<span> in which a large nucleus splits into two smaller nuclei with the release of energy. </span>

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2 years ago