What influences the path that surface currents take?

jasenka [17]

Answer:

I think its A plz tell me if im right

7 0

3 years ago

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0

otez555 [7]

The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

$F_{1}$ = 20 N, $F_{2}$ = 25 N, a = -0.9 $m/s^{2}$

W = 83 N

m = $\frac{83}{9.81}$

= 8.46

Now, we will balance the forces along the y-component as follows.

N = W + $F_{2}$

= 83 + 25 = 108 N

Now, balancing the forces along the x component as follows.

$F_{1} - F_{r}$ = ma

$20 - F_{r} = 8.46 \times (-0.9)$

$F_{r}$ = 7.614 N

Also, we know that relation between force and coefficient of friction is as follows.

$F_{r} = \mu \times N$

$\mu = \frac{F_{r}}{N}$

= $\frac{7.614}{108}$

= 0.0705

Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.

7 0

3 years ago

Read 2 more answers

A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet. Compared to the a

Whitepunk [10]

Answer:

C) one-half as great

Explanation:

We can calculate the acceleration of gravity in that planet, using the following kinematic equation:

$\Delta x=v_0t+\frac{gt^2}{2}$

In this case, the sphere starts from rest, so $v_0=0$ . Replacing the given values and solving for g':

$g'=\frac{2\Delta x}{t^2}\\g'=\frac{2(22m)}{(3s)^2}\\g'=4.89\frac{m}{s^2}$

The acceleration due to gravity near Earth's surface is $g=9.8\frac{m}{s^2}$ . So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.

5 0

3 years ago

What relationship between the sun and earth did copernicus formulate?

Dmitrij [34]

Answer:

The astronomical model created and published by Nicholas Copernicus in the year 1543 is called Copernican heliocentrism. The model set the Sun in immobile position near the center of the solar system with Earth, as well as the other planets, spherical, epicycled and at consistent frequencies around it.

5 0

3 years ago

On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on t

kozerog [31]

Answer:

a) 6 times farther. b) 6 times longer.

Explanation:

Once released, in the horizontal direction, no other forces act on the ball, so it continues moving at the same initial velocity, which is given by the projection of the velocity vector in the horizontal direction, as follows:

vₓ = v* cos (25º) = 23 m/s * 0.906 = 20.8 m/s

In the vertical direction, the initial velocity is the projection of the velocity vector along the vertical axis, as follows:

vy = v* sin (25º) = 23 m/s * 0.422 = 9.72 m/s

Assuming that the acceleration is constant, and equal to 1/6*g, we can calculate the total time of flight, with the following kinematic equation for the vertical displacement:

$y = voy*t - (\frac{1}{2}*\frac{g}{6} * t^{2} )$

If the total displacement in the vertical direction is 0 (which means that the time if the total time of flight), we can solve for t, as follows:

$t = \frac{voy*12}{g} = \frac{9.72 m/s*12}{9.8m/s2} = 11. 9 s$

On earth, this time could be calculated in the same way:

$t = \frac{voy*12}{g} = \frac{9.72 m/s*2}{9.8m/s2} = 1.98 s$

As the time is defined by the vertical movement, we can find the horizontal distance travelled on the moon, as follows:

Δx = v₀ₓ * t = 20.8 m/s * 11. 9 s = 248.1 m

On earth, the distance travelled had been as follows:

Δx = v₀ₓ * t = 20.8 m/s * 1.98 s = 41.3 m

⇒ Δx(moon) / Δx(earth) = 248.1 / 41.3 = 6.00

b) As we have just found, the time of flight on the moon and on the earth are as follows:

tmoon = 11. 9 s

tearth = 1.98 s

⇒ t(moon) / t(earth) = 11.9 / 1.98 = 6.0

8 0

3 years ago