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3 years ago

Winds are named based on different factors. How are winds usually named?

1 answer:

eduard3 years ago

4 0

Winds are named based on which compass direction the wind is blowing. For example some common ones are NE or N or SE or SW. NE stands for Northeast, N for North, SE for South East and SW for Southwest.

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Gravity is the force that keeps us on the Earth. It pulls us towards the center of the Earth. If you were to move from the surfa

Nikitich [7]

<h2>Answer: B. Gravitational potential energy </h2>

Explanation:

<em>The gravitational potential energy is the energy that a body or object possesses, due to its position in a gravitational field. </em>

That is why this energy depends on the relative height of an object with respect to some point of reference and associated with the gravitational force.

In the case of the <u>Earth</u>, in which <u>the gravitational field is considered constant</u>, the value of the gravitational potential energy $U_{p}$ will be:

$U_{p}=mgh$

Where $m$ is the mass of the object, $g$ the acceleration due gravity and $h$ the height of the object.

As we can see, the value of $U_{p}$ is directly proportional to the height.

6 0

3 years ago

In a lab, a student drags a shoe across the floor at constant speed. If the coefficient of static friction between the floor and

Svetllana [295]

<span>B) 0.6 N
I suspect you have a minor error in your question. Claiming a coefficient of static friction of 0.30N is nonsensical. Putting the Newton there is incorrect. The figure of 0.25 for the coefficient of kinetic friction looks OK. So with that correction in mind, let's solve the problem. The coefficient of static friction is the multiplier to apply to the normal force in order to start the object moving. And the coefficient of kinetic friction (which is usually smaller than the coefficient of static friction) is the multiplied to the normal force in order to keep the object moving. You've been given a normal force of 2N, so you need to multiply the coefficient of static friction by that in order to get the amount of force it takes to start the shoe moving. So: 0.30 * 2N = 0.6N And if you look at your options, you'll see that option "B" matches exactly.</span>

7 0

3 years ago

Read 2 more answers

If the distance between the Earth and Moon were half what it is now, by what factor would the force of gravity between them be c

hichkok12 [17]

Answer:

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

$m_1$ = Mass of Earth

$m_2$ = Mass of Moon

r = Distance between Earth and Moon

Old gravitational force

$F_o=\dfrac{Gm_1m_2}{r^2}$

New gravitational force

$F_n=\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}$

Dividing the equations

$\dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{\dfrac{1}{4}r^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=4$

The ratio is $\dfrac{F_n}{F_o}=4$

The new force would be 4 times the old force

7 0

3 years ago

011 10.0 points

Ulleksa [173]

Answer:

2.47 m

Explanation:

Let's calculate first the time it takes for the ball to cover the horizontal distance that separates the starting point from the crossbar of d = 52 m.

The horizontal velocity of the ball is constant:

$v_x = v cos \theta = (25)(cos 35.9^{\circ})=20.3 m/s$

and the time taken to cover the horizontal distance d is

$t=\frac{d}{v_x}=\frac{52}{20.3}=2.56 s$

So this is the time the ball takes to reach the horizontal position of the crossbar.

The vertical position of the ball at time t is given by

$y=u_y t - \frac{1}{2}gt^2$

where

$u_y = v sin \theta =(25)(sin 35.9^{\circ})=14.7 m/s$ is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

And substituting t = 2.56 s, we find the vertical position of the ball when it is above the crossbar:

$y=(14.7)(2.56) - \frac{1}{2}(9.8)(2.56)^2=5.52 m$

The height of the crossbar is h = 3.05 m, so the ball passes

$h' = 5.52- 3.05 = 2.47 m$

above the crossbar.

8 0

3 years ago

A dog is walking at 2m/s and then begins to run at a speed of 6m/s. What is his acceleration if his total travel time is 2 secon

fiasKO [112]

The formula for velocity vf = vi + at

First list your given information

2m/s Is your initial velocity (vi)
6m/s is you final velocity (vf)
2 seconds is your time (t)

Since you want the a for acceleration get a by itself

a = (vf-vi)/t

So a= (6-2)/2

a= 4/2

a=2

Now units

the units for acceleration are m/s $x^{2}$

2m/s $x^{2}$

7 0

3 years ago