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3 years ago

Wind direction is noted based on what direction the wind is coming FROM?

Physics

1 answer:

Katen [24]3 years ago

4 0

Answer:

yes, or it can be noted to where the wind if going to

Explanation:

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A filing cabinet weighing 556 N rests on the floor. The coefficient of static friction between it and the floor is 0.68, and the

zvonat [6]

Explanation:

"Static friction is a force that keeps an object at rest. It must be overcome to start moving the object."

(556 x 0.68) = static friction of 378.08N. before movement occurs.

The forces (a) and (b) will not move it.

Each will incur a frictional force preventing movement equal to itself, = 222N. and 334N. respectively.

Forces (c) and (d) will move it, and accelerate it.

Forces (c) and (d) will both encounter friction of (556 x 0.56) = 311.36N. when the cabinet is moving.

5 0

3 years ago

Describe a procedure that would increase the potential energy of two magnets if like poles are used. Explain why the energy of t

zalisa [80]

Answer:

If you apply a force to separate 2 opposite poles, the potential energy of the system increases.

5 0

2 years ago

3) If a ball launched at an angle of 10.0 degrees above horizontal from an initial height of 1.50 meters has a final horizontal

Irina-Kira [14]

Answer:

35.6 m

Explanation:

3 0

3 years ago

A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li

Kitty [74]

Answer:

$E_{net} = 6.44 \times 10^5 N/C$

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

$E = \frac{2k \lambda}{r}$

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

$E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}$

now plug in all data

$\lambda_1 = 4.68 \muC/m$

$\lambda_2 = 2.48 \mu C/m$

$r = 0.200 m$

now we have

$E_{net} = \frac{2k}{r}(4.68 + 2.48)$

$E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})$

$E_{net} = 6.44 \times 10^5 N/C$

8 0

3 years ago

Brutus, a champion weight lifter, raises 220 kg a distance of 3.10 m. (a) how much work is done by brutus lifting the weights? j

Nana76 [90]

A) work = force * distance
mass is not a force, weight is, so we have to find the weight of the block.
Weight = mg
Weight = (220kg)(9.8)
Weight = 2156N
Work = 2156N * 3.10m
work = 6683.6J
b) Since he is holding the weights, it's not moving, therefore, he doesn't do any work
c) The answer is still the same amount of work when he lifted them.
d) The answer is no since when he let go the weight, he doesn't apply any force to the weight.
e) P = work/time
P = 6683.6J / 2.1s
P = 3182.67 watts

3 0

3 years ago