True or False: Global wind patterns have less to do with air temperature than local winds.

Do longitude lines run horizontally (east-west) or vertically (north-south)?

nlexa [21]

Longitude- Horizontal (East West)
Latitude- Vertical (North South)

3 0

3 years ago

a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the

ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: $F=m*a$ with this we can get both accelerations; solving for acceleration $a=\frac{F}{m}$ . Now $a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2})$ and $a_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2})$ . Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, $x=\frac{1}{2}*a_{girl}*t^{2}$ and $15.0-x=\frac{1}{2}*a_{sled}*t^{2}$ , solving for the time we get: $t=\sqrt{\frac{2x}{a_{girl} } }$ and $t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } }$ with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get: $\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }$ . Finally we get: $\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} }$ and replacing the values we have got: $\frac{x}{0.14} =\frac{(15.0-x)}{0.73}$ so $5.33*x=15-x$ so x=2.37 (meters).

5 0

3 years ago

Witch of the following is the correct definition of convention

iogann1982 [59]

Answer:

A is the answer .(A) is correct

3 0

2 years ago

A cat, walking along the window ledge of a new york apartment, knocks off a flower pot, which falls to the street 280 feet below

Harlamova29_29 [7]

H = 280 ft, the height of the flower pot.
g = 32 ft/s²

Neglect air resistance.
Note that 1 ft/s = 15/22 mi/h

The initial vertical velocity is zero.
Let v = the velocity with which the flower pot hits the ground.
Then
v² = 2gh
= 2*(32 ft/s²)*(280 ft)
= 17920 (ft/s)²
v = 133.866 ft/s

Also,
v = (133.866 ft/s)*(15/22 (mi/h)/(ft/s)) = 91.272 mi/h

Answer: 133.9 ft/s or 91.3 mi/h

5 0

3 years ago

The density of a block of wood is 694 kg/m3. Its mass is 689 g. We tie the block to the bottom of a swimming pool using a single

Serhud [2]

Answer:

<em>The tension in the string = 2.065 N</em>

Explanation:

From Archimedes principle,

R.d = density of the wood block/density of water = weight of the wood block/Upthrust of the wood block in water.

R.d = D₁/D₂ = W/U

W/U =D₁/D₂.................................. Equation 1

Where W = weight of the wood block, U = upthrust of the wood block in water, D₁ = Density of the wood block, D₂ = Density of water.

Making U the subject of the equation,

U = WD₂/D₁........................... Equation 2

Given: W = mg = (689/1000)9.8 = 6.75 N, D₁ = 694 kg/m³, D₂ = 1000 kg/m³.

Substituting these values into equation 2,

U = 6.75(694)/1000

U = 4684.5/1000

U = 4.685 N.

Note: Three forces act on the wood block in the pool. and they are

(i) The weight(W) acting downs

(ii) The upthrust (U) acting upwards,

(iii) The Tension (T) in the string, acting upwards.

Thus,

W = U+T

T = W - U ................................. Equation 3

Where W = 6.75 N, U = 4.685 N

T = 6.75 - 4.685

T = 2.065 N.

T = 2.065 N

<em>Thus the tension in the string = 2.065 N</em>

7 0

2 years ago