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2 years ago

Phases of the moon. Help!???????

Physics

1 answer:

Veseljchak [2.6K]2 years ago

6 0

Answer:

The Lunar Month.

New Moon.

Waxing Crescent Moon.

First Quarter Moon.

Waxing Gibbous Moon.

Full Moon.

Waning Gibbous Moon.

Third Quarter Moon.

Explanation:

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An airplane flies at 40 m/s at an altitude of 50 meters. The pilot drops a heavy package which falls to the ground. Where, appro

igor_vitrenko [27]

Answer:

128 m

Explanation:

From the question given above, the following data were obtained:

Horizontal velocity (u) = 40 m/s

Height (h) = 50 m

Acceleration due to gravity (g) = 9.8 m/s²

Horizontal distance (s) =?

Next, we shall determine the time taken for the package to get to the ground.

This can be obtained as follow:

Height (h) = 50 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

50 = ½ × 9.8 × t²

50 = 4.9 × t²

Divide both side by 4.9

t² = 50 / 4.9

t² = 10.2

Take the square root of both side

t = √10.2

t = 3.2 s

Finally, we shall determine where the package lands by calculating the horizontal distance travelled by the package after being dropped from the plane. This can be obtained as follow:

Horizontal velocity (u) = 40 m/s

Time (t) = 3.2 s

Horizontal distance (s) =?

s = ut

s = 40 × 3.2

s = 128 m

Therefore, the package will land at 128 m relative to the plane

6 0

2 years ago

A piece of metal will feel colder than a piece of wood at the same temperature. Why?

alexira [117]

Metals in general, are good heat conductors

7 0

2 years ago

3. A certain wire, 3 m long, stretches by 1.2 mm when under tension of 200 N. By how much does

nikitadnepr [17]

Answer:

The extension of the second wire is $e_2 = 0.0024 \ m = 2.4 mm$

Explanation:

From the question we are told that

The length of the wire is $L = 3 \ m$

The elongation of the wire is $e = 1.2mm = \frac{1.2}{1000} = 0.0012 m$

The tension is $F = 200 \ N$

The length of the second wire is $L_2 = 6 \ m$

Generally the Young's modulus(Y) of this material is

$Y = \frac{stress}{strain }$

Where $stress = \frac{F}{A}$

Where A is the area which is evaluated as

$A = \pi r^2$

and $strain = \frac{extention}{length} = \frac{e}{L}$

$Y = \frac{\frac{F}{\pi r^2 } }{ \frac{e}{L} }$

Since the wire are of the same material Young's modulus(Y) is constant

So we have

$\frac{F * L }{r^2 e} = \pi * Y = constant$

$F * L = constant * r^2 e$

Now the ration between the first and the second wire is

$\frac{F_1}{F_2} * \frac{L_1}{L_2} = \frac{r*2_1}{r^2} * \frac{e_1}{e_2}$

Since tension , radius are constant

We have

$\frac{L_1}{L_2} = \frac{e_1}{e_2}$

substituting values

$\frac{3}{6} = \frac{0.0012}{e_2}$

$0.5 e_2 = 0.0012$

$e_2 = \frac{ 0.0012 }{0.5}$

$e_2 = 0.0024 \ m = 2.4 mm$

3 0

3 years ago

When the speed of the bottle is 2 m/s, the KE is kg m2/s2. When the speed of the bottle is 3 m/s, the KE is kg m2/s2. When the s

d1i1m1o1n [39]

mass of the bottle in each case is M = 0.250 kg

now as per given speeds we can use the formula of kinetic energy to find it

1) when speed is 2 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(2)^2 = 0.5 J$

2) when speed is 3 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(3)^2 = 1.125 J$

3) when speed is 4 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(4)^2 = 2 J$

4) when speed is 5 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(5)^2 = 3.125 J$

5) when speed is 6 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(6)^2 = 4.5 J$

3 0

3 years ago

Read 2 more answers

From an h = 53 feet observation tower on the coast, a Coast Guard officer sights a boat in difficulty. The angle of depression o

maksim [4K]

Answer:

757,93 feets

Explanation:

We can make a right triangle between the boat (A), the Coast Guard officer (B) and the base of the observation tower (C), like in the graph attached. Now, you could also made a rectangle, adding the horizontal at the height of the Coast Guard, starting in B and ending in D, the vertex opossing C.

The angle of depression, its O in the graph.

Now, as we got an rectangle, of course, the segment AD its the same length as CB, and CA, the distance from the boat to shoreline, its the same length as DB.

ADB its an right triangle, with AB, the hypothenuse, and BD and DA, the catheti (or <em>legs</em>).

Now, we know the lenght BC, the height of the tower, 53 feets, so we also know the lenght of DA. DA its the opposite cathetus to the angle O. We wish to know the length AC, equal to the lenght DB, the adjacent cathetus of the angle O.

Know, the trigonometric function that connects the adjacent cathetus with the opossite cathetus its the tangent.

$tangent( O ) = \frac{opposite}{adjacent}$

We can take that the angle O = 4 °, and knowing that the opossite cathetus its 53 feets, we got:

$tangent( 4) = \frac{53 feets}{DB}$

$DB= \frac{53 feets}{tangent( 4)}$

$DB= 757,93 feets$

This its equal to the distance from the boat to the shoreline.

4 0

2 years ago