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3 years ago

5

you would most likely find a weather Ballon an it's highest altitude in which layer above earth (A) Exosphere (B) Mesospher (C)

Stratosphere (D) Thermosphere

2 answers:

laila [671]3 years ago

7 0

U will find it in the Exosphere

Trava [24]3 years ago

6 0

The answer to this question is

C. stratosphere...

this picture should help you

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A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann

Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/ $s^{2}$

Bring out the given parameters from the question:

Initial Velocity ( $V_{1}$ ) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) = 800 m

Projection angle ∅ = ?

Horizontal distance = $V_{1x}$ tcos ∅ .......................... Equation 1

where $V_{1x}$ = velocity in the X - direction

t = Time taken

Vertical Distance = y = $V_{1y}$ t - $\frac{1}{2}$ g $t^{2}$ ................... Equation 2

Where $V_{1y}$ = Velocity in the Y- direction

t = Time taken

$V_{1y}$ = $V_{1}$ sin∅

Making time (t) subject of the formula in Equation 1

t = d/( $V_{1x}$ cos ∅)

t = $\frac{2000}{1000coso}$ = $\frac{2}{cos0}$ = $\frac{d}{cos o}$ ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = $V_{1y}$ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = dtan∅ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Applying geometry

$\frac{1}{cos o}$ = $tan^{2} o$ + 1

Vertical Distance = h = d tan∅ - 2 g ( $tan^{2} o$ + 1)

substituting the given parameters

800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1) Equation 4

Replacing tan ∅ = Q .....................Equation 5

In order to get a quadratic equation that can be easily solve.

800 = 2000 Q - 19.6 $Q^{2}$ + 19.6

Rearranging 19.6 $Q^{2}$ - 2000 Q + 780.4 = 0

$Q_{1}$ = 101.6291

$Q_{2}$ = 0.411

Inserting the value of Q Into Equation 5

tan ∅ = 101.63 or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

∅ = 89.44° ∅ = 22.37°

4 0

3 years ago

Please select the word from the list that best fits the definition.

poizon [28]

The true answer is the choice <span>B.Anton van Leeuwenhoek
this one was a Dutch biologist, inventor and physicist (1632, 1723). he has discoverd the moving objects from pond water undr his microscope, the animalcules.</span>

4 0

3 years ago

A wire 40cm long and of diameter 0.60mm has a resistance of 1.5 ohm what is the resistivity of the material of which it is made

zavuch27 [327]

Answer:

The resistivity of the wire is $1.05\times 10^{-6}\ \Omega-m$

Explanation:

We have,

Length of a wire is 40 cm or 0.4 m

Diameter of a wire is 0.60 mm

Radius is 0.3 mm or 0.0003 m

Resistance of a wire is 1.5 ohm

Now we need to find the resistivity of the material of which it is made. The resistance of a wire in terms of its resistance, length and area is given by :

$R=\rho \dfrac{l}{A}$

$\rho$ = resistivity

$\rho=R \dfrac{\pi r^2}{l}\\\\\rho=1.5\times \dfrac{3.14\times (0.0003 )^2}{0.4}\\\\\rho=1.05\times 10^{-6}\ \Omega-m$

So, the resistivity of the wire is $1.05\times 10^{-6}\ \Omega-m$ .

3 0

3 years ago

4. For the following questions, use a periodic table and your atomic calculations to find the unknown information about each iso

natita [175]

The answer is 7 (its like the only question thats easy so far lol)

4 0

3 years ago

A particle with charge q = +5e and mass m = 8.2×10-26 kg is injected horizontally with speed 1.1×106 m/s into the region between

loris [4]

Answer:

d = 1.27m

Explanation:

Given m = 8.2×10-26kg, v = 1.1×10⁶m/s, q = +5e = 5×1.6×10‐¹⁹ C.

E = 49kN/C = 49000N/C

The displacement is given by

d = 1/2× mv²/qE = 1/2 × 8.2×10-²⁶ × (1.1×10⁶)²/(5×1.6×10-¹⁹ ×49000) = 1.27m

3 0

3 years ago

Read 2 more answers