A blank is a quality that has magnitude and direction

masya89 [10]

Yes I believe in god.I believe in him because there is huge gaps in science that can not be solved. I also believe in him because I was also taught about him when I was little by family & my mom so I grew up learning about him which made my beliefs stronger of course.

3 0

3 years ago

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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

2 years ago

Find the distance in nm between two slits that produces the first minimum for 410-nm violet light at an angle of 14.5°.

Galina-37 [17]

Answer:

820 nm

Explanation:

We are given that

Wavelength= $\lambda=410 nm$

$\lambda=410\times 10^{-9} m$

$1nm=10^{-9} m$

$\theta=14.5^{\circ}$

For first minimum therefore

m=0

We know that for destructive interference

$(m+\frac{1}{2})\lambda=dsin\theta$

Substitute the values

$(0+\frac{1}{2})\times 410\times 10^{-9}=dsin 14.5$

$d=\frac{410\times 10^{-9}}{2\times sin 14.5}$

$d=820\times 10^{-9} m=820 nm$

Hence, the distance between two slits that produces the first minimum=820 nm

5 0

2 years ago

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How fast must an object move before its length appears to be contracted to one-fourth its proper length? (Give your answer in te

Tresset [83]

Answer:

0.97c

Explanation:

From the relativistic equation for length contraction, we have

$l$ = $l_{0}\sqrt{1 - \beta }$

where

$l$ is the final length of the object

$l_{0}$ is the original length of the object before contraction

β = $v^{2} /c^2$

where v is the speed of the object

c is the speed of light in free space = 3 x 10^8 m/s

The equation can be re-written as

$l$ / $l_{0}$ = $\sqrt{1 - \beta }$

For the length to contract to one-fourth of the proper length, then

$l$ / $l_{0}$ = 1/4

substituting into the equation, we'll have

1/4 = $\sqrt{1 - \beta }$

substituting for β, we'll have

1/4 = $\sqrt{1 - v^2/c^2 }$

squaring both side of the equation, we'll have

1/16 = 1 - $v^2/c^2$

$v^2/c^2$ = 1 - 1/16

$v^2/c^2$ = 15/16

square root both sides of the equation, we have

v/c = 0.968

v = 0.97c

3 0

3 years ago

Which animal in the rainforest is not nocturnal?

Stels [109]

Answer:

tree frog i believe hope its right

5 0

3 years ago