Is a star's emission spectrum unique?

A runner went from 6 m/s and two seconds what was his acceleration

Radda [10]

Answer:

is it 3?

Explanation:

Im taking a guess and just dividing 6 and 2

8 0

3 years ago

A hunter aims at a deer which is 40 yards away. Her cross- bow is at a height of 5ft, and she aims for a spot on the deer 4ft ab

shutvik [7]

Answer:

a) θ₁ = 0.487º , b) t = 0.400 s , x = 11.73 ft

Explanation:

For this exercise let's use the projectile launch relationships.

The initial height is I = 5 ft and the final height y = 4 ft

y = y₀ + $v_{oy}$ t - ½ g t²

The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft

x = v₀ₓ t

t = x / v₀ₓ

We replace

y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²

v_{oy} = v₀ sin θ

v₀ₓ = vo cos θ

y –y₀ = x tan θ - ½ g x² / v₀² cos² θ

5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)

1 = 120 tan θ - 0.0213 sec² θ

Let's use the trigonometry relationship

Sec² θ = 1 - tan² θ

1 = 120 tan θ - 0.0213 (1 –tan²θ)

0.0213 tan²θ + 120 tanθ -1.0213 = 0

We change variables

u = tan θ

u² + 5633.8 u - 48.03 = 0

We solve the second degree equation

u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2

u = [- 5633.8 ± 5633.82] / 2

u₁ = 0.0085

u₂= -5633.81

u = tan θ

θ = tan⁻¹ u

For u₁

θ₁ = tan⁻¹ 0.0085

θ₁ = 0.487º

For u₂

θ₂ = -89.99º

The launch angle must be 0.487º

b) let's look for the time it takes for the arrow to arrive

x = v₀ₓ t

t = x / v₀ cos θ

t = 120 / (300 cos 0.487)

t = 0.400 s

The deer must be at a distance of

v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s

x = v t

x = 29.33 0.4

x = 11.73 ft

3 0

3 years ago

A monkey running through the jungle goes 0.198 km straight to the East, then turns 15.8° deflection from straight East toward th

inn [45]

Answer:

$|\vec r|=339.82\ m$

$\theta=-6.67^o$

Explanation:

<u>Displacement </u>

It's a vector magnitude that measures the space traveled by a particle between an initial and a final position. The total displacement can be obtained by adding the vectors of each individual displacement. In the case of two displacements:

$\vec r=\vec r_1+\vec r_2$

Given a vector as its polar coordinates (r,\theta), the corresponding rectangular coordinates are computed with

$x=rcos\theta$

$y=rsin\theta$

And the vector is expressed as

$\vec z==$

The monkey first makes a displacement given by (0.198 km,0°). The angle is 0 because it goes to the East, the zero-reference for angles. Thus the first displacement is

$\vec r_1==\ km=\ m$