Choose the four major environmental

After observing a moth that is camouflaged against dark-colored bark, a scientist asks a question. The scientist discovers that

Rasek [7]

Because of how it's worded the answer would most likely be number four

6 0

3 years ago

Read 2 more answers

The table shows data for the planet Uranus. A 2 column table with 4 rows. The first column is labeled Quantity with entries, Esc

prohojiy [21]

Answer:

The answer is 218

Explanation:

Weight = mass * gravitational acceleration

weight is represented by F

F = 25kg (8.7)

(I'm pretty sure that you don't have to include the meters per second/per second thing)

4 0

2 years ago

Space pilot Mavis zips past Stanley at a constant speed relative to him of 0.800c. Mavis and Stanley start timers at zero when t

inna [77]

Answer:

Explanation:

a. The equation of Lorentz transformations is given by:

x = γ(x' + ut')

x' and t' are the position and time in the moving system of reference, and u is the speed of the space ship. x is related to the observer reference.

x' = 0

t' = 5.00 s

u =0.800 c,

c is the speed of light = 3×10⁸ m/s

Then,

γ = 1 / √ (1 - (u/c)²)

γ = 1 / √ (1 - (0.8c/c)²)

γ = 1 / √ (1 - (0.8)²)

γ = 1 / √ (1 - 0.64)

γ = 1 / √0.36

γ = 1 / 0.6

γ = 1.67

Therefore, x = γ(x' + ut')

x = 1.67(0 + 0.8c×5)

x = 1.67 × (0+4c)

x = 1.67 × 4c

x = 1.67 × 4 × 3×10⁸

x = 2.004 × 10^9 m

x ≈ 2 × 10^9 m

Now, to find t we apply the same analysis:

but as x'=0 we just have:

t = γ(t' + ux'/c²)

t = γ•t'

t = 1.67 × 5

t = 8.35 seconds

b. Mavis reads 5 s on her watch which is the proper time.

Stanley measured the events at a time interval longer than ∆to by γ,

such that

∆t = γ ∆to = (5/3)(5) = 25/3 = 8.3 sec which is the same as part (b)

c. According to Stanley,

dist = u ∆t = 0.8c (8.3) = 2 x 10^9 m

which is the same as in part (a)

7 0

2 years ago

A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle

tatiyna

Answer:

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°

sinA = 0.8

cosA = 0.6

width of the river,d = 40.0 m,

the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,

The river itself was 100 m below the ramp H = 100 m,

(a) find speed v

vertical displacement

$-h= vsinA\times t-gt^2/2$

putting values h=15 m, v=0.8

$-15 = 0.8vt - 4.9t^2$ ............. (1)

horizontal displacement d = vcosA×t = 0.6×v ×t

so v×t = d/0.6 = 40/0.6

plug it into (1) and get

$-15 = 0.8\times40/0.6 - 4.9t^2$

solving for t we get

t = 3.734 s

also, v = (40/0.6)/t = 40/(0.6×3.734) = 17.8 m/s

(b) If his speed was only half the value found in (a), where did he land?

v = 17.8/2 = 8.9 m/s

vertical displacement = $-H =v sinA t - gt^2/2$

⇒ $4.9t^2 - 8.9\times0.8t - 100 = 0$

t = 5.30 s

then

d =v×cosA×t = 8.9×0.6×5.30= 28.3 m

3 0

2 years ago

What is the speed of light?

Zielflug [23.3K]

The speed of light is 299,792,458 meters per second in vacuum,
somewhat less in any material. How much less depends on the material.

6 0

3 years ago