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3 years ago

Can prokaryotes have more than one cell

Physics

2 answers:

Westkost [7]3 years ago

5 0

I think the answer is yes, but i might be wrong

GaryK [48]3 years ago

3 0

I think the answer is yes

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Amanda spent 2/5 of her time after school doing homework and ¼ of her remaining time riding her bike. If she rode her bike for 4

Doss [256]

Answer:120 min

Explanation:

Given

Amanda spent $\frac{2}{5}$ of her time after school doing Home work

And $\frac{1}{4}$ of her remaining time riding her bike

It is given that she rode her bike for 45 minutes in a week

Let t be the time after school

therefore Amanda spend $\frac{2t}{5}$ in home work and $\frac{3t}{5}$ time is left

From remaining $\frac{3t}{5}$ time she spends $\frac{1}{4}$ time riding her bike

therefore $\frac{3t}{5}\times \frac{1}{4}=45$

thus $t=300 min$

therefore time spent on home work is $\frac{2}{5}\times 300=120 min$

6 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

two engines are turned on for 763 s at a moment when the velocity of the craft has x and y components of v0x = 6380 m/s and v0y

svet-max [94.6K]

Answer:

Explanation:

Given

initial velocity component of engines is

$v_0_x=6380 m/s$

$v_0_y=6770 m/s$

time period of engine running=763 s

Displacement in $x=4.50\times 10^6$

$y=7.27\times 10^6$

Using $s=ut+\frac{at^2}{2}$ in x and y direction

$x=v_0_x\times t+\frac{at^2}{2}$

$4.50\times 10^6=6380\times 763+\frac{a\times 763^2}{2}$

$4.50\times 10^6-4.86\times 10^6=\frac{a\times 763^2}{2}$

$a=-1.23 m/s^2$

In y direction

$y=v_0_y\times t+\frac{a't^2}{2}$

$7.27\times 10^6=6770\times 763+\frac{a\times 763^2}{2}$

$7.27\times 10^6-5.16\times 10^6=\frac{a\times 763^2}{2}$

$a=7.24 m/s^2$

x component $=-1.23 m/s^2$

y component $=7.24 m/s^2$

3 0

2 years ago

Read 2 more answers

Q|C A string with linear density 0.500 g/m is held under tension 20.0 N. As a transverse sinusoidal wave propagates on the strin

julia-pushkina [17]

A string with linear density 0.500 g/m.

Tension 20.0 N.

The maximum speed $v_{y, max}$

The energy contained in a section of string 3.00 m long as a function of $v_{y, max}$ .

We are given following data for string with linear density held under tension :

μ = 0.5 $\frac{g}{m}$

= 0.5 x 10⁻³ $\frac{kg}{m}$

T = 20 N

If string is L = 3m long, total energy as a function of $v_{y, max}$ is given by:

E = 1/2 x μ x L x ω² x A²

= 1/2 x μ x L x $v^{2} _{y, max}$

= 7.5 x 10⁻⁴ $v^{2} _{y, max}$

So, The total energy as a function of $v^{2} _{y, max}$ = 7.5 x 10⁻⁴ $v^{2} _{y, max}$

Learn more about linear density problem here:

brainly.com/question/17190616

#SPJ4

3 0

1 year ago

The turbines can be seen inside this hydroelectric dam. Why are they located at that particular height?

Yakvenalex [24]

Answer:

Explanation:

the answer is number three

5 0

3 years ago

Read 2 more answers