1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
DedPeter [7]
3 years ago
14

g As observed on earth, a certain type of bacteria is known to double in number every 24 hours. Two cultures of these bacteria a

re prepared, each consisting initially of one bacterium. One culture is left on earth and the other placed on a rocket that travels at a speed of 0.893c relative to the earth. At a time when the earthbound culture has grown to 256 bacteria, how many bacteria are in the culture on the rocket, according to an earth-based observer
Physics
1 answer:
tangare [24]3 years ago
8 0

Answer:

86.4 hrs

Explanation:

The amount of bacteria is initially 1

It doubles every 24 hrs.

After first 24 hrs, the amount = 2

After next 24 hrs = 4

After next 24 hrs = 8

After next 24 hrs = 16

After next 24 hrs = 32

After next 24 hrs = 64

After next 24 hrs = 128

After next 24 hrs = 256

Total time taken to reach 256 = 24 x 8 = 192 hrs

For the bacteria culture on the rocket that travels at a speed of 0.893c relative to the earth, this time is contracted by the relationship

t = t'(1 - ¥^2)^0.5

Where t is the contracted time =?

t' is the time on earth

¥ = v/c

Where v is the speed of the rocket

c is the speed of light

since v = 0.893c

¥ = 0.893

Substituting, we have

t = 192 x (1 - 0.893^2)^0.5

t = 192 x 0.2025^0.5

t = 192 x 0.45 = 86.4 hrs

You might be interested in
The cornea behaves as a thin lens of focal lengthapproximately 1.80 {\rm cm}, although this varies a bit. The material of whichi
Keith_Richards [23]

Answer:

Explanation:

  a )

from lens makers formula

\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})

f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face

putting the values

\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})

1.462 = 2 - 1 / r₂

1 / r₂ = .538

r₂ = 1.86 cm .

= 18.6 mm .

b )

object distance u = 25 cm

focal length of convex lens  f  = 1.8 cm

image distance  v   = ?

lens formula

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}

\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}

.5555 - .04

= .515

v = 1.94 cm

c )

magnification = v / u

= 1.94 / 25

= .0776

size of image = .0776 x size of object

= .0776 x 10 mm

= .776 mm

It will be a real image and it will be inverted.

 

5 0
2 years ago
A proton moving at 3.0 × 10^4 m/s is projected at an angle of 30° above a horizontal plane. If an electric field of 400 N/C is a
GuDViN [60]

Answer:

The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s

Explanation:

From Newton's second law, F = mg and also from coulomb's law F= Eq

Dividing both equations by mass;

F/m = Eq/m = mg/m, then

g = Eq/m --------equation 1

Again, in a projectile motion, the time of flight (T) is given as

T = (2usinθ/g) ---------equation 2

Substitute in the value of g into equation 2

T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}

Charge of proton = 1.6 X 10⁻¹⁹ C

Mass of proton = 1.67 X 10⁻²⁷ kg

E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°

Solving for T;

T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}

T = 7.83 X10⁻⁷ s

6 0
2 years ago
A position vector with magnitude 10 m points to the right and up. its x-component is 6.0 m. part a what is the value of its y-co
lidiya [134]

The position vector can be transcribed as:

A<span> = 6 i + y j                           </span>

i <span>points in the x-direction and j points in the y-direction.</span>

The magnitude of the vector is its dot product with itself:

<span>|A|2 = A·A</span>

<span>102  = (6 i + y j)•(6 i+ y j)            Note that i•j = 0, and  i•i  = j•j = 1 </span>

<span>100  = 36 + y2       </span>

<span>64    = y2</span>

<span>get the square root of 64 = 8</span>

<span>The vertical component of the vector is 8 cm.</span>

3 0
2 years ago
In 1958 a large earthquake in Alaska produced a tsunami. What was the approximate height of the tsunami?
lilavasa [31]
The approximate height of the tsunami in Alaska in 1958 is 1720ft
6 0
3 years ago
Read 2 more answers
How do I solve the following problem?
lukranit [14]
Use the impulse-momentum theorem.

Ft = mv

Substitute your known values:

(120kg)(20m/s) = F(1.5)&#10;&#10;F= 1600N

Hope this helps!
7 0
3 years ago
Other questions:
  • Which is the symbol for gold?
    11·2 answers
  • The Hubble constant, $H_{0}$, is the ratio of the recessional velocity of a galaxy to its distance. The Hubble constant is estim
    10·1 answer
  • What is the term that we use to describe two species living together?
    7·1 answer
  • A uniform electric field of magnitude 6.8 × 10 5 N/C points in the positive x direction. (a) Find the electric potential differe
    6·1 answer
  • What is the amount of a certain isotope that occurs in a natural sample of an element
    6·2 answers
  • An upward force act on a proton as it moves with a speed of 2.0 x 10^5 meters/seconds through a magnetic field of 8.5 x 10^2
    7·1 answer
  • What are the starting materials for nuclear fission
    16·2 answers
  • 3) A driver in a 1000-kg car traveling at 24 m/s slams on the brakes and skids to a stop. If the coefticient of friction between
    12·1 answer
  • the length of iron rod at 100 C is 300.36 cm and at 159 C is 300.54 cm.Calculate its length at 0 c and coefficient of linear exp
    12·1 answer
  • Demarcus launches a small weight into the air. The weight takes 6.4 seconds to reach the ground again. At what time did the weig
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!