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
saul85 [17]
3 years ago
13

A rocket passes you at a speed of 0.85c, and you measure its length to be 35.0 m. What is its measured length when at rest?

Physics
1 answer:
viktelen [127]3 years ago
8 0

Answer:

66.4 m

Explanation:

To solve the problem, we can use the length contraction formula, which states that the length observed in the reference frame moving with the object (the rocket) is given by

L=L_0 \sqrt{1-(\frac{v}{c})^2}

where

L_0 is the proper length (the length measured from an observer at rest)

v is the speed of the object (the rocket)

c is the speed of light

Here we know

v = 0.85c

L = 35.0 m

So we can re-arrange the equation to find the length of the rocket at rest:

L_0 = \frac{L}{\sqrt{1-(\frac{v}{c})^2}}=\frac{35.0}{\sqrt{1-(\frac{0.85c}{c})^2}}=66.4 m

You might be interested in
19. What gives an atomic nucleus its
lana [24]

Answer: The correct choice is, B: protons

Explanation:

Protons have a positive charge whereas neutrons have no charge and electrons have a negative charge. Due to positive charge of protons, nucleus of an atom acquires a positive charge. Therefore, we can conclude that protons are the particles give the nucleus its positive charge.

3 0
2 years ago
The distance from Earth to the star Epsilon Eridani is about 10.5 light years. Which of the following statements is true?
Simora [160]
<span>The correct answer is B. - It would take a ray of light 10.5 light years to travel from Earth to Epsilon Eridani, or vice-versa. Using our current technology it would take far longer than 21.0 years for a space ship from Earth to travel that far - I would have to guess many hundreds of years.</span>
4 0
3 years ago
Read 2 more answers
Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's v
mihalych1998 [28]

Answer:

a.

\displaystyle a(0 )=8.133\ m/s^2

\displaystyle a(2)=2.05\ m/s^2

\displaystyle a(4)=0.52\ m/s^2

b.\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15

c. t=9.9 \ sec

Explanation:

Modeling With Functions

Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

\displaystyle V(t)=a(1-e^{bt})

For Carl Lewis's run at the 1987 World Championships, the values of a and b are

\displaystyle a=11.81\ ,\ b=-0.6887

Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

\displaystyle V(t)=11.81(1-e^{-0.6887t})

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})

\displaystyle a(t)=8.133547\ e^{-0.6887t}

For t=0

\displaystyle a(0)=8.133547\ e^o

\displaystyle a(0 )=8.133\ m/s^2

For t=2

\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}

\displaystyle a(2)=2.05\ m/s^2

\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}

\displaystyle a(4)=0.52\ m/s^2

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C

\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

\displaystyle x(0)=0=>11.81\times1.45201+C=0

Solving for C

\displaystyle c=-17.1482\approx -17.15

Now we complete the equation for the distance

\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15

c. Find the time Lewis needed to sprint 100.0 m.

The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100

Rearranging

\displaystyle t+1.45\ e^{-0.6887t}=9.92

We define an auxiliary function f(t) to help us find the value of t.

\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92

Let's try for t=9 sec

\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92

Now with t=9.9 sec

\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184

That was a real close guess. One more to be sure for t=10 sec

\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).  

At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}

7 0
3 years ago
The specific gravity of mantle rock is about 3.3. True False
Alla [95]
True is the correct answer
5 0
3 years ago
which of these changes does a submarine encounter as it returns from the bottom of the ocean to the surface of the ocean
fiasKO [112]

a.the amount of sunlight increases.

Explanation:

As a submarine rises to the surface, the change it encounters that is true from the given options is that the amount of sunlight increases.

The bottom of the ocean is dark and receives little to no sunlight due to the scattering of the rays by ocean water.

  • As the submarine rises, the volume of water column on it decreases and the pressure on it decreases too.
  • Also, the temperature rises steadily to the surface.

learn more:

Heat and temperature brainly.com/question/914750

#learnwithBrainly

7 0
3 years ago
Other questions:
  • A car weighing 11.1 kN and traveling at 13.4 m/s without negative lift attempts to round an unbanked curve with a radius of 61.0
    13·1 answer
  • Which one of the following statements regarding strength training workouts is most accurate
    6·1 answer
  • Arthur and Betty start walking toward each other when they are 100 m apart. Arthur has a speed of 3.0 m/s and Betty has a speed
    13·2 answers
  • An elastic loaded balloon launcher fires balloons at an angle of [38.0o above horizontal] from the surface of the ground. if the
    10·1 answer
  • __________ is the gradual increase in the temperature of earth’s atmosphere. greenhouse effect air pollution global warming temp
    12·2 answers
  • A cart moving at 2.7 m/s travels for 2 minutes, How far did it go?​
    6·1 answer
  • Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example
    7·1 answer
  • 2. True or false. Gravity is the strongest non-contact force on Earth.
    6·2 answers
  • What is (9.8) / (6.75x10-6)
    15·1 answer
  • What direction does tangential velocity point?
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!