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
bearhunter [10]
3 years ago
10

A train whose proper length is 1200 m passes at a high speed through a station whose platform measures 900 m, and the station ma

ster observes that when the train passes it occupies exactly the entire length of the platform. What is the speed of the train?
Physics
1 answer:
TiliK225 [7]3 years ago
6 0

Answer:

0.66c

Explanation:

Use length contraction equation:

L = L₀ √(1 − (v²/c²))

where L is the contracted length,

L₀ is the length at 0 velocity,

v is the velocity,

and c is the speed of light.

900 = 1200 √(1 − (v²/c²))

3/4 = √(1 − (v²/c²))

9/16 = 1 − (v²/c²)

v²/c² = 7/16

v = ¼√7 c

v ≈ 0.66 c

You might be interested in
What is the displacement from a starting position of (14.0, 3.0) m to a final position of (−3.0, −4.0) m?
dem82 [27]

Answer:

change in y = -7

change in x = -17

magnitude of displacement = sqrt(7^2+17^2)

tan of angle below -x axis = 7/17

because in third quadrant where x and y are negative

3 0
3 years ago
A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 19
Hunter-Best [27]

The car will take 300 m before it stops due to applying break.

<h3>What's the relation between initial velocity, final velocity, acceleration and distance?</h3>
  • As per Newton's equation of motion, V² - U² = 2aS
  • V= final velocity velocity of the object, U = initial velocity velocity of the object, a= acceleration, S = distance covered by the object
  • Here, U = 60 ft/sec, V = 0 m/s, a= -6 ft/sec²
  • So, 0² - 60² = 2×6× S

=> -3600 = -12S

=> S = 3600/12 = 300 m

Thus, we can conclude that the distance covered by the car is 300 m before it stopped.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 6 ft/sec². How long will it take before the car stops?

Learn more about the Newton's equation of motion here:

brainly.com/question/8898885

#SPJ1

7 0
2 years ago
The centers of two 15.0 kg spheres are separated by 3.00 m. The magnitude of the gravitational force between the two spheres is
kompoz [17]
 we have to use newtons law of gravitation which is
F=GMm/r^2 
G=6.67 x 10^<span>-11N kg^2/m^2
</span>M=<span>(15kg)
</span>m=15 kg
r=(3.0m)^2<span> 
</span>putting values we have 
<span>=(6.67 x 10^-11N kg^2/m^2)(15kg)(15kg)/(3.0m)^2 </span>
=1.67 x 10^-9N 
7 0
3 years ago
What are the basic si units for the speed of light?.
DanielleElmas [232]
C=meters/second or C=m/s
7 0
2 years ago
If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h
raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory (u_x) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

u^2=u_x^2+u_y^2

Where, u_x\ and\ u_y are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1

Now, we know that, for a projectile motion, the maximum height is given as:

H=\frac{u_y^2}{2g}

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m

Therefore, the maximum height reached is 4.63 m.

3 0
3 years ago
Other questions:
  • The allowed energies of a simple atom are 0.0 eV, 3.0 eV, and 4.0 eV. An electron traveling at a speed of 1.3*10^6 m/s collision
    5·1 answer
  • In most cases, what happens to a liquid when it cools?
    11·1 answer
  • A 21.0 kg shopping cart is moving with a velocity of 4.0 m / s. It strikes a 7.0 kg box that is initially at rest. They stick to
    10·1 answer
  • Gravitational pull is determined by ______?
    13·1 answer
  • Name the three different types of radiation and describe how they are different in their penetrating abilities. Explain.
    6·2 answers
  • Que es patria , Que es civimos , Cuales son los simbolos patrios , Que es ley , Que es la vandera , y como esta conformada
    13·1 answer
  • Describe binary fission with amoeba.​
    11·1 answer
  • Materials A, B, and C are solids that are at their melting temperatures. Material A requires 200 J to melt 4 kg, material B requ
    13·1 answer
  • Which of the following defines heredity?
    5·1 answer
  • Explain the meaning of the error​
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!