PHYSICS COLLEGE

Question 7 of 10A railroad freight car with a mass of 32,000 kg is moving at 2.0 m/s when it
runs into an at-rest freight car with a mass of 28,000 kg. The cars lock
together. What is their final velocity?
A.1.1 m/s
B. 2.2 m/s
C. 60,000 kg•m/s
D. 0.5 m/s

Answers

Answer 1
Answer:

Answer:

a

Explanation:

you take 32,000kg ÷2.0m

Related Questions

To remove 800j of heat the compressor in the fridge does 500j of work. how much heat is released into the room?
A solid 0.6950 kg ball rolls without slipping down a track toward a vertical loop of radius ????=0.8950 m . What minimum translational speed ????min must the ball have when it is a height H=1.377 m above the bottom of the loop in order to complete the loop without falling off the track? Assume that the radius of the ball itself is much smaller than the loop radius ???? . Use ????=9.810 m/s2 for the acceleration due to gravity.
A suspended platform of negligible mass is connected to the floor below by a long vertical spring of force constant 1200 N/m. A circus performer of mass 70 kg falls from rest onto the platform from a height of 5.8 m above it. Find the maximum spring compression
What is the frequency of a photon that has the same momentum as a neutron moving with a speed of 1.90 × 103 m/s?
2. Annealed low-carbon steel has a flow curve with strength coefficient of 75000 psi and strain-hardening exponent of 0.25. A tensile test specimen with a gauge length of 2 in. is stretched to a length of 3.3 in. Determine the flow stress and the average flow stress that the metal experienced during this deformation.

Overnight a thin layer of ice forms on the surface of a 40-ft-wide river that is essentially of rectangular cross-sectional shape. Under these conditions, the flow depth is 3 ft. During the following day the sun melts the ice cover. Determine the new depth if the flowrate remains the same and the surface roughness of the ice is essentially the same as that for the bottom and sides of the river.

Answers

Answer:

the new depth is 2.3 ft

Explanation:

the solution is in the attached Word file

Which of the following statements are true?A. The decrease in the amplitude of an oscillation caused by dissipative forces is called damping. B. The increase in amplitude of an oscillation by a driving force is called forced oscillation. C. In a mechanical system, the amplitude of an oscillation diminishes with time unless the lost mechanical energy is replaced. D. An oscillation that is maintained by a driving force is called forced oscillation.

Answers

Statements that are right as regards oscillation are:

A. The decrease in the amplitude of an oscillation caused by dissipative forces is called damping.

B. The increase in amplitude of an oscillation by a driving force is called forced oscillation.

C. In a mechanical system, the amplitude of an oscillation diminishes with time unless the lost mechanical energy is replaced.

D. An oscillation that is maintained by a driving force is called forced oscillation.

  • Amplitude can be regarded as magnitude of change that is been experienced by oscillating variable with each oscillation.

  • When there is a decrease in the amplitude of an oscillation as a result dissipative forces, then it is regarded as damping.

  • When there is increase in amplitude of an oscillation as a result of driving force then it is termed  forced oscillation.

Therefore, the options are correct.

Learn more at:

brainly.com/question/15272453?referrer=searchResults

Answer:

right A, B, C, D

Explanation:

They ask which statements are true

A) Right. The decrease in amplitude is due to the dissipation of energy by friction and is called damping

B) Right. In resonant processes the amplitude of the oscillation increases, being a forced oscillation

C) Right. In a system with energy loss, the amplitude must decrease, therefore energy must be supplied to compensate for the loss.

D) Right. It is a resonant process the driving force keeps the oscillation of the system

There is a person who throws a coin vertically downward with an initial speed of 11.8 m/s from the roof of a building, 34.0 m above the ground. How long does it take the coin to reach the ground? Answer in s.

Answers

Answer:

Time taken by the coin to reach the ground is 1.69 s

Given:

Initial speed, v = 11.8 m/s

Height of the building, h = 34.0 m

Solution:

Now, from the third eqn of motion:

v'^(2) = v^(2) + 2gh

v'^(2) = 11.8^(2) + 2* 9.8* 34.0 = 805.64

v' = √(805.64) = 28.38 m/s

Now, time taken by the coin to reach the ground is given by eqn (1):

v' = v + gt

t = (v' - v)/(g) = (28.38 - 11.8)/(9.8) = 1.69 s

The Golden Gate Bridge in San Francisco has a main span of length 1.28 km, one of the longest in the world. Imagine that a steel wire with this length and a cross-sectional area of 3.10 ✕ 10^−6 m^2 is laid on the bridge deck with its ends attached to the towers of the bridge, on a summer day when the temperature of the wire is 43.0°C. When winter arrives, the towers stay the same distance apart and the bridge deck keeps the same shape as its expansion joints open. When the temperature drops to −10.0°C, what is the tension in the wire? Take Young's modulus for steel to be 20.0 ✕ 10^10 N/m^2. (Assume the coefficient of thermal expansion of steel is 11 ✕ 10−6 (°C)−1.)

Answers

Answer:

361.46 N

Explanation:

\alpha = Coefficient of thermal expansion = 11* 10^(-6)\ /^(\circ)C

Y = Young's modulus for steel = 20* 10^(10)\ Pa

A = Area = 3.1* 10^(-6)\ m^2

L_0 = Original length = 1.28 km

\Delta T = Change in temperature = 45-(-10)

Length contraction is given by

\Delta L=\alpha L_0\Delta T

Also,

\Delta L=(L_0T)/(YA)

\alpha L_0\Delta T=(L_0T)/(YA)\n\Rightarrow T=\alpha \Delta TYA\n\Rightarrow T=11* 10^(-6)* (43-(-10))* 20* 10^(10) * 3.1* 10^(-6)\n\Rightarrow T=361.46\ N

The tension in the wire is 361.46 N

(PLEASE HELP ITS DUE SOON ILL MARK BRAINLIEST AND 5 STARS & PLEASE SHOW WORK!!)(And the answer is not 44 I already tried that and it doesn’t start with 4 either)

Answers

Lol I would help you but I have no clue

A spring oscillator is designed with a mass of 0.231 kg. It operates while immersed in a damping fluid, selected so that the oscillation amplitude will decrease to 1.00% of its initial value in 9.43 s. Find the required damping constant for the system.

Answers

Answer:

.487 s⁻¹

Explanation:

Let damping constant be τ . The equation of decreasing amplitude can be written as

A = A₀ e^{-\tau t

A / A₀ = e^{-\tau t

At t = 9.43 s , A / A₀ = .01

.01 = [e^{-\tau*9.43

ln.01 = - 9.43 τ

-4.6 = -9.43τ

τ = .487 s⁻¹

Answer:

0.05508 kg/sec

Explanation:

mass of the oscillator m= 0.231 Kg

amplitude of oscillation given by

A=A_0e^(-It)

Ao= maximum amplitude

t= time and  1.00% of its initial value in t= 9.43 s.

A= 0.01Ao

⇒0.01=e^(-I×9.43)

ln100= 9.43×l

l=0.4883

we know that l= c/2m

c= damping constant

c= 2ml

=2×0.231×0.4883

=0.05508 kg/sec
Other Questions
What are 4 ways individuals can influence the government?
Which is true about inelastic collisions: a. An inelastic collision does not obey conservation of energy. b. An inelastic collision conserves kinetic energy. c. Objects will stick together upon collision. d. Momentum is not conserved in inelastic collisions..
Two point charges have a total electric potential energy of -24 J, and are separated by 29 cm.If the total charge of the two charges is 45 μC, what is the charge, in μC, on the positive one? What is the charge, in μC, on the negative one?
A crate of eggs is located in the middle of the flatbed of a pickup truck as the truck negotiates a curve in the flat road. The curve may be regarded as an arc of a circle of radius 35.0 m. If the coefficient of static friction between crate and truck is 0.600, how fast can the truck be moving without the crate sliding?