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
Simora [160]
3 years ago
15

Two equal mass carts approach each other with velocities equal in magnitude but opposite in direction. Friction can be neglected

. If the carts collide completely inelastically, what will be the final velocity of the combined system?
To the left with half the initial velocity
Not enough information!
0
To the right with half the intial velocity
Physics
1 answer:
Viefleur [7K]3 years ago
7 0

Answer:

0

Explanation:

Since momentum is given by:

\vec{p}=m.\vec{v}

Therefore momentum is a vector quantity in the direction of velocity.

  • Inelastic collision is the type of collision in which the masses combine after collision and start moving in the direction of the greater momentum but here in this case we have the momentum equal and opposite in direction.

<u>This case can be described mathematically as:</u>

m_1.\vec{v_1}=m_2.\vec{v_2}

since the mass and velocity are equal:

m.\vec{v}=m.\vec{v}

net momentum:

p_{net}=0

You might be interested in
For which group of people might brain-powered cars allow the greatest increase in mobility?
dolphi86 [110]
Brain-powered cars?
The answer would be people who lack motor skills in their legs.
8 0
3 years ago
Read 2 more answers
A car moving at 10.0 m/s encounters a depression in the road that has a circular cross-section with a radius of 30.0 m. What is
Dennis_Churaev [7]

Answer:

F = 789 Newton

Explanation:

Given that,

Speed of the car, v = 10 m/s

Radius of circular path, r = 30 m

Mass of the passenger, m = 60 kg

To find :

The normal force exerted by the seat of the car when the it is at the bottom of the depression.

Solution,

Normal force acting on the car at the bottom of the depression is the sum of centripetal force and its weight.

N=mg+\dfrac{mv^2}{r}

N=m(g+\dfrac{v^2}{r})

N=60\times (9.81+\dfrac{(10)^2}{30})

N = 788.6 Newton

N = 789 Newton

So, the normal force exerted by the seat of the car is 789 Newton.

6 0
3 years ago
A mechanical wave is a wave that ____ it’s engery
vesna_86 [32]

Does require a medium to transfer

6 0
3 years ago
Read 2 more answers
What (in N/C) is the electric field 3.00 m from the center of the terminal of a Van de Graaff with an 8.00 mC charge, noting tha
mixer [17]

To solve this problem we will apply the concept related to the electric field defined from the laws of Coulomb. For this purpose we will remember that the electric field is equivalent to the product of the Coulomb constant due to the change of the charge over the squared distance, mathematically this is

E = \frac{kq}{r^2}

Here,

k = Coulomb's constant

r = Distance from center of terminal to point where electric field is to found

q = Excess charge placed on the center of terminal of Van de Graff's generator

Replacing we have that,

E = \frac{(9*10^{9})(8*10^{-3})}{3^2}

E = 8*10^6N/C

Therefore the electric field is 8*10^6N/C

6 0
3 years ago
A gas contained within a piston-cylinder assembly undergoes two processes, A and B, between the same end states, 1 and 2, where
Sidana [21]

Answer:

a) 90 kJ

b) 230.26 kJ

Explanation:

The pressure at the first point  P_{1}= 10 bar —> 10 x 102 = 1020 kPa

The volume at the first point  V_{1}= 0.1 m^3  

The pressure at the second point P_{2}= 1 bar —> 1 x 102 = 102 kPa

The volume at the second point V_{2} = 1 m^3  

Process A.

constant volume V = C from point (1) to P = 10 bar.

Constant pressure P = C to the point (2).  

Process B.

The relation of the process is PV = C  

Required  

For process A & B

(a) Sketch the process on P-V coordinates

(b) Evaluate the work W in kJ.  

Assumption  

Quasi-equilibrium process

Kinetic and potential effect can be ignored.  

Solution

For process A.

V=C  

There is no change in volume then

W_{a(1)}= 0\\P=10^{2}

The work is defined by  

W_{a(2)}=\int\limits^V_V {P} \, dV

W_{A(2)} =║10^{2} V║limit 1--0.1

W_{A(2)} = 90 kJ

Process B  

PV=C  

By substituting with point (1) C = 10^2 x 1= 10^2  

The work is defined by

W_{b}=\int\limits^V_V {P} \, dV\\P=10^{2} V^{-1}\\ W_{b}=\int\limits^V_V {10^{2} V^{-1}} \, dV\\\\

W_{A(2)} = ║10^{2} ln(V)║limit 1--0.1

         =230.26 kJ

3 0
3 years ago
Other questions:
  • The pituitary gland releases chemicals known as?
    8·1 answer
  • What did the romans build as a step toward environment management
    11·1 answer
  • What the electron and outermost energy level of periods 2and3,4,5 and 6
    12·1 answer
  • I need help with 3 question.
    5·1 answer
  • A rock exerts 5000 Pa of pressure on the ground. If the rock weighs 250 N, how much area is in contact with the ground?
    15·2 answers
  • Torque can cause the angular momentum vector to rotate in UCM. This motion is called ___________.
    5·1 answer
  • Metamorphic rocks are formed at high temperature and medium to low pressure
    6·1 answer
  • One of the predicted problems due to global warming is that ice in the polar ice caps will melt and raise sea level everywhere i
    13·1 answer
  • what output force is generated when an input force of 630 n is applied to a machine with a mechanical advantage of 3
    8·2 answers
  • How are wavelength and frequency related for a wave moving at a constant speed?
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!