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sesenic [268]
3 years ago
7

A 2000 kg car is moving at 15 m/sec when it collides with a 1200 kg car sitting still. Briefly compare the impulse imparted on t

he 1200 kg car by the 2000 kg car to the impulse imparted on the 2000 kg car by the 1200 kg car. Which car undergoes the greater change in momentum? Briefly explain your response.
Physics
1 answer:
Roman55 [17]3 years ago
4 0

Answer:

The 1200 kg car.

Explanation:

The 1200 kg car had bo momentum at the initial stage of the journey because it wasn't on motion. But after collision with the 2000 kg car it gained alot of momentum.

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2 Which is true of a parallel circuit?
mars1129 [50]

Answer:

fxb

c

Explanation:bfffffffff

8 0
3 years ago
A circular ring with area 4.45 cm2 is carrying a current of 13.5 A. The ring, initially at rest, is immersed in a region of unif
Gwar [14]

Answer:

a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m

b) ΔU = -0.000747871 J

c)  w = 47.97 rad / s

Explanation:

Given:-

- The area of the circular ring, A = 4.45 cm^2

- The current carried by circular ring, I = 13.5 Amps

- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)

- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)  

- The magnetic moment final orientation, vec ( μf ) = -μ k^

- The inertia of ring, T = 6.50×10^−7 kg⋅m2

Solution:-

- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:

                    μ = N*I*A

Where,

           N: The number of turns

           I : Current in coil

           A: the cross sectional area of coil

- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):

                    μ = 1*( 13.5 ) * ( 4.45 / 100^2 )

                    μ = 0.0060075 A-m^2

- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:

             vec ( τi ) = vec ( μi ) x vec ( B )

              = 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )

              = ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

- Perform cross product:

          \left[\begin{array}{ccc}i&j&k\\-0.004806&0.0036045&0\\0.126&0.0315&-0.042\end{array}\right]  = \left[\begin{array}{ccc}-0.00015139\\-0.00020185\\-0.00060556\end{array}\right] \\\\

- The initial torque ( τi ) is written as follows:

           vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )

           

- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):

- The initial potential energy stored in the circular ring ( Ui ) is:

          Ui = - vec ( μi ) . vec ( B )

          Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

          Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]

          Ui = - [(-0.000605556 + 0.00011)]

          Ui = 0.000495556 J

- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :

          Uf = - vec ( μf ) . vec ( B )

          Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

          Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]

          Uf = -0.000252315 J

- The decrease in magnetic potential energy of the ring is arithmetically determined:

          ΔU = Uf - Ui

          ΔU = -0.000252315 - 0.000495556  

          ΔU = -0.000747871 J

Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.

- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.

- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:

                Ui + Eki = Uf + Ekf

Where,

             Eki : The initial kinetic energy ( initially at rest ) = 0

             Ekf : The final kinetic energy at second position

- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.    

               -ΔU = Ekf

                0.5*T*w^2 = -ΔU

                w^2 = -ΔU*2 / T

Where,

                w: The angular speed at second position

               w = √(0.000747871*2 / 6.50×10^−7)

              w = 47.97 rad / s

6 0
3 years ago
A planet's temperature depends on its distance from the Sun as well as the strength of its greenhouse effect. Let's remove the e
pishuonlain [190]

Answer:

Smallest Effect

Mars

Earth

Venus

Largest Effect

5 0
3 years ago
A pendulum has 294 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o
baherus [9]

Newton's law of conservation states that energy of an isolated system remains a constant. It can neither be created nor destroyed but can be transformed from one form to the other.


Implying the above law of conservation of energy in the case of pendulum we can conclude that at the bottom of the swing the entire potential energy gets converted to kinetic energy. Also the potential energy is zero at this point.


Mathematically also potential energy is represented as


Potential energy= mgh


Where m is the mass of the pendulum.


g is the acceleration due to gravity


h is the height from the bottom z the ground.


At the bottom of the swing,the height is zero, hence the potential energy is also zero.


The kinetic energy is represented mathematically as


Kinetic energy= 1/2 mv^2


Where m is the mass of the pendulum


v is the velocity of the pendulum


At the bottom the pendulum has the maximum velocity. Hence the kinetic energy is maximum at the bottom.


Energy can neither be created e destroyed. It can only be transferred from one form to another. Implying this law and the above explainations we conclude that at the bottom of the pendulum,the potential energy=0 and the kinetic energy=294J as the entire potential energy is converted to kinetic energy at the bottom.



6 0
3 years ago
Suppose that a public address system emits sound uniformly in all directions and that there are no reflections . The intensity a
Marina86 [1]

Answer:

I_{2}=2.39*10^{-5} W/m^{2}

Explanation:

The definition of the intensity in terms of power is given by:

I=\frac{P}{A}

Where:

  • P is the power
  • A is the area

If the sound emits uniformly in all directions and that there are no reflections, we can assume the geometry of the wave sound is spherical.

Let's recall the area of a sphere is A = 4\pi R^{2}

To the first location we have:

I_{1}=3*10^{-4} W/m^{2}=\frac{P}{4\pi 22^{2}}

and to the second location we have:

I_{2}=\frac{P}{4\pi 78^{2}}

Now, we can divide each intensity to find the second intensity.

\frac{I_{2}}{I_{1}}=\frac{\frac{P}{4\pi 78^{2}}}{\frac{P}{4\pi 22^{2}}}

I_{2}=I_{1}* \frac{22^{2}}{78^{2}}

I_{2}=3*10^{-4}\frac{22^{2}}{78^{2}}

I_{2}=2.39*10^{-5} W/m^{2}

I hope it helps you!

     

5 0
3 years ago
Read 2 more answers
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