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NNADVOKAT [17]
3 years ago
15

Please help! I'm not sure what equation or the process to do this question.

Physics
1 answer:
lions [1.4K]3 years ago
5 0

Answer:

The momentum is 1.94 kg m/s.

Explanation:

To solve this problem we equate the potential energy of the spring with the kinetic energy of the ball.

The potential energy U of the compressed spring is given by

U = \dfrac{1}{2} kx^2,

where x is the length of compression and k is the spring constant.

And the kinetic energy of the ball is

K.E = \dfrac{1}{2}mv^2.

When the spring is released all of the potential energy of the spring goes into the kinetic energy of the ball; therefore,

\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2,

solving for v we get:

v = x \sqrt{\dfrac{k}{m} }.

And since momentum of the ball is p=mv,

p =mx \sqrt{\dfrac{k}{m} }.

Putting in numbers we get:

p =(0.5kg)(0.25m) \sqrt{\dfrac{(120N/m)}{0.5kg} }.

\boxed{p=1.94kg\: m/s}

You might be interested in
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
To determine the height of a tall building such as Sears Tower in Chicago, Illinois a ball was dropped from the top of the build
Darya [45]

Answer:

The height of Sears Tower is 1448.5 feet.

Explanation:

<h3>We apply the free fall formula to the ball: </h3><h3>y=v_{o} *t+\frac{1}{2} *g*t^{2}</h3><h3>y: The vertical distance the ball moves at time t  </h3><h3>v_{o}i: Initial speed </h3><h3>g=Gravity acceleration=9.8*(\frac{\frac{1ft}{0.305m} }{s^{2} } )</h3>

Known information

We know that the vertical distance (y) that the ball moves in 9,5s  is equal to height of Sears Tower (h).  

Too we know that the ball is released from rest, then,v_{0}=0

Height of Sears Tower calculation:

We replace  in the equation 1 the data following;

y=h

v_{o} =0

g=32,1\frac{ft}{s^{2} }

t= 9,5s

h=0*9.5+\frac{1}{2} *32.1*9.5^{2}

h=1448.5 ft

Answer: The height of Sears Tower is 1448.5 ft

6 0
3 years ago
A block of wood is 4 cm wide, 5 cm long, and 10 cm high. It weighs 100 grams. Calculate its volume. Calculate its density. Will
77julia77 [94]

Answer:

V=200cm^3\\\\\rho =0.500g/cm^3

It will float.

Explanation:

Hello.

In this case, given the width, length and height, we can compute the volume as follows:

V=W*L*H\\\\V=4cm*5cm*10cm\\\\V=200cm^3

Moreover, since the density is computed via the division of the mass by the volume:

\rho =\frac{m}{V}

We obtain:

\rho =\frac{100g}{200cm^3} \\\\\rho =0.500g/cm^3

In such a way, since the solid has a lower density than the water, we infer it will float.

Best regards.

3 0
3 years ago
A stone is thrown horizontally at 8.0 m/s from a cliff 78.4 m high. How far from the base of the cliff does the stone strike the
Ivenika [448]
64 meters from the base of the cliff.
4 0
3 years ago
Read 2 more answers
If NEPA charges 5k per kWh, what is the cost of
tiny-mole [99]

Answer:

24k

Explanation:

We multiply by 200V by 24

8 0
2 years ago
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