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zzz [600]
3 years ago
8

A 55.0-kg lead ball is dropped from the Leaning Tower of Pisa. The tower is 55.0 m high. What is the speed of the ball after it

has traveled 4.20 m downward
Physics
1 answer:
PolarNik [594]3 years ago
4 0

Answer:

The speed of the ball is 9.07 m/s.

Explanation:

Given that,

Mass of the lead ball, m = 55 kg

Height of the tower, h = 55 m

We need to find the speed of the ball it has traveled 4.20 m downward, x = 4.2 meters

The initial speed of the ball will be 0 as it was at rest initially. Let v is the speed of the ball after it has traveled 4.20 m downward. It is a case of equation of motion such that :

v^2-u^2=2ax

v^2=2ax

Here, a = g

v^2=2\times 9.8\times 4.2

v = 9.07 m/s

So, the speed of the ball is 9.07 m/s. Therefore, this is the required solution of given condition.

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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
Explain Thomsons model of an atom<br><br><br>please its aurgent fast​
Aleks04 [339]

Answer:

Thomson's model showed an atom that had a positively charged medium, or space, with negatively charged electrons inside the medium. After its proposal, the model was called a "plum pudding" model because the positive medium was like a pudding, with electrons, or plums, inside.

3 0
2 years ago
A swing has a period of 10 seconds. What is its frequency ?
kow [346]

Frequency = 1 / (period)

Frequency = 1 / (10 seconds) = (1/10) ( / second) = 0.1 per second = <em>0.1 Hz</em>.


4 0
3 years ago
A leaf is floating on the Surface of the water,what will happen to its movement? Explain
docker41 [41]

Answer:

MRCORRECT has answered the question

Explanation:

surface tension of water helps creatures(mostly of insecta class such as water striders) to walk on water. . it also helps water to move up the xylem tissue ofhigher plants without breaking up

4 0
3 years ago
A girl playing tug-of-war with her dog pulls the dog a distance of 8.0m by exerting a force at an angle of 18° with the horizont
AnnZ [28]

Answer:

25 N

Explanation:

Work is a product of force and perpendicular distance moved.

W=Fd where F is force exerted and d is perpendicular distance.

However, for this case, the distance is inclined hence resolving it to perpendicular so that it be along x-axis we have distance as dcos\theta

Therefore, W=Fdcos\theta

Making F the subject of the formula then

F=\frac {W}{dcos\theta} where \theta is the angle of inclination. Substituting 190 J for W then 18 degrees for \theta and 8 m for d then

F=\frac {190}{8cos18^{\circ}}\approx 25N

3 0
2 years ago
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