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sesenic [268]
3 years ago
13

A baseball player hits a ball at an angle of 52 degrees and at a height of 4.8 ft. if the ball's initial velocity after being hi

t is 150 and if no one catches the ball when will it hit the ground?
Physics
2 answers:
Natalija [7]3 years ago
8 0
The vertical velocity of the ball is: (150\frac{ft}{s})(sin52) = 118.2\frac{ft}{s}. Since gravity acts at -32\frac{ft}{s^2}, this means that the time of flight up to the maximum height (at which vertical velocity is 0) would be:

(0 - 118.2)=(-32)t \\ t = 3.69 s
This maximum height is:
4.8 + 118.2t + 0.5(-32)t^2 \\ = 4.8 + 118.2(3.69) - 16(3.69)^2  \\ = 223.1 ft
Then the time for it to fall down will be:

0.5(-32)(t_{2})^2=-223.1 \\ t_2 = 3.73s
The total time will be 3.69 + 3.73 = 7.42 seconds.
enot [183]3 years ago
7 0

Answer:

C. 7.43

Explanation:

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A bowler who always left the same three pins standing could be considered a(n) ____ bowler.
RideAnS [48]
A bowler who always left the same 3 pins standing could be considered a C. Precise bowler as from bowling countless number of times he has observed the same amount of pins knocked down each time.
4 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
What are two examples of goods and two examples of services
julsineya [31]

The goods and the services make up the basis of every economy. The goods can simply be defined as merchandise or possessions. The services can be defined as the actions through which help is provided, or work is done for someone else. Example of goods are the food and furniture, with the food being crucial for the survival of the people, while the furniture is an essential part of every home and its practicality and decor. Examples of services are teaching and car repairing. The teaching is crucial for the development of the societies, as through it the people get education, while the repairing of cars is very important as lot of people have them, can not afford to buy new ones all the time, and they need for their daily movement over longer distances.

8 0
3 years ago
John pushes a box with a constant force as shown in the graph below.
andrew11 [14]
From the graph, it can be seen that the constant force that John exerted in order to move the object is 14N. Work is calculated by multiplying the force with the distance to which the object moves in parallel with the direction of the force. 
                                      Work = Force x displacement
                                      Work = (14 N) x (8 m)
                                        Work = 112 J
The closest value is 110J. Thus, the answer to this item is the second choice. 
4 0
3 years ago
A player at first base catches a throw traveling 38 m/s. The baseball, which has a mass of 0.145 kg, comes to a complete stop in
mixas84 [53]

Answer:

F = 39.36 N

Explanation:

given,

initial speed, u = 38 m/s

final speed, v = 0 m/s

mass of ball = 0.145 Kg

time, t = 0.14 s

Force = ?

using impulse formula

J = change in momentum

J = F x t

m(v - u) = F x t

0.145 x (0 - (-38)) = F x 0.14

F x 0.14 = 5.51

F = 39.36 N

force exerted by the ball is equal to 39.36 N.

5 0
3 years ago
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