A skydiver has jumped out of an aircraft but has not yet opened her

During a game the same batter swings at a ball thrown by the pitcher and hits a line drive. Just before the ball is hit it is tr

sergeinik [125]

Answer:

The total work on the ball is 36.25 Joules

Explanation:

There is an important principle on classical mechanics that is the work-energy principle it states that the total work on an object is equal the change on its kinetic energy, mathematically expressed as:

$W_{net}=\Delta K = K_f -K_i$ (1)

With W net the total work, Kf the final kinetic energy and Ki the initial kinetic energy. We're going to use this principle to calculate the total work on the baseball by the force exerted by the bat.

Kinetic energy is the energy related with the movement of an object and every classical object with velocity has some kinetic energy, it is defined as:

$K=\frac{mv^2}{2}$

With m the mass of the object and v its velocity, knowing this we can use on:

$W_{net}= \frac{mv_f^2 -mv_i^2}{2}=\frac{m(v_f^2 -v_i^2)}{2}$

In our case vf is the velocity just after the hit and vi the velocity just before the hit. For an average baseball its mass is 145g that is 0.145 kg, then

$W_{net}=\frac{0.145*(30.0^2 -20.0^2)}{2}$

$W_{net}=36.25 J$

8 0

3 years ago

A car of mass 960 kg is free-wheeling down an incline at a constant speed of 9.0 m s–1.

Orlov [11]

Need to draw it first.....
but there is no option here to draw .. why?
anyway , here the car is going with a uniform velocity..
so the acceleration of the car=0, a=0
so resistive force on the car=component along the inclined plane of the weight of the car=mg*cos(90-15)=960*9.8*cos 75=?

4 0

3 years ago

A straight wire lies along the y-axis initially carrying a current of 10 A in the positive y-direction. The current decreases an

Elan Coil [88]

Answer:

Explanation:

The magnetic field due to straight wire is into the square coil.

As the current in straight wire decreases the magnetic flux in the coil decreases . The induced magnetic field is into the coil.The induced current is along +y direction

8 0

4 years ago

When a metal rod is heated, its resistance changes both because of a change in resistivity and because of a change in the length

monitta

Answer:

$3.34\Omega$

Explanation:

The resistance of a metal rod is given by

$R=\frac{\rho L}{A}$

where

$\rho$ is the resistivity

L is the length of the rod

A is the cross-sectional area

The resistivity changes with the temperature as:

$\rho(T)=\rho_0 (1+\alpha (T-T_0))$

where in this case:

$\rho_0$ is the resistivity of silver at $T_0=21.0^{\circ}C$

$\alpha=6.1\cdot 10^{-3} ^{\circ}C^{-1}$ is the temperature coefficient for silver

$T=180.0^{\circ}C$ is the current temperature

Substituting,

$\rho(180^{\circ}C)=\rho_0 (1+6.1\cdot 10^{-3}(180-21))=1.970\rho_0$

The length of the rod changes as

$L(T)=L_0 (1+\alpha_L(T-T_0))$

where:

$L_0$ is the initial length at $21.0^{\circ}C$

$\alpha_L = 18\cdot 10^{-6} ^{\circ}C^{-1}$ is the coefficient of linear expansion

Substituting,

$L(180^{\circ}C)=L_0(1+18\cdot 10^{-6}(180-21))=1.00286L_0$

The cross-sectional area of the rod changes as

$A(T)=A_0(1+2\alpha_L(T-T_0))$

So, substituting,

$A(180^{\circ}C)=A_0(1+2\cdot 18\cdot 10^{-6}(180-21))=1.00572A_0$

Therefore, if the initial resistance at 21.0°C is

$R_0 = \frac{\rho_0 L_0}{A_0}=1.70\Omega$

Then the resistance at 180.0°C is:

$R(180^{\circ}C)=\frac{\rho(180)L(180)}{A(180)}=\frac{(1.970\rho_0)(1.00285L_0)}{1.00572A_0}=1.9644\frac{\rho_0 L_0}{A_0}=1.9644 R_0=\\=(1.9644)(1.70\Omega)=3.34\Omega$

7 0

3 years ago

What is the car’s average velocity (in m/s) in interval between t=1.0s to t=1.5s?

Lemur [1.5K]

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, the formula for velocity is;

Velocity (in m/s) = distance/time

The distance the car covered in the completed question is divided by the difference in the time interval

The difference in the time interval will be = 1.5s - 1.0s = 0.5s

NOTE: the distance must be in meters or be converted to meters

7 0

3 years ago