HELP ME PLEASEE!!!THIS IS SCIENCE

A 0.140-kg baseball is dropped from rest from a height of 1.8 m above the ground. It rebounds to a height of 1.4 m. What change

aliina [53]

Answer:

$\Delta p=-1.56\ kg-m/s$

Explanation:

It is given that,

Mass of the baseball, m = 0.14 kg

It is dropped form a height of 1.8 m above the ground. Let u is the velocity when it hits the ground. Using the conservation of energy as :

$u=\sqrt{2gh}$

h = 1.8 m

$u=\sqrt{2\times 9.8\times 1.8}$

u = 5.93 m/s

Let v is the speed of the ball when it rebounds. Again using the conservation of energy to find it :

$v=\sqrt{2gh'}$

h' = 1.4 m

$v=-\sqrt{2\times 9.8\times 1.4}$

v = -5.23 m/s

The change in the momentum of the ball is given by :

$\Delta p=m(v-u)$

$\Delta p=0.14(-5.23-5.93)$

$\Delta p=-1.56\ kg-m/s$

So, the change in the ball's momentum occurs when the ball hits the ground is 1.56 kg-m/s. Hence, this is the required solution.

4 0

3 years ago

Steam enters an adiabatic turbine steadily at 7 MPa, 5008C, and 45 m/s, and leaves at 100 kPa and 75 m/s. If the power output of

marusya05 [52]

Answer:

a) $\dot m = 6.878\,\frac{kg}{s}$ , b) $T = 104.3^{\textdegree}C$ , c) $\dot S_{gen} = 11.8\,\frac{kW}{K}$

Explanation:

a) The turbine is modelled by means of the First Principle of Thermodynamics. Changes in kinetic and potential energy are negligible.

$-\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0$

The mass flow rate is:

$\dot m = \frac{\dot W_{out}}{h_{in}-h_{out}}$

According to property water tables, specific enthalpies and entropies are:

State 1 - Superheated steam

$P = 7000\,kPa$

$T = 500^{\textdegree}C$

$h = 3411.4\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

State 2s - Liquid-Vapor Mixture

$P = 100\,kPa$

$h = 2467.32\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

$x = 0.908$

The isentropic efficiency is given by the following expression:

$\eta_{s} = \frac{h_{1}-h_{2}}{h_{1}-h_{2s}}$

The real specific enthalpy at outlet is:

$h_{2} = h_{1} - \eta_{s}\cdot (h_{1}-h_{2s})$

$h_{2} = 3411.4\,\frac{kJ}{kg} - 0.77\cdot (3411.4\,\frac{kJ}{kg} - 2467.32\,\frac{kJ}{kg} )$

$h_{2} = 2684.46\,\frac{kJ}{kg}$

State 2 - Superheated Vapor

$P = 100\,kPa$

$T = 104.3^{\textdegree}C$

$h = 2684.46\,\frac{kJ}{kg}$

$s = 7.3829\,\frac{kJ}{kg\cdot K}$

The mass flow rate is:

$\dot m = \frac{5000\,kW}{3411.4\,\frac{kJ}{kg} -2684.46\,\frac{kJ}{kg}}$

$\dot m = 6.878\,\frac{kg}{s}$

b) The temperature at the turbine exit is:

$T = 104.3^{\textdegree}C$

c) The rate of entropy generation is determined by means of the Second Law of Thermodynamics:

$\dot m \cdot (s_{in}-s_{out}) + \dot S_{gen} = 0$

$\dot S_{gen}=\dot m \cdot (s_{out}-s_{in})$

$\dot S_{gen} = (6.878\,\frac{kg}{s})\cdot (7.3829\,\frac{kJ}{kg\cdot K} - 6.8000\,\frac{kJ}{kg\cdot K} )$

$\dot S_{gen} = 11.8\,\frac{kW}{K}$

4 0

3 years ago

After our sun leaves the main sequence it will most likely

meriva

Our Sun will probably become a red giant before collapsing into a dwarf star.

7 0

3 years ago

A student squeezes a clothespin 82 times in a minute. Then, using the same hand and the same clothespin, he squeezes the clothes

tia_tia [17]

Could have gotten tired after the first trial. ? i’m not sure if that’s it

3 0

3 years ago

A solid sphere of uniform density starts from rest and rolls without slipping down an inclined plane with angle θ = 30o. The sph

kondaur [170]

M*g*sin20-f=m*a
and the rotational frame 
f*r=I*a/r 
where f is the force of friction, a is the translational acceleration and I is the moment of inertia of the sphere 

combine and solve for f 
a=g*sin20-f/m 
and 
f*r^2/I=g*sin20-f/m 
or 
f=g*sin20/(r^2/I+1/m) 

I=2*m*r^2/5 
therefore 
f=2*m*g*sin20/7

8 0

3 years ago