What is the kinectic energy of 150kg object that is moving at 72 m/s

Neon, helium, methane, krypton, xenon, hydrogen, and ozone make up & what of the atmosphere

Alinara [238K]

Part of the upper atmosphere

6 0

3 years ago

. A 1.00 kg rock is thrown up into the air from ground level at a speed of 8.00 m/s. The ball travels up

Liula [17]

Answer:

p = -8 kg-m/s

Explanation:

Given that,

Initial speed of the rock, u = 8 m/s

Mass of the rock, m = 1 kg

The ball travels up to a maximum height, then returns to the ground.

We need to find the rock's momentum as it strikes the ground. Let v be the final speed of the rock. Its final speed is as same as initial speed i.e. 8 m/s but in negative direction. So

p = mv

p = 1 kg × (-8 m/s)

= -8 kg-m/s

So, the rock's momentum as it strikes the ground is (-8 kg-m/s).

5 0

2 years ago

[1] The assembly starts from rest and reaches an angular speed of 150 rev/min under the action of a 20-N force T applied to the

ExtremeBDS [4]

Answer:

t = 5.89 s

Explanation:

To calculate the time, we need the radius of the pulley and the radius of the sphere which was not given in the question.

Let us assume that the radius of the pulley ( $r_p$ ) = 0.4 m

Let the radius of the sphere (r) = 0.5 m

w = angular speed = 150 rev/min = (150 × 2π / 60) rad/s = 15.708 rad/s

Tension (T) = 20 N

mass (m) = 3 kg each

$\int\limits^0_t {Tr_p} \, dt=H_2-H_1\\( Tr_p)t=4rm(rw)\\( Tr_p)t=4r^2mw$

$t = \frac{4r^2mw}{Tr_P}$

Substituting values:

$t = \frac{4r^2mw}{Tr_P}= \frac{4*(0.5)^2*3*15.708}{20*0.4}=5.89s$

7 0

3 years ago

Read 2 more answers

efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.

PolarNik [594]

Answer:

Inlet : $v_i=0.0646\frac{m}{s}$

Outlet: $v_o=0.171\frac{m}{s}$

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

$\dot{m}=\frac{\dot{V}}{\upsilon}$ , where $\dot{V}$ represent the flow rate and $\upsilon$ the specific volume at the pressure and temperature given.

$A_i=0.5m^2$ is the inlet area

$P_i=600Kpa$ pressure at the inlet area

$T_i=70C$ temperature at the inlet area

$A_o=1m^2$ is the outlet area

$P_o=100Kpa$ pressure at the outlet area

$T_o=C$ temperature at the outlet area

$\dot{m}=0.75\frac{kg}{s}$ represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

$\upsilon_i =0.04304\frac{kg}{m^3}$ and the entropy is $h_i=1.0645\frac{KJ}{KgK}=h_o$

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy $h_o=1.0645\frac{KJ}{KgK}$