What is the formula for calculating the efficiency of a heat engine

red dwarf stars (stars between 0.08 and 0.5 solar masses) evolve very differently than other stars as they age because ____.

Alex_Xolod [135]

Red dwarf stars evolve very differently than other stars as they age because their interiors are well mixed, through strong convection.

<h3>What are red dwarf stars?</h3>

Red dwarf stars are the smallest and coolest kind of stars on the main sequence.

Red dwarf stars (stars between 0.08 and 0.5 solar masses) evolve very differently than other stars as they age because their interiors are well mixed, through strong convection.

Learn more about Red dwarf stars here: brainly.com/question/3151458

#SPJ11

6 0

2 years ago

Which is the correct equation for the elastic potential energy stored in a spring

salantis [7]

(B)

Explanation:

$pe = \frac{1}{2} k {x}^{2}$

3 0

3 years ago

A car travels 90 km per hour how many how many Kilometres of distance does it travel in one second

gavmur [86]

Answer:

Its Moving at a constant Speed of 90km/hr

You can convert this speed to Meters/Second or Change it to Km/Second(Because we need our distance in Km)

So 90km/hr x 1/60 x1/60

=90km/3600sec

v=0.025km/sec

When an Object Moves at Constant Speed... Its displacement(distance) is given by

d=vt

given v=0.025km/sec

t=1sec

d=0.025km/sec x 1sec

d=0.025km.

7 0

3 years ago

Read 2 more answers

James threw a ball vertically upward with a velocity of 41.67ms-1 and after 2 second David threw a ball vertically upward with a

Reptile [31]

Answer:

When have passed 3.9[s], since James threw the ball.

Explanation:

First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.

We'll use the kinematics equations to find these two unknowns.

$y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can find the velocity after 2 seconds.

$v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.

$y=y_{o} +v_{o} *t-\frac{1}{2} *g*t^{2} \\y=63.72 +22.05 *t-\frac{1}{2} *(9.81)*t^{2} \\\\y=63.72 +22.05 *t-4.905*t^{2} \\$

As we can see the equation is based on Time (t).

Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t

$y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]$