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3 years ago

Which of the following correctly describes the first animals that appeared?

Physics

1 answer:

Anettt [7]3 years ago

5 0

A these were large creatures similar to dinosaurs

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Light with wavelength in air ( lambdaair ) is incident on a oil slick ( noil = 1. 25) floating on the ocean ( nwater = 1. 33). w

crimeas [40]

26mm is the thinnest thickness of oil that will brightly reflect the light.

What is wavelength ?

The distance over which a periodic wave's shape repeats is known as the wavelength in physics. It is a property of both traveling waves and standing waves as well as other spatial wave patterns. It is the distance between two successive corresponding locations of the same phase on the wave, such as two nearby crests, troughs, or zero crossings. The spatial frequency is the reciprocal of wavelength. The Greek letter lambda () is frequently used to represent wavelength. The term wavelength is also occasionally used to refer to modulated waves, their sinusoidal envelopes, or waves created by the interference of several sinusoids.

To learn more about wavelength visit:

brainly.com/question/16051869

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4 0

1 year ago

f the pressure of a sample of gas is doubled at constant temperature, what happens to the volume of the gas?

vaieri [72.5K]

it is halved.

p1v1=p2v2

2p1x1/2xv1=p1v1

4 0

2 years ago

Where are the magnetic fields strongest near a bar magnet?

mixer [17]

<h2>ANSWER:</h2>

The north pole

The magnetic field of a bar magnet is strongest at either pole of the magnet. It is equally strong at the north pole when compared with the south pole. The force is weaker in the middle of the magnet and halfway between the pole and the center.

CarryOnLearning(◍•ᴗ•◍)

6 0

2 years ago

A steam Rankine cycle operates between the pressure limits of 1500 psia in the boiler and 2 psia in the condenser. The turbine i

AlladinOne [14]

Answer:

a. Mass flow rate through the boiler = 5.462lbm/s

b. Power produced by the turbine = 2525.8kW

c. The rate of heat supply in the boiler = 6901.42Btu/s

d. Thermal efficiency of the cycle = 34.3%

Explanation:

In order to provide a solution, we must assume that ;

- The system is operating at a steady condition

- Kinetic and potential energy changes are negligible

Now from steam tables, we calculate specific volume $v$ and enthalpy $h$ as,

$h_1 = 95.96Btu/lb$ ( $h_1 = h_f$ at $2psia$ )

$v_1 = 0.016238ft^3/lb$ ( $v_1 = v_f$ at $2psia$ )

$w_{p,in} = v_1(P_2-P_1) = 0.016238(1500-2) * \frac{1}{5.404} = 4.501 Btu/lb$

$w_p = h_2 - h_1\\h_2 = w_p+h_1=4.501+95.96=100.461Btu/lb$

$h_3 = 1364.0Btu/lb$

$s_3 = 1.5073Btu/lb.R$

( at $P_3 = 1500psia$ & $T_3 = 800^0F$ )

$P_4 = 2psia\\S_4 = S_3\\x_4S = \frac{S_4-S_f}{S_{fg}}=\frac{1.5073-0.1783}{1.7374}=0.765$

( $S_f$ & $S_{fg}$ when pressure is 2psia)

$h_4S = h_f+x_4S*h_{fg}=95.96+(0.765)(1021.0)=877.025Btu/lb$

$n_T= \frac{h_3-h_4}{h_3-h_4S}\\ h_4=h_3-n_T(h_3-h_4S)=1364.0-0.90(1364.0-877.025)=925.7Btu/lb$

Therefore,

$q_{in}=h_3-h_2=1364.0-100.461=1263.54Btu/lb\\q_{out}=h_4-h_1=925.7-95.96=829.74Btu/lb\\w_{net}=q_{in}-q_{out}=1263.54-829.74=433.8Btu/lb$

To calculate the mass flow rate of steam in the cycle, we use the formula

$W_{net}=mw_{net}\\m=\frac{W_{net}}{w_{net}} =\frac{2500}{433.8}=5.763*(\frac{0.94782Btu}{1Kj} )=5.462lb/s$

where $1Kj = 0.947817 Btu$

The power output and the rate of heat addition are calculated thus,

$W_{T,out}=m(h_3-h_4)=(5.462lb/s)*(1364-925.7)Btu/lb*(\frac{1Kj}{0.94782Btu} )\\=5.462*438.3*1.055=2525.8KW$

$Q_{in}=mq_{in}=5.462(1263.54)=6901.46Btu/s$

The thermal efficiency of the cycle can be found thus;

$n_{th}=\frac{W_{net}}{Q_{in}} =\frac{2500}{6901.46}*(\frac{0.94782Btu}{1Kj} ) =0.343$

= 34.3%

5 0

2 years ago

Pls help me solve this

musickatia [10]

Answer:

a) The uniform velocity travelled by the car is 10 meters per second.

(Point b has been erased by the user)

c) The distance travelled by the car with uniform velocity is 100 meters.

Explanation:

a) Calculate the uniform velocity travelled by the car:

The uniform velocity is the final velocity ( $v$ ), in meters per second, of the the uniform accelerated stage:

$v = v_{o} + a\cdot t$ (1)

Where:

$v_{o}$ - Initial velocity, in meters per second.

$a$ - Acceleration, in meters per square second.

$t$ - Time, in seconds.

If we know that $v_{o} = 0\,\frac{m}{s}$ , $a = 2\,\frac{m}{s^{2}}$ and $t = 5\,s$ , then the uniform velocity is:

$v = 0\,\frac{m}{s} + \left(2\,\frac{m}{s^{2}} \right)\cdot (5\,s)$

$v = 10\,\frac{m}{s}$

The uniform velocity travelled by the car is 10 meters per second.

(Point b has been erased by the user)

c) The distance travelled by the car ( $\Delta x$ ), in meters, with uniform velocity is calculated by the following kinematic expression:

$\Delta x = v\cdot t$ (2)

If we know that $v = 10\,\frac{m}{s}$ and $t = 10\,s$ , then the distance travelled is:

$\Delta x = \left(10\,\frac{m}{s} \right)\cdot (10\,s)$

$\Delta x = 100\,m$

The distance travelled by the car with uniform velocity is 100 meters.

4 0

2 years ago