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pogonyaev
4 years ago
9

From what characteristic do vertebrates get their name

Physics
1 answer:
viva [34]4 years ago
6 0
 The organisms in which  notochord (backbone) is present are called vertebrates
You might be interested in
Why is the sky blue in colour?​
Svetllana [295]

\huge\mathrm{Answer࿐}

Our Atmosphere is full of freely subtended particles, and when light from sun strikes the particles then scattering occurs, in which blue colour is most scattered and hence causing Sky to appear blue .

Hope it helps !!

_____________________________

\mathrm{ \#TeeNForeveR}

3 0
3 years ago
Read 2 more answers
Can someone help me please?
ipn [44]

Answer:

The starting position is 3 m and the object is traveling at 3 m/s

Explanation:

4 0
4 years ago
In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is lau
mestny [16]

Answer:

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

         Em₀ = U = m g y₁

final point. At the exit of the ramp

         Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

          Em₀ = Em_f

         m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

        v = w r

the moment of inertia of a sphere is

         I = \frac{2}{5} m r²

we substitute

         m g (y₁ - y₂) = ½ m v² + ½ (\frac{2}{5} m r²) (\frac{v}{r})²

         m g h = ½ m v² (1 + \frac{2}{5})

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

         mg h = \frac{7}{5} (\frac{1}{2} m v²)

         v = √5/7  √2gh

This is the exit velocity of the vertical movement of the sphere

         v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

   starting point

          Em₀ = U = mg y₁

   final point

          Em_f = K + U = ½ m v² + m g y₂

          Em₀ = Em_f

          mg y₁ = ½ m v² + m g y₂

          m g (y₁ -y₂) = ½ m v²

          v = √2gh

this is the speed of the box

          v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

          v² = v₀² - 2 g y

at the highest point v = 0

           y = vo₀²/ 2g

for the sphere

           y_sphere = 5/7 2gh / 2g

           y_esfera = 5/7 h

for the block

           y_block = 2gh / 2g

            y_block = h

       

therefore the block reaches higher than the sphere

         \frac{y_{sphere}} {y_bolck} = 5/7

3 0
3 years ago
An air-conditioning system is to be filled from a rigid container that initially contains 5 kg of liquid R-134a at 24°C. The val
OverLord2011 [107]

Answer:

Answer

The Final Quality of teh R-134a in the container  is  0.5056

The Total Heat transfer is Q_{in} = 22.62 KJ

Explanation:

Explanation is  in the following  attachments

3 0
3 years ago
Some hydrogen gas is enclosed within a chamber being held at 200^\ { C} with a volume of 0.025 \rm m^3. The chamber is fitted wi
vlada-n [284]

Answer:

The final volume is 0.039 m^3

Explanation:

<u>Data:</u>

Initial temperature: T1=200C

Final temperature: T2=200C

Initial pressure: P1=1.50 \times10^6 Pa

Final pressure: P2=0.950 \times10^6 Pa

Initial volume: V1=0.025m^{3}

Final volume: V2=?

Assuming hydrogen gas as a perfect gas it satisfies the perfect gas equation:

\frac{PV}{T}=nR (1)

With P the pressure, V the volume, T the temperature, R the perfect gas constant and n the number of moles. If no gas escapes the number of moles of the gas remain constant so the right side of equation (1) is a constant, that allows to equate:

\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

Subscript 2 referring to final state and 1 to initial state.

solving for V2:

V_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}}=\frac{(1.50 \times10^6)(0.025)(200)}{(200)(0.950 \times10^6)}

V_{2}=0.039 m^3

3 0
4 years ago
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