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STatiana [176]
3 years ago
7

Nodes give a vertebrate’s back flexibility. A. True B. False

Physics
2 answers:
Elis [28]3 years ago
8 0
B- False it
invertebrates have flexibility ( octopuses and squids are good example of invertebrates)
Lapatulllka [165]3 years ago
5 0
No B. its false<span> they do not. Lyphnodes are a permant part of the body and you have them in your Neck</span>
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A 120 kg lineman moving west at 2 m/s tackles an 80 kg football fullback moving east at 8 m/s. After the collision both players
Alinara [238K]
The answer is <span>Vf = -2 m/s</span>
3 0
4 years ago
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a 3.18-kg rock is released from rest at a height of 26.6 m. ignore air resistance and determine (a) the kinetic energy at 26.6 m
antiseptic1488 [7]

Answer:

(a) 0 J

(b) 828.96 J

(c) 828.96 J

(d) 828.96 J

(e) 0 J

(f) 828.96 J

Explanation:

Given:

Mass of the rock is, m=3.18\ kg

Initial height of the rock is, h_{i}=26.6\ m

Final height of the rock is, h_f=0\ m

Initial velocity of the rock is, v_i=0\ m/s(Rest)

Acceleration due to gravity is, g=9.8\ m/s^2

(a)

Initial kinetic energy is given as:

K_i=\frac{1}{2}mv_i^2

Plug in the given values and solve for 'K_i'. This gives,

K_i=\frac{1}{2}\times 3.18\times 0^2=0

Therefore, initial kinetic energy is 0 J.

(b)

Initial potential energy is given as:

U_i=mgh_i\\U_i=3.18\times 9.8\times 26.6=828.96\ J

Therefore, initial potential energy is 828.96 J.

(c)

Mechanical energy is equal to the sum of kinetic and potential energy. It is always a constant value. Therefore,

ME=K_i+P_i\\ME=0+828.96=828.96\ J

Therefore, mechanical energy at 26.6 m is 828.96 J.

(d)

Now, at the final height of 0 m, the decrease in potential energy will be equal to the increase in the kinetic energy. In other words, the potential energy at the start will be converted to kinetic energy at the bottom.

Therefore, kinetic energy at the 0 m is given as:

K_f=U_i=828.96\ J

(e)

Potential energy at the final height is given as:

P_f=mgh_f=3.18\times 9.8\times 0=0\ J

Therefore, final potential energy is 0 J.

(f)

Total mechanical energy is a constant and doesn't depend on the height.

So, total mechanical energy at 0 m is same as that at 26.6 m.

Total mechanical energy is 828.96 J.

8 0
4 years ago
The helium used to fill birthday balloons doesn’t come out of the air, but from. out of the ground. Some of the gas might escape
Vladimir79 [104]
Well , i think helium gas is present in the Earth Crust because metals found in the crust have nuclei that decay into another elements
One of this element that formed is Helium

hope this helps
3 0
4 years ago
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The radioactive element radium (Ra) decays by a process known asalpha decay,in which the nucleus emits a helium nucleus. (Factoi
tamaranim1 [39]

<u>Answer:</u> The energy released in the decay process is 4.6800\times 10^{11}J

<u>Explanation:</u>

The equation for the alpha decay of Ra-226 follows:

_{88}^{226}\textrm{Ra}\rightarrow _{2}^{4}\textrm{He}+_{86}^{222}\textrm{Rn}

To calculate the mass defect, we use the equation:

Mass defect = Sum of mass of product - Sum of mass of reactant

\Delta m=(m_{Rn}+m_{He})-(m_{Ra})

We know that:

m_{Rn}=222.0176u\\m_{Ra}=226.0254u\\m_{He}=4.0026u

Putting values in above equation, we get:

\Delta m=(222.0176+4.0026)-(226.0254)=-0.0052g=-5.2\times 10^{-6}kg

(Conversion factor: 1 kg = 1000 g )

To calculate the energy released, we use Einstein equation, which is:

E=\Delta mc^2

E=(-5.2\times 10^{-6}kg)\times (3\times 10^8m/s)^2

E=-4.6800\times 10^{11}J

Hence, the energy released in the decay process is 4.6800\times 10^{11}J

8 0
3 years ago
How much heat would be needed to completely evaporate 31.5 g of boiling water at a temperature of 100 "C? Express your answer in
BaLLatris [955]

Answer:

Heat needed = 71.19 J

Explanation:

Here heat required can be calculated by the formula

            H = mL

M is the mass of water and L is the latent heat of vaporization.

Mass of water, m = 31.5 g = 0.0315 kg

Latent heat of vaporization of water = 2260 kJ/kg

Substituting

            H = mL = 0.0315 x 2260 = 71.19 kJ

Heat needed = 71.19 J

7 0
3 years ago
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