Apart from cutaneous respiration<span> present in all </span>species<span>, most lissamphibians are born in an aquatic larval stage with gills. After metamorphosis, they develop lungs to breathe on land. The larvae of urodeles and apods present external, filamentous and highly branched gills which allow them to breathe underwater.
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Answer: 4.9 x 10^6 joules
Explanation:
Given that:
mass of boulder (m) = 2,500 kg
Height of ledge above canyon floor (h) = 200 m
Gravita-tional potential energy of the boulder (GPE) = ?
Since potential energy is the energy possessed by a body at rest, and it depends on the mass of the object (m), gravitational acceleration (g), and height (h).
GPE = mgh
GPE = 2500kg x 9.8m/s2 x 200m
GPE = 4900000J
Place result in standard form
GPE = 4.9 x 10^6J
Thus, the gravita-tional potential energy of the boulder-Earth system relative to the canyon floor is 4.9 x 10^6 joules
Answer:
2a) x = 32 [mil/h]; 2b) t = 0.5[h]; 3a) t = 2.5 [h]; 3b) x = 185[mil]
Explanation:
2a)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 32 [\frac{mil}{h}] \\t=time = 1 [h]\\x=v*t\\x=32[\frac{mil}{h} ]*1[h]\\x=32[mil}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Cv%3Dvelocity%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%20%3D%2032%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%5D%20%5C%5Ct%3Dtime%20%3D%201%20%5Bh%5D%5C%5Cx%3Dv%2At%5C%5Cx%3D32%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%2A1%5Bh%5D%5C%5Cx%3D32%5Bmil%7D)
2b)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{420}{840}\\ t=0.5[h]](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Ct%3D%5Cfrac%7Bx%7D%7Bv%7D%20%5C%5Ct%3D%5Cfrac%7B420%7D%7B840%7D%5C%5C%20t%3D0.5%5Bh%5D)
3a)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{35}{14}\\ t=2.5[h]](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Ct%3D%5Cfrac%7Bx%7D%7Bv%7D%20%5C%5Ct%3D%5Cfrac%7B35%7D%7B14%7D%5C%5C%20t%3D2.5%5Bh%5D)
3b)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 74 [\frac{mil}{h}] \\t=time = 2.5 [h]\\x=v*t\\x=74[\frac{mil}{h} ]*2.5[h]\\x=185[mil}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Cv%3Dvelocity%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%20%3D%2074%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%5D%20%5C%5Ct%3Dtime%20%3D%202.5%20%5Bh%5D%5C%5Cx%3Dv%2At%5C%5Cx%3D74%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%2A2.5%5Bh%5D%5C%5Cx%3D185%5Bmil%7D)
Answer:
40 cm
Explanation:
We are given that
Load=800 N
Effort=200 N
Load distance=10 cm
We have to find the effort distance.
We know that

Using the formula

Effort distance=
Effort distance=
Effort distance=40 cm
Hence, the effort distance will be 40 cm.