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3 years ago

Ravens eat meat but they don’t kill it themselves. What kind of animal does this make them?

Physics

1 answer:

pogonyaev3 years ago

7 0

That means a raven is a scavenger. A scavenger is an animal that feeds on carrion, dead plant material, or refuse. :-)

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Please help on this one?

vlada-n [284]

Answer:

idk

Explanation:

7 0

2 years ago

Guys helppppp pleaseeeeeeeee

yan [13]

2 pencils, because that would equal left grams then the others

4 0

2 years ago

A vw bug is driving north on a straight two lane road at a constant 54 mph and chevy impala south at 47 on the same road (but di

mario62 [17]

Velocity of vw bug is given as

$v_1 = 54 mpH$ towards north

velocity of imphala is given as

$v_2 = 47 mph$ towards south

now we need to find velocity of imphala relative to bug

$v_{21} = v_2 - v_1$

$v_{21} = 47 - (-54) = 101 mpH$

so the speed is 101 mpH towards south so correct option is C

5 0

2 years ago

A pressure cylinder has a diameter of 150-mm and has a 6-mm wall thickness. What pressure can this vessel carry if the maximum s

myrzilka [38]

Answer:

p = 8N/mm2

Explanation:

given data ;

diameter of cylinder = 150 mm

thickness of cylinder = 6 mm

maximum shear stress = 25 MPa

we know that

hoop stress is given as = $\frac{pd}{2t}$

axial stress is given as = $\frac{pd}{4t}$

maximum shear stress = (hoop stress - axial stress)/2

putting both stress value to get required pressure

$25 = \frac{ \frac{pd}{2t} -\frac{pd}{4t}}{2}$

$25 = \frac{pd}{8t}$

t = 6 mm

d = 150 mm

therefore we have pressure

p = 8N/mm2

7 0

3 years ago

Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900

Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron $m_{e}=9.11\times10^{-31}\ kg$

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

$E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}$

$E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}$

$E=0.582\ Mev$

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

$E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}$

$E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}$

$E=0.350\ Mev$

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0

3 years ago