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4 years ago

7

PLEASEE HELP ME ❤️❤️

1 answer:

BartSMP [9]4 years ago

3 0

1.)Multi-cellular
2.)Organism
3.)Cells
4.)Uni-cellular
5.)Homeostasis
6.)Reproduction
7.)Response
8.)Stimulus
9.)Energy
10.)Photosynthesis
11.)Plants
12.)Glucose
13.)Reproduction
I pretty sure these are correct,
Hope this helps

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With a bar magnet where are the lines of force closest together

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A playground merry-go-round of radius 2.00 m has a moment of inertia I = 275 kg · m2 and is rotating about a frictionless vertic

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Answer:

0.2932 rad/s

Explanation:

r = Radius = 2 m

$I_i$ = Initial angular momentum = $275\ kgm^2$

$\omega_i$ = Initial angular velocity = 14 rev/min

$I_f$ = Final angular momentum

$\omega_f$ = Final angular velocity

Here the angular momentum of the system is conserved

$I_i\omega_i=I_f\omega_f\\\Rightarrow 275\times 14\times \dfrac{2\pi}{60}=(275+275(2)^2)\omega_f\\\Rightarrow \omega_f=\dfrac{275\times 14\times \dfrac{2\pi}{60}}{275+275(2)^2}\\\Rightarrow \omega_f=0.2932\ rad/s$

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3 years ago

A real power supply can be modeled as an ideal EMF of 60 Volts in series with an internal resistance. The voltage across the ter

lara31 [8.8K]

Answer:

5 ohms

Explanation:

Given:

EMF of the ideal battery (E) = 60 V

Voltage across the terminals of the battery (V) = 40 V

Current across the terminals (I) = 4 A

Let the internal resistance be 'r'.

Now, we know that, the voltage drop in the battery is given as:

$V_d=Ir$

Therefore, the voltage across the terminals of the battery is given as:

$V= E-V_d\\\\V=E-Ir$

Now, rewriting in terms of 'r', we get:

$Ir=E-V\\\\r=\frac{E-V}{I}$

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

$r=\frac{60-40}{4}\\\\r=\frac{20}{4}\\\\r=5\ ohms$

Therefore, the internal resistance of the battery is 5 ohms.

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4 years ago

If the specific surface energy for magnesium oxide is 1.0 J/m2 and its modulus of elasticity is (225 GPa), compute the critical

alexdok [17]

Answer:

The critical stress required for the propagation of an initial crack $\sigma_{c}$ = 21.84 M pa

Explanation:

Given data

Modulus of elasticity E = 225 × $10^{9}$ $\frac{N}{m^{2} }$

Specific surface energy for magnesium oxide is $\gamma_{s}$ = 1 $\frac{J}{m^{2} }$

Crack length (a) = 0.3 mm = 0.0003 m

Critical stress is given by $\sigma_{c}^{2} }$ = $\frac{2 E \gamma}{\pi a}$ -------- (1)

⇒ 2 E $\gamma_{s}$ = 2 × 225 × $10^{9}$ × 1 = 450 × $10^{9}$

⇒ $\pi$ a = 3.14 × 0.0003 = 0.000942

⇒ Put these values in equation 1 we get

⇒ $\sigma_{c}^{2} }$ = $\frac{450 }{0.000942} 10^{9}$

⇒ $\sigma_{c}^{2} }$ = 4.77 × $10^{14}$

⇒ $\sigma_{c}$ = 2.184 × $10^{7}$ $\frac{N}{m^{2} }$

⇒ $\sigma_{c}$ = 21.84 $\frac{N}{mm^{2} }$

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This is the critical stress required for the propagation of an initial crack.

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3 years ago

Second and third class levers are differentiated by __________.

ira [324]

Answer:

i think it is b

hope this helps

Explanation:

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