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3 years ago

Which would be an adaptation for living in the tundra?

Physics

2 answers:

Harlamova29_29 [7]3 years ago

7 0

Answer:A hibernation to lower metabolism in the winter

Explanation:

the others would not be beneficial in the tundra since the tundra is cold adaptations that retain heat would be best

nevsk [136]3 years ago

3 0

<h2>Answer:</h2><h2></h2>

C. Hibernation to lower metabolism

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The induced magnetic field at radial distance 4.0 mm from the central axis of a circular parallel-plate capacitor is 1.8 ✕ 10−7

wel

Answer:

$\frac{dE}{dt}=2.07*10^{13}\frac{V/m}{s}$

Explanation:

According to Gauss's law, the electric flux through the circular plates is defined as the electric field multiplied by its area:

$\Phi=EA=E(\pi R^2)(1)$

The magnetic field around the varying electric field of the circular plates is given by:

$B=\frac{\epsilon_0 \mu_o}{2\pi r}\frac{d\Phi}{dt}(2)$

Replacing (1) in (2) and solving for $\frac{dE}{dt}$ :

$B=\frac{\epsilon_0 \mu_o\pi R^2}{2\pi r}\frac{dE}{dt}\\\frac{dE}{dt}=\frac{2rB}{\epsilon_0 \mu_o R^2}\\\frac{dE}{dt}=\frac{2(4*10^{-3}m)(1.8*10^{-7}T)}{(8.85*10^{-12}\frac{C^2}{N\cdot m^2})(4\pi *10{-7}\frac{Tm}{A})(2.5*10^{-3}m)^2}\\\\\frac{dE}{dt}=2.07*10^{13}\frac{V/m}{s}$

3 0

3 years ago

The angular velocity of a process control motor is (13−12t2) rad/s, where t is in seconds. Part A At what time does the motor re

mihalych1998 [28]

Answer:

Explanation:

Given

$\omega =13-\frac{1}{2}\cdot t^2$

Motor reverse its direction when \omega =0

$13-0.5t^2=0$

$26=t^2$

$t=\sqrt{26}=5.099\approx 5.1 s$

(b)

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega$

$\int d\theta =\int_{0}^{5.1}\omega dt$

$\int d\theta =\int_{0}^{t}(13-.05t^2)dt$

$\theta =(13t-0.1667\times t^3)_0^{5.1}$

$\theta =44.192^{\circ}$

4 0

3 years ago

Two resistors, A and B, are connected in series to a 6.0 V battery. A voltmeter connected across resistor A measures a potential

mestny [16]

Answer:

Resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

Explanation:

When the two resistors are in series, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B.

Given that V₁ + V₂ = 6.0 V and V₁ = 4.0 V,

V₂ = 6.0 V - V₁ = 6.0 V - 4.0 V = 2.0 V

Also, let the current in series be I.

So, V₁ = IR₁ and V₂ = IR₂

I = V₁/R₁ and I = V₂/R₂

equating both expressions, we have

V₁/R₁ = V₂/R₂

4.0 V/R₁ = 2.0 V/R₂

dividing through by 2.0 V, we have

2/R₁ = 1/R₂

taking the reciprocal, we have

R₂ = R₁/2

R₁ = 2R₂

From the parallel connection, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B. Since it is parallel, V₁ = V₂ = V = 6.0 V

Also, V₂ = I₂R₂ where I₂ = current in resistor B = 2.0 A and R₂ = resistance of resistor B

So, R₂ = V₂/I₂

= 6.0 V/2.0 A

= 3.0 Ω

R₁ = 2R₂

= 2(3.0 Ω)

= 6.0 Ω

So, resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

6 0

3 years ago

Which planet has a storm called the Great Dark Spot swirling in its atmosphere?

Viktor [21]

Im thinking D. Neptune..

3 0

4 years ago

Read 2 more answers

Help please help help help

poizon [28]

Mass (m) of the object is 1kg
Velocity (v) is 2m/s
and the height is 0
when the mechanical energy M.E is solved for m*v= 1(2)=2 so the M.E=2

3 0

3 years ago