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2 years ago

How are traits influenced by the environment? (No links)

Physics

1 answer:

Ganezh [65]2 years ago

8 0

Answer:

Environmental factors influence traits in plants and animals. These traits include things like weight, height, size, and color. If a dog does not get proper nutrition and exercise, their weight will be negatively impacted. The dog is likely to be overweight if the environmental factors are not changed.

Explanation:

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A potential difference of 24 V is applied to a 150-ohm resistor. How much current flows through the resistor?

Lady_Fox [76]

Given :- A resistor of 150 ohm, hence Resistance (R) = 150 ohm

Potential Difference (v) = 24 V

Current (I) = ?

V = IR

24 = I × 150

I = 24/150

I = 0.16 ampere

hope it helps!

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3 years ago

Jenny loves puffed rice cereal. As she is pouring the cereal, two little

Ede4ka [16]

Answer:

4.9 x 10-23 N

Explanation:

I took the test

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3 years ago

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Does most psychologists believe that ESP exist

vodomira [7]

Answer:

Explanation:

most psychologists are skeptical of ESP and believe it's not real. There actually more skeptical of ESP than other scientists.

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3 years ago

Non examples of gravity ?

Afina-wow [57]

Floating. When you have no gravity you have nothing to be pushing you down to the floor so that would be an example of no gravity pushing on you.

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3 years ago

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A rectangular coil with 50 turns of conducting wire and a total resistance of 10.0 Ω initially lies in the yz-plane at time t =

castortr0y [4]

Answer:

a) 43.20V

b) 2.71W/s

c) 40.25s

d) 7.77Nm

Explanation:

(a) The emf of a rotating coil with N turns is given by:

$emf=NBA\omega sin(\omega t)$

N: turns

B: magnitude of the magnetic field

A: area

w: angular velocity

the emf max is given by:

$emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V$

(b) the maximum rate of change of the magnetic flux is given by:

$\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}$

(c) $emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V$

(d) The torque is given by:

$\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm$

3 0

3 years ago