5) A net force of 12 Newtons acting north on an object for 4.0 seconds will produce an impulse of

Circular coil 1 has N turns and circular coil 2 has 6N. Coil 1 has area A and coil 2 has area A/4. If Coil 1 has current i runni

solmaris [256]

Answer:

$\frac{L_1}{L_2} =\frac{1}{3}$

Explanation:

Recall that the formula for an inductance (L) for coil on N turns, are A and current I is given by:

$L=\frac{\mu_0\,N^2\,A}{I}$

Then, for the first coil we have;

$L_1=\frac{\mu_0\,N^2\,A}{I}$

and for coil 2 we have:

$L_2=\frac{\mu_0\,(6\,N)^2\,(A/4)}{3\,I}$

then, the quotient L1/L2 can be written as:

$\frac{L_1}{L_2} =\frac{\mu_0\,N^2\,A\,3\,I}{\mu_0\,(6\,N)^2\,(A/4)\,I}=\frac{12}{36} =\frac{1}{3}$

7 0

3 years ago

A uniform electric ﬁeld exists in the region between two oppositely charged plane-parallel plates. An electron is released from

Zigmanuir [339]

Answer:

Explanation:

given S = distance from the first = 3.20cm = 0.032m, t = 1.30×10−8 s

q = 1.6 x 10_19C

using S = at^2/2

acceleration = 0.032 X 2 /(1.30×10−8)^2

a = 3.79 x 10^14m/s^2

From F = ma

F = qE

ma = qE

E = ma /q = 9.11 x 10^-31 x 3.79 x 10^14 / 1.6 x 10^-19

E = magnitude of this electric ﬁeld. = 2156.3N/C

b) Find the speed of the electron when it strikes the second plate ; V^2 = 2as

= 2 X 3.79 x 10^14 X 0.032

= 4.92 X 10^6m/s

5 0

3 years ago

Design an inverting amplifier that has a gain of -47 (this gain is negative). Pick resistor values that you have in the lab kit.

Nostrana [21]

Answer:

R2= 470 ohms R1= 10ohms. Schematic is attached

Explanation:

Av=-(R2/R1)

-47=-(470/10)

-47=-47

7 0

3 years ago

Listed following are three possible models for the long-term expansion (and possible contraction) of the universe in the absence

sineoko [7]

Answer:

From smallest ratio to the largest ratio:

Coasting Universe - Critical Universe - Recollapsing Universe(From left to right)

Explanation:

The coasting universe is one that expands at a constant rate given by the Hubble constant throughout all of cosmic time. It has a ratio of actual density to critical density that is less than 1

The critical universe is one that is at balance with no expansion .I.e. the actual density and the critical density are equal, which makes the ratio of actual density to critical density to be equal to 1

Recollapsing Universe: The expansion of the universe reverses in the future and the universe eventually recollapses. The recollapsing universe has the ratio of the actual density to the critical density to be greater than 1

7 0

3 years ago

A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly

never [62]

Answer:

1) Vf = 3.36 m/s

2) v = 6.86 m/s

3) s = 92.3 m

4) a = - 0.23 m/s²

Explanation:

1)

We use first equation of motion in this case:

Vf = Vi + at

where,

Vf = Final velocity = ?

Vi = Initial velocity = 0 m/s

a = acceleration = 1.4 m/s²

t = time = 2.4 s

Therefore,

Vf = 0 m/s + (1.4 m/s²)(2.4 s)

Vf = 3.36 m/s

2)

We again use first equation of motion but with t= 4.9 s now:

Vf = 0 m/s + (1.4 m/s²)(4.9 s)

Vf = 6.86 m/s

Now, this velocity remained constant for net 11 seconds. Hence, the velocity of hare after 8.9 s is:

v = 6.86 m/s

3)

First we use second equation of motion to find distance covered in accelerated motion:

s₁ = Vi t + (0.5)at²

s₁ = (0 m/s)(4.9 s) + (0.5)(1.4 m/s²)(4.9 s)²

s₁ = 16.8 m

Now, we calculate the distance covered in uniform motion:

s₂ = vt

s₂ = (6.86 m/s)(11 s)

s₂ = 75.5 m

Now, total distance covered before slowing down is given as:

s = s₁ + s₂ = 16.8 m + 75.5 m

s = 92.3 m

3)

Using third equation of motion for the decelerated motion:

2as = Vf² - Vi²

where,

a = deceleration = ?

s = distance covered = 102 m

Vf = Final Speed = 0 m/s

Vi = Initial Speed = 6.86 m/s

Therefore,

(2)(a)(102 m) = (0 m/s)² - (6.86 m/s)²

a = - (47.06 m²/s²)/(204 m)

a = - 0.23 m/s²

3 0

4 years ago