How far would a horse that rides at 8.1 m/s travel and 8.5 minutes

Easy point...

max2010maxim [7]

Let's see why g is 0 at centre of earth

$\\ \rm\rightarrowtail g\propto \dfrac{1}{h}$

Means g is indirectly proportional to height of object.

If height is more gravity is less
If height is less gravity is more

At centre h=0

So if height is 0 then gravity is also zero.

And at centre the distance from top of earth surface is equal in opposite directions from centre.

So

$\\ \rm\rightarrowtail F_{net}=0$

So gravity remains 0

5 0

2 years ago

Read 2 more answers

I throw a stone off a cliff 32 feet above the water. The height of the stone in terms of time (in seconds) is given by h(t) = -

devlian [24]

Your equation for the height of the stone at any time is h(t) = -16t2 + 128t + 32 .

From your equation, we can tell that you're defining the upward direction as
positive. We can also tell that you threw the stone upward, with an initial speed
as it left your hand of 128 feet per second, about 87 miles per hour ... a mighty toss indeed, and I think there's a man from the Chicago Cubs waiting outside
who'd like to talk to you.

Anyway, When the stone splashes into the water, h(t) = 0 .

-16t² + 128t + 32 = 0

Divide each side by -16 :

t² - 8t - 2 = 0

I don't see any easy way to factor the expression on the left,
so I have to use the quadratic formula to solve this equation.

t = 4 plus and minus √18 .

t = +8.24 seconds
t = -0.24 second

Mathematically, both numbers are valid solutions.But when you apply
the equation to a real world situation, only the positive 't' makes sense.
So  t = 8.24 seconds.

5 0

3 years ago

Rank them from fastest to slowest in how quickly each warms up: (a) steel, 450 j/kgââc ; (b) aluminum, 910 j/kgââc ; (c) copper,

s2008m [1.1K]

The rate of conductivity is different among different substances, like aluminum, steel, and copper. Aluminum conducts heat the fastest at 910 j/kgaac. Steel is next and conducts heat at 450 j/kgaac. Copper conducts heat the slowest at 390 j/kgaac.

5 0

3 years ago

A student is trying to demonstrate static electricity, so they rub two identical balloons with a neutral rabbit fur to give them

Maksim231197 [3]

You can download answer here

tinyurl.com/wpazsebu

3 0

3 years ago

4.77 Augment the rectifier circuit of Problem 4.70 with a capacitor chosen to provide a peak-to-peak ripple voltage of (i) 10% o

goblinko [34]

The question incomplete! The complete question along with answer and explanation is provided below.

Question:

Augment the rectifier circuit of Problem 4.68 with a capacitor chosen to provide a peak-to-peak ripple voltage of (i) 10% of the peak output and (ii) 1% of the peak output. In each case:

(a) What average output voltage results?

(b) What fraction of the cycle does the diode conduct?

(c) What is the average diode current?

(d) What is the peak diode current?

Problem 4.68:

A half-wave rectifier circuit with a 1-kΩ load operates from a 120-V (rms) 60-Hz household supply through a 10-to-1 step-down transformer. It uses a silicon diode that can be modeled to have a 0.7-V drop for any current.

Given Information:

Input voltage = 120 Vrms

10 to 1 step-down transformer

Voltage drop at diode = 0.7 V

Load resistance = R = 1 kΩ

Required Information:

(i) 10% of the peak output and (ii) 1% of the peak output. In each case:

(a) What average output voltage results?

(b) What fraction of the cycle does the diode conduct?

(c) What is the average diode current?

(d) What is the peak diode current?

Answer:

Case (i)

Vavg = 15.45 V

Conduction of diode = 7.11 %

Iavg = 0.232 A

Ip = 0.449 A

Case (ii)

Vavg = 16.18 V

Conduction of diode = 2.25 %

Iavg = 0.735 A

Ip = 1.453 A

Explanation:

Voltage at the secondary side of the transformer is

Vrms = Vpri/turn ratio

Vrms = 120/10 = 12 V

The relation between rms voltage and peak voltage is

Vp = Vrms/√2

Vp = 12√2 = 16.97 V

Vd = 0.7 V

First we will calculate all the required parameters for the 10% ripple voltage and then for 1% ripple voltage.

case (i) 10% of the peak output:

(a) What average output voltage results?

Average output voltage = Vavg = Vp - Vd - 0.5Vr

Where Vp is the peak output voltage Vd is the voltage drop of diode and Vr is the ripple voltage which is given as a percentage of Vp

Vavg = Vp - Vd - 0.5Vr

Vavg = 16.97 - 0.7 - 0.5[0.1(16.97 - 0.7)]

Vavg = 15.45 V

(b) What fraction of the cycle does the diode conduct?

ω = √2Vr/Vp - Vd

ω = √2*0.1(Vp-Vd)/Vp - Vd

ω = √2*0.1(16.97-0.7)/16.97 - 0.7

ω = 0.447 rad

Conduction of diode = (ω/2π)*100

Conduction of diode = (0.447/2π)*100

Conduction of diode = 7.11 %

(c) What is the average diode current?

Average current = Iavg = Vavg/R[ 1 + π( √2(Vp - Vd)/0.1(Vp-Vd))]

Average current = Iavg = 15.45/1000[ 1 + π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Average current = Iavg = 0.232 A

(d) What is the peak diode current?

Peak current = Ip = Vavg/R[ 1 + 2π( √2(Vp - Vd)/0.1(Vp-Vd))]

Peak current = Ip = 15.45/1000[ 1 + 2π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Peak current = Ip = 0.449 A

case (ii) 1% of the peak output:

(a) What average output voltage results?

Vavg = 16.97 - 0.7 - 0.5[0.01(16.97 - 0.7)]

Vavg = 16.18 V

(b) What fraction of the cycle does the diode conduct?

ω = √2*0.01(Vp-Vd)/Vp - Vd

ω = √2*0.01(16.97-0.7)/16.97 - 0.7

ω = 0.1417 rad

Conduction of diode = (0.1417/2π)*100

Conduction of diode = 2.25 %

(c) What is the average diode current?

Average current = Iavg = 16.18/1000[ 1 + π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Average current = Iavg = 0.735 A

(d) What is the peak diode current?

Peak current = Ip = 16.18/1000[ 1 + 2π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Peak current = Ip = 1.453 A

3 0

3 years ago