Generally natural shape of stone is in shaped as (a)angular (b)irregular (c)cubical cone shape (d)regular

Suppose you have a 9.00 V battery, a 2.00 μF capacitor, and a 7.40 μF capacitor. (a) Find the charge and energy stored if the ca

Andru [333]

Answer:

$Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J$

$Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J$

Explanation:

a)

Identify the unknown: 

The charge and energy stored if the capacitors are connected in series

List the Knowns:

Capacitance of the first capacitor: $C_{1}$ = 2цF = 2 x 10-6 F

Capacitance of the second capacitor $C_{2}$ = 7.4цF = 7.4 x 10-6 F

Voltage of battery: V = 9 V

Set Up the Problem:

Capacitance of a series combination:

$\frac{1}{C_{s} } =\frac{1}{C_{1} } +\frac{1}{C_{2} } +\frac{1}{C_{3} }$ +............

$\frac{1}{C_{s} } =\frac{1}{2} +\frac{1}{ 7.4} \\C_{s} =\frac{2*7.4}{2+7.4}=1.575 *10^-6 F\\$

Capacitance of a series combination is given by:

$C_{s}=\frac{Q}{V}$

Then the charge stored in the series combination is:

$Q=C_{s} V$

Energy stored in the series combination is:

$U_{c}=\frac{1}{2} V^{2} C_{s}$

Solve the Problem: 

$Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J$

b)

Identify the unknown:

The charge and energy stored if the capacitors are connected in parallel

Set Up the Problem: 

Capacitance of a parallel combination:

$C_{p} =C_{1} +C_{2} +C_{3}$

$C_{p} =2+7.4=9.4*10^-6F$

Capacitance of a parallel combination is given by

$C_{p} =\frac{Q}{V}$

Then the charge stored in the parallel combination is

$Q=C_{p} V$

Energy stored in the parallel combination is:

$U_{c}=\frac{1}{2} V^2C_{p}$

Solve the Problem:

$Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J$

5 0

3 years ago

Read 2 more answers

Impact strips may be designed into a bumper cover. True False

d1i1m1o1n [39]

Answer:

true I think

Explanation:

true I think

5 0

3 years ago

A car is traveling at sea level at 78 mi/h on a 4% upgrade before the driver sees a fallen tree in the roadway 150 feet away. Th

Dmitrij [34]

Answer: V = 47.7 mi/hr

Explanation:

first we calculate elements of aero-dynamic resistance

Ka = p/2 * CD * A.f

p is the density of air(0.002378 slugs/ft^3) for zero altitude, CD is the drag coefficient(0.35) and A.f is the front region of the vehicle

so we substitute

Ka = 0.002378/2 * 0.35 * 18

Ka = 0.00749

Now we calculate the final speed of the vehicle (V2) using the relation;

S = (YbW/2gKa)In[ (UW + KaV1^2 + FriW ± Wsinθg) / (UW + KaV2^2 + FriW ± Wsinθg)

so

WE SUBSTITUTE

150 = (1.04 * 2700 / 2 * 32.2 * 0.0075) In [(0.8 * 2700 + 0.0075 *(78mil/hr * 5280ft/1min * 1hr/3600s)^2 + 0.017 * 2700 ± 2700 * 0.04) / (0.8 * 2700 + 0.0075 * V2^2 + 0.017 * 2700 ± 2700 * 0.04)]

150 = (2808/0.483) In [(2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

150 = 5813.66 In [ (2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

divide both sides by 5813.66

0.0258 = In [ (2412.06) / ( 0.0075V2^2 + 2313.9)]

take the e^ of both side

e^0.0258 = (2412.06) / ( 0.0075V2^2 + 2313.9)

1.0261 = (2412.06) / ( 0.0075V2^2 + 2313.9)]

(0.0075V2^2 + 2313.9) = 2412.06 / 1.0261

(0.0075V2^2 + 2313.9) = 2350.7

0.0075V2^2 = 2350.7 - 2313.9

0.0075V2^2 = 36.8

V2^2 = 36.8 / 0.0075

V2^2 = 4906.6666

V2 = √4906.6666

V2 = 70.0476 ft/s

converting to miles per hour

V2 = 70.0476 ft/s * 1 mil / 5280 ft * 3600s / 1hr

V = 47.7 mi/hr

8 0

3 years ago

A heavy ball with a weight of 150 N is hung from the ceiling of a lecture hall on a 4.0-m-long rope. The ball is pulled to one s

shusha [124]

Answer:

The tension in the rope at the lowest point is 270 N

Explanation:

Given;

weight of the ball, W = 150 N

length of the rope, r = 4 m

velocity of the ball, v = 5.6 m/s

When the ball passes through the lowest point, the tension on the rope is the sum of weight of the ball and centripetal force.

T = W + F

Centripetal force, F = mv²/r

where;

m is the mass of the ball

m = W/g

m = 150 / 9.8 = 15.306 kg

Centripetal force, F = mv²/r

F = (15.306 x 5.6²)/4

F = 120 N

T = W + F

T = 150 + 120

T = 270 N

Therefore, the tension in the rope at the lowest point is 270 N

6 0

3 years ago

6. What are the potential consequences if a business does not follow regulations? (1 point)

dimaraw [331]

Answer:

all of the above

Explanation:

6 0

3 years ago