Most of the symbols used to indicate elements are from an elements name in which language

How much charge is on each plate of a 3.00-Î¼F capacitor when it is connected toa 15.0-V battery? b) If this same capacitor is c

Sauron [17]

Answer:

(a) 45 micro coulomb

(b) 6 micro Coulomb

Explanation:

C = 3 micro Farad = 3 x 10^-6 Farad

V = 15 V

(a) q = C x V

where, q be the charge.

q = 3 x 10^-6 x 15 = 45 x 10^-6 C = 45 micro coulomb

(b)

V = 2 V, C = 3 micro Farad = 3 x 10^-6 Farad

q = C x V

where, q be the charge.

q = 3 x 10^-6 x 2 = 6 x 10^-6 C = 6 micro coulomb

6 0

3 years ago

HELP ME PLEASE!!!!!!!!!

Stolb23 [73]

As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards

So here we know that

$F = \frac{kq_1q_3}{d_{13}^2}$

now from the above equation

$F = \frac{(9\times 10^9)(2\times 10^{-6})(4 \times 10^{-6})}{0.5^2}$

$F = 0.288 N$

so both of the charges will apply 0.288 N force on q3 charge along the line joining them

now the net force due to vector sum is given by

$F_{net} = 2Fcos\theta$

here we know that angle is

$\theta = 37 degree$

now we have

$F_{net} = 2\times 0.288 cos37$

$F_{net} = 0.46 N$

so net force on q3 is 0.46 N vertically upwards along +Y axis

6 0

3 years ago

A seagull flies at a velocity of 9.00 m/s straight into the wind.

RideAnS [48]

a)If it takes the bird 18.0 minutes to fly 6 km away from the earth, the wind's speed will be 4 m/s.

b) The bird would need 7 minutes and 42 seconds to fly back 6 kilometers if he turned around and flew with the wind.

c)Compared to the 133.33 seconds it would take without the wind, the overall round-trip time is affected by the wind.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

A seagull flies at a velocity, $\rm v_{SA} = 9 \ m/sec$

The time the bird takes,t=18.0 min

The distance traveled relative to the earth = 6.00 km

a)

The seagull's relative velocity with reference to the ground as;

$\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec$

Air velocity with reference to the ground is;

$\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec$

b)

If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;

$\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec$

The time the bird takes;

$\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7 \ min \ and \ 42 \ sec$

c)\

The total round-trip time compared to what it would be with no wind. is;

$\rm t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec$

The time for the round trip is;

$\rm t = \frac{12 \times 10^ 3 }{ 9 \ m/sec } \\\\ t = 1333.33 \ sec$

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be 4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

4 0

2 years ago

Wil-E-Coyote drops a bowling ball off a cliff to try to catch the Roadrunner. The cliff is

PtichkaEL [24]

Answer:

t = 5.19 s

Explanation:

We have,

Height of the cliff is 132 m

It is required to find the time taken by the ball to fall to the ground. Let t is the time taken. So, using equation of kinematics as :

$y=ut+\dfrac{1}{2}gt^2\\\\\text{since}\ u=0\\\\y=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2y}{g}}\\\\t=\sqrt{\dfrac{2\times 132}{9.8}}\\\\t=5.19\ s$

So, it will take 5.19 seconds to fall to the ground.

8 0

3 years ago

Given the two sets, which statement is true?<br><br> A = {1, 2}<br> B = {1, 2, 3, 4}

Mnenie [13.5K]

What are the two sets? I only see the two answers?

3 0

3 years ago