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2 years ago

What type of meter would be connected on both sides of the resistor in a circuit

Physics

2 answers:

WITCHER [35]2 years ago

8 0

the answer IS NOT "An ammeter, which measures current"

AleksandrR [38]2 years ago

7 0

Ammeter in series at any side as required since it is bilateral and voltmeter is connected in parallel to measure voltage drop across it.

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A 4.0-mF capacitor initially charged to 50 V and a 6.0-mF capacitor charged to 30 V are connected to each other with the positiv

Juli2301 [7.4K]

Answer:

<em>The final charge on the 6.0 mF capacitor would be 12 mC</em>

Explanation:

The initial charge on 4 mF capacitor = 4 mf x 50 V = 200 mC

The initial Charge on 6 mF capacitor = 6 mf x 30 V =180 mC

Since the negative ends are joined together the total charge on both capacity would be;

q = $q_{1} -q_{2}$

q = 200 - 180

q = 20 mC

In order to find the final charge on the 6.0 mF capacitor we have to find the combined voltage

q = (4 x V) + (6 x V)

20 = 10 V

V = 2 V

For the final charge on 6.0 mF;

q = CV

q = 6.0 mF x 2 V

q = 12 mC

Therefore the final charge on the 6.0 mF capacitor would be 12 mC

5 0

2 years ago

Read 2 more answers

Un pendule est constitue par une masse ponctuelle m= 0,1kg accrocher a un fil sans masse de longueur L = 0,4 m on ecarte ce pend

BabaBlast [244]

I could help
If I know what I was reading I’m sorry I need to translate

3 0

2 years ago

A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by

Delicious77 [7]

Answer:

W = -510.98J

Explanation:

Force = 43N, 61° SW

Displacement = 12m, 22° NE

Work done is given as:

W = F*d*cosA

where A = angle between force and displacement.

Angle between force and displacement, A = 61 + 90 + 22 = 172°

W = 43 * 12 * cos172

W = -510.98J

The negative sign shows that the work done is in the opposite direction of the force applied to it.

6 0

2 years ago

three neutral metal cans mounted on isulating stands are touching a negatively charge ballon is brought near can a can b is then

Nimfa-mama [501]

Charge on can A is positive.
Charge on can C is negative.
Punctuation and capitalization are very useful things to pay attention to and this question would be a lot easier to understand if you had actually used both capitalization and punctuation. If I'm understanding the question, you have 3 metal can that are insulated from the environment and initially touching each other in a straight line. Then a negatively charged balloon is brought near, but not touching one of the cans in that line of cans. While the balloon is near, the middle can is removed. Then you want to know the charge on the can that was nearest the balloon and the charge on the can that was furthermost from the balloon.
As the balloon is brought near to can a, the negative charge on the balloon repels some of the electrons from can a (like charges repel). Some of those electrons will flow to can b and in turn flow to can c. Basically you'll have a charge gradient that's most positive on that part of the can that's closest to the balloon, and most negative on the part of the cans that's furthest from the balloon. You then remove can B which causes cans A and C to be electrically isolated from each other and prevents the flow of elections to equalize the charges on cans A and C when the balloon is removed. So you're left with a deficiency of electrons on can A, so can A will have a positive overall charge, and an excess of electrons on can C, so can C will have a negative overall charge.

7 0

2 years ago

A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, th

ziro4ka [17]

Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

In the first question, we have to caculate the constant of the spring with this equation:

$m*g=k*x$

Getting the k:

$k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]$

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

$x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\$

The energy of a spring:

$E=\frac{1}{2}*k*x^{2}$

For the first case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]$

For the second case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]$

If you take the relation E2/E1 = 4.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

$E=\frac{1}{2}*k*x^{2}$ ⇒ $k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]$

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

$E=W*h=500[N]*10[m]=5000[J]$

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J

7 0

2 years ago