Ask question

Ask question

All categories

3 years ago

14

A circuit has resistors R 1 = 468 Ω and R 2 = 125 Ω R1=468 Ω and R2=125 Ω , and two batteries V 1 = 12.0 V and V 2 = 3.00 V V1=1

2.0 V and V2=3.00 V . The variable resistor R is adjusted until the galvanometer reads 0 A 0 A . Calculate the unknown resistance R R .

1 answer:

Greeley [361]3 years ago

7 0

Answer:

0.0192A

Explanation:

Since, the reading of the galvanometer is 0 A, the voltage across the resistance R will be:

Step 1

VR = V2

VR = 3.00v

Step 2

Calculating the current through the resistance R as below,

IR = V1 - V2 /R1

IR = 12 - 3 /468

IR =0.0192A

You might be interested in

Nicholas sends 20.0 minutes ironing shirts with his 1800-watt irn . How many joules of energy were usedby the iron

Nuetrik [128]

1 watt = 1 joule/sec
1800 watts = 1800 joules/sec

   (1800 J/sec) x (20 min) x (60 sec/min)

   = (1800 x 20 x 60) joules = 2,160,000 joules .

   ( 2.16 megajoules )

6 0

3 years ago

How to solve 2.14 using calculus

trasher [3.6K]

Answer:

The acceleration is $6.42\frac{m}{s^2}$

Explanation:

Given the velocity function:

$v=0.86\frac{m}{s^3}t^2$

you can obtain the instantaneous acceleration "a" as its first derivative:

$a=\dot{v}=2\cdot0.86\frac{m}{s^3}\cdot t=1.72\frac{m}{s^3}t$

To determine the value of "a" when the velocity was 12m/s, you need to figure out the value for "t" when this happens. At what time t is the velocity 12m/s?

$12.0\frac{m}{s}=0.86\frac{m}{s^3}t^2\implies t=3.74s$

This value of t is less than the 5 seconds mentioned in the text - so that is a good sign that the formula is valid for this value. And so you can use t=3.47s in the derivative (acceleration) above:

$a=1.72t=1.72\frac{m}{s^3}\cdot 3.74s = 6.42\frac{m}{s^2}$

3 0

2 years ago

How do people in your community deal with loud sound

Anna35 [415]

Answer: Nothing they mind thier own business instead of being really noisy.

Explanation

Now if someone was screaming they probley would go investagate investigate but something else nah.

8 0

2 years ago

The 5-lb collar is released from rest at A and travels along the frictionless guide. Determine the speed of the collar when it s

Sholpan [36]

Answer:

Explanation:

Stiffness of spring k equal to 4 lb/ft.

Unstretched length of the spring L is equal to 0.5 feet.

Weight of the collar W is 15lb

Radius of curvature of curve guide is 1 feet

length of vertical rod is 1.5 feet

Initial speed of collar when released from rest at A is 0 feet per seconds

use the energy conservation equation

$P_A +K_A=P_B+K_B$

Estimate the potential energy , component as position B as below

$P_A=Wh_1+\frac{1}{2} ks^2_1\\\\=5\times(1.5+1)+\frac{1}{2} \times 4 \times(1.5+1-0.5)^2\\\\=20.5lb.ft$

Estimate the kinetic energy , component as position A as below

$K_A=\frac{1}{2} \frac{W}{g} V^2_1\\\\=\frac{1}{2} \frac{5}{32.2} \times0^2\\\\=0lb.ft$

Estimate the kinetic energy , component as position B as below

$K_A=\frac{1}{2} \frac{W}{g} V^2_2\\\\=\frac{1}{2} \frac{5}{32.2} \times V^2_2\\\\=00777V^2_2$

Substitute 20.5lb- ft for $P_A$

0.5lb-ft for $P_B$

0lb -ft for $K_A$

$0.0777V_2^2 for K_B$

$20.5+0=0.5+0.777V^2_2\\\\V^2_2=257.6\\\\V_2=16.05$

= 16.05ft/sec

7 0

2 years ago

A light ray in air hits piece of quartz at a 39.0 deg angle. Inside the quartz, the light slows down to 2.06*10^8 m/s. At what a

Naddika [18.5K]

Answer:

Angle of light at which it will travel in the quartz is given as

$\theta = 25.6^o$

Explanation:

As we know that the refractive index of a medium is the ratio of speed of light in air and speed of light in the medium

So we will have

$\mu = \frac{c}{v}$

here we have

$\mu = \frac{3 \times 10^8}{2.06 \times 10^8}$

$\mu = 1.456$

now by Snell's law we have

$n_1 sin\theta_1 = n_2 sin\theta_2$

$1 sin39 = 1.456 sin\theta$

$\theta = 25.6^o$

3 0

3 years ago