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When a 20.0-ohm resistor is connected across the terminals of a 12.0-V battery, the voltage across the terminals of the battery falls by 0.300 V. What is the internal resistance of this battery

Answers

Answer 1

Answer:

The internal resistance of the battery is 0.5 ohms.

To calculate the internal resistance of the battery, we use the formula below

Formula:

(V/R)r = V'............. Equation 1

Where:

V = Voltage across the terminal of the battery
R = Resistance connected across the battery
r = internal resistance of the battery
V' = voltage drop of the battery.

Make r the subject of the equation

r = V'R/V............ Equation 2

From the question,

Given:

V = 12 V
R = 20 ohms
V' = 0.3 V

Substitute these values into equation 2

r = (0.3×20)/12
r = 6/12
r = 0.5 ohms.

Hence, The internal resistance of the battery is 0.5 ohms.

Learn more about internal resistance here: brainly.com/question/14883923

Answer 2

Answer:

Answer:

The internal resistance is $r = 0.5 \ \Omega$

Explanation:

From the question we are told that the resistance of

The resistance of the resistor is $R = 20.0\ \Omega$

The voltage is $V = 12.0 \ V$

The magnitude of the voltage fall is $e = 0.300\ V$

Generally the current flowing through the terminal due to the voltage of the battery is mathematically represented as

$I = (V)/(R)$

substituting values

$I = (12.0 )/(20 )$

$I = 0.6 \ A$

The internal resistance of the battery is mathematically represented as

$r = (e)/(I)$

substituting values

$r = (0.300)/( 0.6 )$

$r = 0.5 \ \Omega$

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The radius of Earth is 6370 km in the Earth reference frame. The cosmic ray is moving at 0.880Co relative to Earth.a. In the reference frame of a cosmic ray how wide does Earth seem along the flight direction?
b. In the reference frame of a cosmic ray how wide does Earth seem perpendicular to the flight direction?
Express your answer with the appropriate units.

Answers

Answer:

6052114.67492 m

$12.742* 10^(6)\ m$

Explanation:

v = Velocity of cosmic ray = 0.88c

c = Speed of light = $3* 10^8\ m/s$

d = Width of Earth = Diameter of Earth = $12.742* 10^(6)\ m$

When the cosmic ray is moving towards Earth then in the frame of the cosmic ray the width of the Earth appears smaller than the original

This happens due to length contraction

Length contraction is given by

$d_e=d\sqrt{1-(v^2)/(c^2)}\n\Rightarrow d_e=12.742* 10^(6)\sqrt{1-(0.88^2c^2)/(c^2)}\n\Rightarrow d_e=6052114.67492\ m$

The Earth's width is 6052114.67492 m

Contraction only occurs in the cosmic ray's frame of reference in the direction of the ray. But in perpendicular direction the width remains unchanged.

Hence, the width is $12.742* 10^(6)\ m$

A lead ball is dropped into a lake from a diving board 6.10 mm above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 4.50 ss after it is released. How deep is the lake?

Answers

Answer:

D=1.54489 m

Explanation:

Given data

S=6.10 mm= 0.0061 m

To find

Depth of lake

Solution

To find the depth of lake first we need to find the initial time ball takes to hit the water.To get the value of time use below equation

$S=v_(1)t+(1/2)gt^(2) \n 0.0061m=(0m/s)t+(1/2)(9.8m/s^(2) )t^(2)\n t^(2)=(0.0061m)/(4.9m/s^(2) )\n t=\sqrt{1.245*10^(-3) }\n t=0.035s$

So ball takes 0.035sec to hit the water

As we have found time Now we need to find the final velocity of ball when it enters the lake.So final velocity is given as

$v_(f)=v_(i)+gt\nv_(f)=0+(9.8m/s^(2) )(0.035s)\n v_(f)=0.346m/s$

Since there are (4.50-0.035) seconds left for (ball) it to reach the bottom of the lake

So the depth of lake given as:

$D=|vt|\nD=|0.346m/s*4.465s|\nD=1.54489m$

Answer: d = 1.54m

The depth of the lake is 1.54m

Explanation:

The final velocity of the ball just before it hit the water can be derived using the equation below;

v^2 = u^2 + 2as ......1

Where ;

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled.

Since the initial velocity is zero, and the acceleration is due to gravity, the equation becomes:

v^2 = 2gs

v = √2gs ......2

g = 9.8m/s^2

s = 6.10mm = 0.0061m

substituting into equation 2

v = √(2 × 9.8× 0.0061)

v = 0.346m/s

The time taken for the ball to hit water from the time of release can be given as:

d = ut + 0.5gt^2

Since u = 0

d = 0.5gt^2

Making t the subject of formula.

t = √(2d/g)

t = √( 2×0.0061/9.8)

t = 0.035s

The time taken for the ball to reach the bottom of the lake from the when it hits water is:

t2 = 4.5s - 0.035s = 4.465s

And since the ball falls for 4.465s to the bottom of the lake at the same velocity as v = 0.346m/s. The depth of the lake can be calculated as;

depth d = velocity × time = 0.346m/s × 4.465s

d = 1.54m

The depth of the lake is 1.54m

*PLEASE HELP*A baseball is pitched with a horizontal velocity of 25.21 m/s. Mike Trout hits the ball, sending it in the opposite direction (back toward the pitcher) at a speed of -50.67 m/s. The ball is in contact with the bat for 0.0014 seconds. What is the
acceleration of the ball?

Answers

Answer:

-54,200 m/s^2

Explanation:

a=(vf-vi)/t

When the sum of all the forces acting on a block on an inclined plane is zero, the blockA) must be at rest
B) must be accelerating
C) may be slowing down
D) may be moving at constant speed

Answers

Answer:

hmmm thats too hard for me.

Explanation:

7: A 2 mA current passes through a 1.4 cm long solenoid producing a magnetic field of .162 G. How many turns are in the solenoid

Answers

Answer:

The number of turns is 64449395

Explanation:

The expression for the solenoid formula is stated below, and it is what we are going to use to solve for the number of turns

B= μ₀nl

where B= magnetic field

μ₀= permeability , 4π × 10⁻⁷ Henry

n= number of turns

l= length of coil

Given data

current I= 2 mA

length L= 1.4 cm to meter we have 0.014

magnetic field B= 0.162 T

From the expression we can make n subject of formula we have

n=B/μ₀l

Substituting we have

n= 0.162/4π × 10⁻⁷*2*10^-3

n= 0.162/8π × 10^⁻10

n= 0.162/2.5136*10^-9

n= 64449395

According to American Heart Association, your target heart rate is__________. Group of answer choices

A .all of the above

B. 85% - 95% of your max heart rate

C. 20%-30% of your max heart rate

D. 50%-85% of your max heart rate

Answers

Answer:

Explanation:

A normal heart rate is from 85-95% the other heart rate is not normal because your heart is beating more than normal

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