a battery of 6v is connected in series with resistors of 0.1 ohm, 0.15 ohm, 0.2 ohm, 0.25 ohm and 6 ohm. how much current would flow through 0.2 ohm resistors ​

Answer:

The significant figures in a measurement consist of all the certain digits in that measurement plus one uncertain or estimated digit. In the ruler illustration below, the bottom ruler gave a length with 2 significant figures, while the top ruler gave a length with 3 significant figures.

Explanation:

Answer:

100 Ω

Explanation:

Given that

Open circuit voltage, V = 1 V

Short circuit current, I = 10 mA

Source resistance R, = ?

This is rather a straight forward question. Remember Ohms Law? Current being directly proportional to the voltage and inversely proportional to the resistance?

Yeah, that's the formula we'd be using.

Ohms Law states that V = IR, and thus, if we make R subject of the formula, we have

R = V / I, on substituting the values, we have

R = 1 / 10*10^-3

R = 1 / 0.01

R = 100 Ω

◈ A particle is moving along the x-axis according to the equation

____________________________

✪ Position at t = 3s :

➝ S = 4 + 6t - 2t²

➝ S = 4 + 6(3) - 2(3)²

➝ S = 4 + 18 - 18

➝ S = 4m

✪ Instantaneous velocity at t = 3s :

➝ v = dx/dt = d(4 + 6t - 2t²)/dt

➝ v = 6 - 4t

➝ v = 6 - 4(3)

➝ v = 6 - [tex]\sf{12}[/tex]

➝ v = -12m/s

✪ Instantaneous acc. at t = 3s :

➝ a = dv/dt = d(6 - 4t)/dt

➝ a = -4m/s²

[Acceleration does not depend on time]

Answer:

   t_total = 23.757 s

Explanation:

This is a kinematics exercise.

Let's start by calculating the distance and has to reach the limit speed of

v = 18.8 m / s

         v = v₀ + a t₁

the elevator starts with zero speed

         v = a t₁

         t₁ = v / a

         t₁ = 18.8 / 2.40

         t₁ = 7.833 s

in this time he runs

         y₁ = v₀ t₁ + ½ a t₁²

         y₁ = ½ a t₁²

         y₁ = ½ 2.40 7.833²

         y₁ = 73.627 m

This is the time and distance traveled until reaching the maximum speed, which will be constant throughout the rest of the trip.

           x_total = x₁ + x₂

           x₂ = x_total - x₁

           x₂ = 373 - 73,627

           x₂ = 299.373 m

this distance travels at constant speed,

           v = x₂ / t₂

           t₂ = x₂ / v

           t₂ = 299.373 / 18.8

           t₂ = 15.92 s

therefore the total travel time is

           t_total = t₁ + t₂

           t_total = 7.833 + 15.92

           t_total = 23.757 s

Answer:

This question has already been answered.

Explanation:

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Answer:

Speed = 10 km/hr

Explanation:

Speed = Distance / Time

Since we have to convert it into minutes to hours ( because it is km/hr )

45 minutes = 0.75 hours

Therefore,

Speed = 7.5 / 0.75

           = 10 km/hr

Hope it helps!

Answer:

10

Explanation:

Answer: Procrastination, Defiance, Laziness

Explanation:

Answer: impulsiveness

Explanation:

Answer:

The  value is  [tex]E =  8.9 *10^{-5} \  J[/tex]

Explanation:

From the question we are told that

   The  area is  [tex]A =  0.700 \  m^2[/tex]

   The  root mean square value  is [tex]E_{rms} =  0.0400 \  V/m[/tex]

   The time taken is [tex]t =  30.0 \  s[/tex]

Generally the energy is mathematically represented as

     [tex]E =  c *  \sepsilon_o *  A  *  t  * E_{rms}^2[/tex]

=>    [tex]E =  3.0*10^{8} *  8.85*10^{-12} *  0.700 *  30 * (0.04)^2[/tex]

=>     [tex]E =  8.9 *10^{-5} \  J[/tex]

Explanation:

Explanation:Those interested in working with trace evidence, such as glass, hairs, and gunshot residue, should focus on instrumentation skills and take courses in geology, soil chemistry, and materials science. If forensic biology, such as DNA analysis, is preferred, take microbiology, genetics, and biochemistry courses. Those interested in the toxicological aspects of this work, such as obtaining and interpreting toxicology reports, should study physiology, biochemistry, and chemistry.

Answer:

Generally it is only possible  to compute the tangent of angle in their various unit but because 5 meter is not an angle then it is impossible  to compute the tangent of 5 meters.

Explanation:

50Kpa

Explanation:

P1V1 = P2V2

Where;

P1= 100. P2= ?

V1 = 500. V2 = 1000

100 × 500 = P2 × 1000

50000 = 1000P2

50000/1000 = P2

50 = P2

P2 = 50Kpa

Answer:

412.5

Explanation:

all you have to do is multiply .15

Answer:

The formula is kelvin = (tempature in farenheit-32) ÷ 1.8 +273.15

So it will be (-22-32) ÷ 1.8 +273.15= 243.15 kelvin

Answer:

T(K) = (T(°F) + 459.67)× 5/9

answer is 243.15

Answer:

30

Explanation:

Answer:

Volume = 3.6 x 10^-12 cm^3

Area = 1.1 x 10^-5 mm^2

Explanation:

The diameter of the bacteria that is in the shape of spherical = 1.9 m  

Now we have to find the volume of the bacteria. So, use the below formula to find the volume of bacteria.

Volume = (4πr^3) / 3

Volume = (4/3)* π*(1.9*10^-6 / 2)^3  

Volume = 3.6 x 10^-18 m^3  

Volume = 3.6 x 10^-12 cm^3

Now find the surface area.

The surface area = 4πr^2

Area = 4  π  (1.9/2)62  

Area = 1.1 x 106-11 m^2  

Area = 1.1 x 10^-5 mm^2

Answer:

Explanation:

Force between two charged particle having charge q₁ and q₂ at distance r

F = k q₁ x q₂ / r²

Putting the values

F = 9 x 10⁹ x 1.6 x 10⁻¹⁹ x 1.6 x 10⁻¹⁹ / 5.29² x 10⁻²²

= .823 x 10⁻⁷

= 823 x 10⁻¹⁰ N

This force will apply on both the charges and will be attractive in nature .

B )

Let speed of electron be v

centripetal force on electron = F

m v² / r = 823 x 10⁻¹⁰

putting the values for electron

9.1 x 10⁻³¹ x v² / 5.29 x 10⁻¹¹ = 823 x 10⁻¹⁰

v² = 478.42 x 10¹⁰

v = 21.87 x 10⁵ m /s

Answer:

The  value is  [tex]\Delta t  =  0.01 \ s[/tex]

Explanation:

From the question we are told that

   The time of flight is  [tex]T  =  0.02 \  s[/tex]

Given that the value of the time of flight is in three decimal place then the error quote is

     [tex]\Delta t  =  0.01 \ s[/tex]

Answer:

The ruler's center of mass is located on the vertical line defined by the 50 cm mark.

Explanation:

Recall that in order to have equilibrium as that described by this example, the net Torque in the ruler should be zero. which means in our case that the weight due to the right half of the ruler, should equal in absolute value of the weight of the left side of the ruler, thus defining where the ruler's center of mass is.

Since this happens at the mark 50 cm, then the center of mass of the ruler is located somewhere inside the ruler somewhere above where the pivot is located (50 cm mark)

The point obviously is in the 3rs quadrant

So

စ= tan^-1( y/x)-180

စ= -89.7°

The angle θ between the radius-vector of the point and the positive x axis is -89.7°

Radius vector is defined as the length of the line that connects a stationary point to a moving point.

It can be defined as a line connecting a circle's center to a particle's location on its circumference.

It can also be defined as the straight line connecting two bodies moving in relative orbits, such as the path a planet takes to reach the sun at any given point along its orbit.

It is also known as position vector or location vector.

The unit vector pointing radially outward from the origin is the gradient of the position vector's length. It is typical for the spheres that are centered on the origin to have level surfaces.

Evidently, the location is in the third quadrant.

So, θ= tan⁻¹ (y / x) - 180

θ= - 89.7°

Thus, the angle θ between the radius-vector of the point and the positive x axis is -89.7°

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Answer:

The torque of the wheel is 0.92 N-m.

Explanation:

Given that,

Radius of wire = 3 cm

Angular acceleration = 92.00 rad/s²

Suppose, Mass of wheel is 4 kg and radius is 5 cm. find the torque of the wheel.

We need to calculate the torque of the wheel

Using formula of torque

[tex]\tau=I\alpha[/tex]

[tex]\tau=MR^2\times\alpha[/tex]

Where, M = mass of wheel

R = radius of wheel

[tex]\alpha[/tex] = angular acceleration

Put the value into the formula

[tex]\tau=4\times(5\times10^{-2})^2\times92[/tex]

[tex]\tau=0.92\ N-m[/tex]

Hence, The torque of the wheel is 0.92 N-m.

Answer:

T2 = 361.15 K

Explanation:

Since pressure is held constant, then we'll use Charles law

Charles law gives that V1/T1 = V2/T2

We are given;

Initial temperature; T1 = 313 K

Initial height; h1 = 23.4 cm

Final height; h2 = 27.0 cm

Now, formula for volume of a cylinder is V = πr²h

V1 = πr²h1 and V2 = πr²h2

Now, since V1/T1 = V2/T2 and we are looking for T2, let's make T2 subject;

T2 = (V2 × T1)/V1

Plugging in the relevant values gives;

T2 = (πr²h2 × 313)/πr²h1

πr² will cancel out to give;

T2 = (h2 × 313)/h1

Putting 23.4 for h1 and 27 for h2 gives;

T2 = (27 × 313)/23.4

T2 = 361.15 K

Answer: I dont see the option choices

Explanation:

Answer:

No its not possible in that yes heat flow is due to temperature different but with time equilibrium in temperature is reached hence there is no temperature difference at both ends at this time

Answer:

Explanation:

We shall apply newton's laws formula

a )

initial velocity in upward direction u = 10 m/s

acceleration due to gravity g = 9.8/ m .s²

Let h be the maximum height where v = o

v² = u² - 2gh

0 = 10² - 2 gh

h = 10² / 2g

= 10² /  2  x 9.8

= 5.10 m

Since the ball was thrown from height of 100 m , total maximum height of ball

= 100 + 5.10

= 105.10 m

Let t be the time taken

v = u - gt

0 = 10 - gt

t = 10 / 9.8

= 1.02 s

b )

when h = 50 on its way downwards , velocity

v² = u² + 2 g s  

v² = 0 + 2 x 9.8 x ( 105.10 - 50 )

[  distance travelled by ball at this point from top = 105.1 - 50 = 55.10 ]

v = 32.86 m / s

Let us find out final velocity of touching the ground . For it distance travelled = 105.10

v² = u² + 2gh

v² = 0 + 2 x 9.8 x 105.1

v = 45.39 m /s

Now velocity at h = 50 is 32.86

velocity at h = 0 is 45.39

time taken to travel fro h = 50 to h = 0

v = u + gt

45 .39 = 32.86 + 9.8 x t

t =  1.28 s .

Answer:

Explanation:

From the question we are told that

 The diameter is [tex]d = 150 \ m[/tex]

  The time taken is  [tex]t = 5 \ min = 300\ s[/tex]

Generally the circumference is mathematically represented as

     [tex]C =  d  *  \pi[/tex]

     [tex]C =  150 *  3.142[/tex]

     [tex]C =  471.3 \  m[/tex]

The average speed is mathematically represented as

      [tex]v  =  \frac{C}{ t}[/tex]

     [tex]v  =  \frac{471.3}{300}[/tex]

    [tex]v =  1.571 \  m/s[/tex]

Explanation:

1) F₁ lies in a plane perpendicular to the xy plane, 60° from the x axis.  The angle between F₁ and the +z axis is 30°.  Therefore, the vector is:

<F₁> = 20 (sin 30° cos 60° i + sin 30° sin 60° j + cos 30° k)

<F₁> = 20 (¼ i + ¼√3 j + ½√3 k)

<F₁> = 5 i + 5√3 j + 10√3 k

F₂ is in the xy plane.  Its slope is -24/7.  The vector is:

<F₂> = 12 (-⁷/₂₅ i + ²⁴/₂₅ j + 0 k)

<F₂> = -3.36 i + 11.52 j

F₃ is parallel to the +x axis.  The vector is:

<F₃> = 17 (i + 0 j + 0 k)

<F₃> = 17 i

F₄ is parallel to the -z axis.  The vector is:

<F₄> = 15 (0 i + 0 j − k)

<F₄> = -15 k

F₅ is in the xy plane.  It forms a 15° angle with the -y axis.  The vector is:

<F₅> = 9 (-sin 15° i − cos 15° j + 0 k)

<F₅> = -9 sin 15° i − 9 cos 15° j

The resultant vector is therefore:

<F> = (5 − 3.36 + 17 − 9 sin 15°) i + (5√3 + 11.52 − 9 cos 15°) j + (10√3 − 15) k

<F> = 16.31 i + 11.49 j + 2.32 k

2) Sum of forces at point B in the x direction:

∑F = ma

Tbc cos 40° − ¹⁵/₁₇ Tab = 0

Tbc cos 40° = ¹⁵/₁₇ Tab

Tbc = 1.15 Tab

Sum of forces at point B in the y direction:

∑F = ma

Tbc sin 40° + ⁸/₁₇ Tab − mAg = 0

Tbc sin 40° + ⁸/₁₇ Tab = (2 kg) (10 m/s²)

(1.15 Tab) sin 40° + ⁸/₁₇ Tab = 20 N

1.21 Tab = 20 N

Tab = 16.52 N

Tbc = 19.02 N

Sum of forces at point C in the x direction:

∑F = ma

Tcd sin 25° − Tbc cos 40° = 0

Tcd sin 25° = Tbc cos 40°

Tcd = 1.81 Tbc

Tcd = 34.48 N

3(a) When the crane is on the verge of tipping, the center of gravity is directly over point F.  Relative to point A:

3.7 m = [ (390 kg) (0.9 m) + (90 kg) (9 m cos θ + 1.7 m) + (80 kg) (9 m cos θ + 2.9 m) ] / (390 kg + 90 kg + 80 kg)

2072 kgm = 351 kgm + 810 kgm cos θ + 153 kgm + 720 kgm cos θ + 232 kgm

1336 kgm = 1530 kgm cos θ

θ = 29.17°

3(b) 3.7 m = [ (390 kg) (0.9 m) + (90 kg) (x + 1.7 m) + (80 kg) (x + 2.9 m) ] / (390 kg + 90 kg + 80 kg)

2072 kgm = 351 kgm + (90 kg) x + 153 kgm + (80 kg) x + 232 kgm

1336 kgm = (170 kg) x

x = 7.86 m

4) Find the lengths of the cables.

Lab = √((2 m)² + (3 m)² + (5 m)²)

Lab = √38 m

Lac = √((2 m)² + (3 m)² + (5 m)²)

Lac = √38 m

Lde = √((2 m)² + (3 m)²)

Lde = √13 m

Sum of forces in the x direction:

∑F = ma

-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0

Sum of forces in the y direction:

∑F = ma

2/√38 Fab − 2/√38 Fac = 0

Fab = Fac

Sum of forces in the z direction:

∑F = ma

3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0

Sum of moments about the y-axis:

∑τ = Iα

(3/√38 Fab) (5 m) + (3/√38 Fac) (5 m) + (3/√13 Fde) (2 m) − (mg) (2 m) = 0

Substitute Fab = Fac and simplify:

6/√38 Fab + 3/√13 Fde − mg = 0

30/√38 Fab + 6/√13 Fde − 2mg = 0

Double first equation:

12/√38 Fab + 6/√13 Fde − 2mg = 0

Subtract from the second equation:

28/√38 Fab = 0

Fab = 0

Fac = 0

Solve for Fde:

3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0

3/√13 Fde = mg

3/√13 Fde = (1.7 kg) (10 m/s²)

Fde = 20.43 N

Solve for Rx:

-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0

Rx = 2/√13 Fde

Rx = 11.33 N

Answer:

Let N = N0 where N0 is the number of atoms originally present.

In 6 days    N = N0 / 2

In 12 days   N = N0 / 4

In 18 days   N = N0 / 8    = .125 N0

So it would take 18 days.

Manganese-52 has a half-life of 6 days. a scientist has to wait 18 days for the radioactivity to be 12.5% of the starting amount

The ability of some unstable atoms to emit nuclear radiation spontaneously, typically in the form of alpha or beta particles frequently accompanied by gamma rays, is known as radioactivity. This radiation is released when a nucleus undergoes radioactive decay and transforms into a different isotope that may be radioactive (unstable) or non-radioactive depending on the number of neutrons and protons in the nucleus (stable).

As given in the problem statement Manganese-52 has a half-life of 6 days. this means after 6 days Manganese-52 will reduce to its half value

similarly, after 12 days it further reduces to 50% of the previous value which would be 25 %

Similarly, after 18 days, it will reduce 50% from the previous value which comes out to be 12.5 %

Thus, The half-life of manganese-52 is 6 days. It takes 18 days for the radioactivity to drop to 12.5% of the initial level.

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