for the long life cells we have to connect them in ____ combination​

Answer:

Explanation:

From the question;

We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.

We are to calculate the following task, i.e. to determine the electric field at the distances:

a)  at 4.75 cm

b)  at 20.5 cm

c) at 125.0 cm

Given that:

the charge (q) = 33.3 nC/m

= 33.3 × 10⁻⁹ c/m

radius of rod = 5.75 cm

a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.

Then, the electric field will be zero.

b) The electric field formula [tex]E = \dfrac{kq }{d}[/tex]

[tex]E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}[/tex]

E = 1461.95 N/C

c) The electric field E is calculated as:

[tex]E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}[/tex]

E = 239.76 N/C

Answer:

the person  is sitting 1.5 m from the left end of the board

Explanation:

Given the data in the question;

Wb = 125 N

Wm = 500 N

T₂ = 250 N

Now, we know that;

T₁ + T₂ = Wb + Wm

T₁ + 250 = 125 + 500

T₁ = 125 + 500 - 250

T₁ = 375 N

so tension of the left chain is 375 N.

Now, taking torque about the left end

500 × d + 125 × 2 = 250 × 4

500d + 250 = 1000

500d = 1000 - 250

500d = 750

d = 750 / 500

d = 1.5 m

Therefore, the person  is sitting 1.5 m from the left end of the board.

Answer:

We want to solve the sum:

6.74*10⁴ + 8.95*10⁴

first, we take the common factor 10⁴ out, so we get:

(6.74 + 8.95)*10⁴

Now we solve the sum:

(15.66)*10⁴

Now we want to rewrite it in exponential form, wo we can rewrite it as:

(15.66)*10⁴ = (1.566*10)*10⁴ = (1.566)*10*10⁴ = (1.566)*10⁴⁺¹ = 1.566*10⁵

k = 5.

Binary code is the answer

Answer:

binary code is the answer of blank

I don't know

because I don't know

Answer:

time of fall is 1.75 s and the velocity with which it strikes the water is 17.15 m/s.

Explanation:

Height, h =  15 m

Newton's second law

Force = mass x acceleration

The unit of gravitational force is Newton and the value is m x g.

where, m is the mas and g is the acceleration due to gravity.  

Let the time of fall is t.

Use second equation of motion

[tex]s= u t +0.5 at^2\\\\15 = 0 +0.5\times 9.8\times t^{2}\\\\t = 1.75 s[/tex]

Let the final speed is v.

Use third equation of motion

[tex]v^2 = u^2 + 2 a s\\\\v^2 = 0 + 2 \times 9.8\times 15\\\\v =17.15 m/s[/tex]

Answer:

120 Newton

Explanation:

Given the following data;

Mass = 12 kg

Angle = 4°

We know that acceleration due to gravity is equal to 10 m/s

To find the minimum force to stop the block from sliding;

Force = mgCos(d)

Where;

m is the mass of an object.

g is the acceleration due to gravity.

d is the angle of inclination (theta).

Substituting into the formula we have;

F = 12*10*Cos(4°)

F = 120 * 0.9976

F = 119.71 ≈ 120 Newton

Answer:

We can also prove the conservation of mechanical energy of a freely falling body by the work-energy theorem, which states that change in kinetic energy of a body is equal to work done on it. i.e. W=ΔK. And ΔE=ΔK+ΔU. Hence the mechanical energy of the body is conserved

Explanation:

If a person wants to calculate the length of time it takes for a ball to fall from a height of 20m, the correct equation that they should use is:

D. Δt= √2Δd/a

The free fall formula should be used to obtain the length of time that it takes for a ball to fall from a given height. This formula also factors the height or distance from which the fall occurred and this is denoted by the letter d. The small letter 'a' is denotative of acceleration due to gravity and this is a constant pegged at -9.98 m/s².

So, the change in height is obtained and multiplied by two. This is further divided by the acceleration and the square root of the derived answer translates to the time taken for the ball to fall from the height of 20m. Of all the options listed, option D represents the correct equation.

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Answer:

Explanation:

Given that:

Focal length for the objective lens = 20 mm, 4 mm, 1.4 mm

For objective lens of focal length f₁ = 20 mm

s₁' = 120 mm + 20 mm = 140 mm

∴

Magnification [tex]m_1 = \dfrac{s'_1}{f_1}[/tex]

[tex]m_1 = \dfrac{140}{20}[/tex]

[tex]m_1 = 7 \ m[/tex]

For objective lens of focal length f₁ = 4 mm

s₁' = 120 mm + 4 mm = 124 mm

[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]

[tex]m_1 = \dfrac{124}{4}[/tex]

[tex]m_1 = 31 \ m[/tex]

For objective lens of focal length f₁ = 1.4 mm

s₁' = 120 mm + 1.4 mm = 121.4 mm

[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]

[tex]m_1 = \dfrac{121.4}{1.4}[/tex]

[tex]m_1 = 86.71 \ m[/tex]

The magnification of the eyepiece is given as:

[tex]m_e = 5X \ and \ m_e = 15X[/tex]

Thus, the largest angular magnification when  [tex]m_1 \ and \ m_e \ are \ large \ is:[/tex]

[tex]M_{large}= (m_1)_{large} \times (m_e)_{large}[/tex]

= 86.71 × 15

= 1300.65

The smallest angular magnification derived when [tex]m_1 \ and \ m_e \ are \ small \ is:[/tex]

[tex]M_{small}= (m_1)_{small} \times (m_e)_{small}[/tex]

= 7 × 5

= 35

The largest magnification will be 1300.65 and the smallest magnification will be 35.

Magnification is defined as the ratio of the size of the image of an object to the actual size of the object.

Now for objective lens and eyepieces, it is defined as the ratio of the focal length of the objective lens to the focal length of the eyepiece.

It is given in the question:

Focal lengths for the objective lens is = 20 mm, 4 mm, 1.4 mm

now we will calculate the magnification for all three focal lengths of the objective lens.

Also, each objective forms an image 120 mm beyond its second focal point.

(1) For an objective lens of focal length   [tex]f_1=20 \ mm[/tex]

[tex]s_1'=120\ mm +20 \ mm =140\ mm[/tex]

Magnification will be calculated as

[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{140}{20} =7[/tex]

(2) For an objective lens of focal length [tex]f_1= \ 4 \ mm[/tex]

s₁' = 120 mm + 4 mm = 124 mm

[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{124}{4} =31[/tex]

(3) For an objective lens of focal length [tex]f_1=1.4\ mm[/tex]

s₁' = 120 mm + 1.4 mm = 121.4 mm

[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{121.4}{1.4} =86.71[/tex]

Now the magnification of the eyepiece is given as:

[tex]m_e=5x\ \ \ & \ \ m_e=15x[/tex]

Thus, the largest angular magnification when  

[tex]m_1 = 86.17\ \ \ \ m_e=15x[/tex]

[tex]m_{large}= (m_1)_{large}\times (m_e)_{large}[/tex]

[tex]m_{large}=86.71\times 15=1300.65[/tex]

The smallest angular magnification derived when

[tex]m_1=7\ \ \ \ m_e=5[/tex]

[tex]m_{small}=(m_1)_{small}\times (m_e)_{small}[/tex]

[tex]m_{small}=7\times 5=35[/tex]

Thus the largest magnification will be 1300.65 and the smallest magnification will be 35.

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Answer:

Explanation:

There are two main ways to de-starch leaves of a plant - the 'Light Exclusion' Method and the 'Carbon Dioxide Deprivation' Method. The 'Light Exclusion' method is a simpler procedure and is used often. Leaves can be destarched by depriving them of light for an extended period of time, usually 24-48 hours.

Answer:

Maria Corazon Sumulong Cojuangco Aquino, popularly known as Cory Aquino, was a Filipino politician who served as the 11th President of the Philippines, the first woman to hold that office.

Answer:

Former President of the Philippines

Explanation:

v1f = -0.16 ms

Explanation:

Use the conservation law of linear momentum:

m1v1i + m2v2i = m1v1f + m2v2f

where

v1i = v2i = 0

m1 = 160 kg

m2 = 0.50 kg

v2f = 50m/s

v1f = ?

So we have

0 = (160 kg)v1f + (0.5 kg)(50 m/s)

v1f = -(25 kg-m/s)/(160 kg)

= -0.16 m/s

Note: the negative sign means that its direction is opposite that of the arrow.

(a) When the spring is compressed 4.5 cm = 0.045 m, it exerts a restoring force on the block of magnitude

F = (1900 N/m) (0.045 m) = 85.5 N

so that at the moment the block is released, this force accelerates the block with magnitude a such that

85.5 N = (1.15 kg) a   ==>   a = (85.5 N) / (1.15 kg) ≈ 74.3 m/s²

The block reaches its maximum speed at the spring's equilbrium point, and this speed v is such that

v ² = 2 (74.3 m/s²) (0.045 m)   ==>   v = √(2 (74.3 m/s²) (0.045 m)) ≈ 2.59 m/s

(b) There is no friction between the block and plane, so the block maintains this speed as it slides over the edge. At that point, it's essentially in free fall, so if y is the height of the plane, then

(7 m/s)² - (2.59 m/s)² = 2gy   ==>   y = ((7 m/s)² - (2.59 m/s)²) / (2g) ≈ 2.16 m

Explanation:

The logarithmic damping decrement of a mathematical pendulum is DeltaT=0.5. How will the amplitude of oscillations decrease during one full oscillation of the pendulum

Answer:

I don't see anything on your question?

Answer:

2.8 MW

Explanation:

There are 7 wind turbines in the wind farm as shown in the diagram. Thus, the energy output by one turbine is 1/7 if the total energy output. So, 19.6/7=2.8MW

Answer:

Answer:

a) S =  63.2 km/h

b) V =  63.2 km/h*(-0.316 , 0.949)

Explanation:

Let's define:

North as the positive y-axis

East as the positive x-axis.

Also, remember the relation:

Distance = Time*Speed

Let's assume that she starts at the position (0km, 0km)

Then she travels due North at 90km/h for two hours, then the displacement is

90km/h*2h = 180km to the north

Then the new position is:

(0km, 180km)

Then she travels West at 60km/h for one hour.

Then the distance traveled to the West (negative x-axis) is:

60km/h*1h = 60km to the west

Then the new position is:

(-60km, 180km).

a) The average speed is defined as the quotient between the displacement and the time.

We know that the total time traveled is 3 hours.

And the displacement is the difference between the final position and the initial position.

this is:

D = √( -60km - 0km)^2 + (180km - 0km)^2)=

D = √( (60km)^2 + (180km)^2) = 189.7 km

Then the average speed is:

S = (189.7 km)/(3 h) = 63.2 km/h

b) Now we want to find the average velocity, this will be equal to the average speed times a versor that points from the origin to the direction of the final position.

So, if the final position is (-60km, 180km)

We need to find a vector that represents the same angle, but that is on the unit circle.

Then, if the module of the final position is 189.7 km (as we found above), then the versor is just given by:

(-60km/ 189.7 km, 180km/ 189.7 km)

(-60/189.7 , 180/189.7)

We can just check that the module of the above versor is 1.

[tex]module = \sqrt{(\frac{-60}{189.7} )^2 + (\frac{180}{189.7} )^2} = \frac{1}{189.7}* \sqrt{(-60 )^2 + (180 )^2} = 1[/tex]

Then the average velocity is:

V = 63.2 km/h*(-60/189.7 , 180/189.7)

We can simplify our versor so the velocity equation is easier to read:

V = 63.2 km/h*(-0.316 , 0.949)

Answer:

Fr = 1.91 * 9*10⁹*q²/L²

Explanation:

Let´s say that the corners of the square are  A B C and D

We are going to find out the force on the charge placed on B  ( the charge placed in the upper right corner.

As all the charges are positive (the same sign), then all the three forces on the charge in B are of rejection.

Force due to charge placed in A

module   Fₓ =  K* q² / L²   in the direction of x

Force due to charge placed in C

module  Fy = K* q²/L²   in the direction of y

Force due to  the charge placed in D

That force will have the direction of the diagonal of the square, and the distance between charges placed in D and A is the length of the diagonal.

d²  =  L²  +  L²  =  2*L²

d  =  √2 * L

The module of the force due to charge place in D

F₄₅ = K*q²/ 2*L²

To get the force we need to add first  Fₓ  and  Fy  

Fx + Fy  =  F₁

module of  F₁ = √ Fx² + Fy²    the direction will be the same as the diagonal of the square then:

F₁   =   √  ( K* q²/L² )²  +   ( K* q²/L² )²

F₁  =  √ 2  *  K*q²/L²

And now we add forces F₁   and F₄₅   to get the net force Fr on charge in point B.

The direction of Fr is the direction of the diagonal and is of rejection

the module is

Fr  =  F₁  *  F₄₅

Fr  =  √ 2  *  K*q²/L²  +   K*q²/ 2*L²

Fr  = ( √ 2 + 0,5 ) * K*q² /L²

K  =  9*10⁹  Nm²C²

Fr = 1.91 * 9*10⁹*q²/L²

We don´t know units of L and q

Answer:

a scale is used to measure the mass of the body

Answer:

[tex]\lambda=1.39\times 10^{-4}\ s^{-1}[/tex]

Explanation:

Given that,

The half-life of Barium-139 is [tex]4.96\times 10^3[/tex]

A sample contains [tex]3.21\times 10^{17}[/tex] nuclei.

We need to find the decay constant for this decay. The formula for half life is given by :

[tex]T_{1/2}=\dfrac{0.693}{\lambda}\\\\\lambda=\dfrac{0.693}{T_{1/2}}[/tex]

Put all the values,

[tex]\lambda=\dfrac{0.693}{4.96\times 10^3}\\\\=1.39\times 10^{-4}\ s^{-1}[/tex]

So, the decay constant is [tex]1.39\times 10^{-4}\ s^{-1}[/tex].

Answer: hello your question is poorly written attached below is the complete question

answer :

TA = 1.6*10^-24 * 60 * 2,  TB = 1.6*10^-24 * ( 60 + 30 ) * 2  -- ( option 1 )

Explanation:

a = 2m/s^2

Ta = m₁ a = 60 * 1.6 * 10^-24 * 2 ц

Tb - Ta = m₂ a

∴ Tb = m₂ a  + Ta

       = ( 30 * 1.6 * 10^-24 * 2 ) +  ( 60 * 1.6 * 10^-24 * 2 )

= ( 30 + 60 ) * 1.6 * 10^-24 * 2 ц

Answer:

0.32 m.

Explanation:

To solve this problem, we must recognise that:

1. At the maximum height, the velocity of the ball is zero.

2. When the velocity of the ball is 2.5 m/s above the ground, it is assumed that the potential energy and kinetic energy of the ball are the same.

With the above information in mind, we shall determine the height of the ball when it has a speed of 2.5 m/s. This can be obtained as follow:

Mass (m) = constant

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) = 2.5 m/s

Height (h) =?

PE = KE

Recall:

PE = mgh

KE = ½mv²

Thus,

PE = KE

mgh = ½mv²

Cancel m from both side

gh = ½v²

9.8 × h = ½ × 2.5²

9.8 × h = ½ × 6.25

9.8 × h = 3.125

Divide both side by 9.8

h = 3.125 / 9.8

h = 0.32 m

Thus, the height of the ball when it has a speed of 2.5 m/s is 0.32 m.

Answer:

The velocity after 1.5 s is 22.7 m/s downwards.

Explanation:

Initial velocity = - 8 m/s

acceleration, a = - 9.8 m/s2

time, t = 1.5 s

Use first equation of motion

v = u + at

v = - 8 - 9.8 x 1.5

v = - 8 - 14.7

v = - 22.7 m/s  

Thus, the velocity after 1.5 s is 22.7 m/s downwards.

Answer:

a) the most probable speed of the molecules is 409.2 m/s

b) required temperature of xenon is 1322 K

Explanation:

Given the data in the question;

a)

Maximum probable speed of hydrogen molecule (H₂)

[tex]V_{H_2[/tex] = √( 2RT / [tex]M_{H_2[/tex] )

where R = 8.314 m³.Pa.K⁻¹.mol⁻¹ and given that T = 20.3 K

molar mass of H₂; [tex]M_{H_2[/tex] = 2.01588 g/mol

we substitute

[tex]V_{H_2[/tex] = √( (2 × 8.314 × 20.3 ) / 2.01588 × 10⁻³  )

[tex]V_{H_2[/tex] = √( 337.5484 / 2.01588 × 10⁻³  )

[tex]V_{H_2[/tex] = 409.2 m/s

Therefore, the most probable speed of the molecules is 409.2 m/s

b)

Temperature of xenon  = ?

Temperature of hydrogen = 20.3 K

we know that;

T = (Vxe² × Mxe) / 2R

molar mass of xenon; Mxe = 131.292 g/mol

so we substitute

T = ( (409.2)² × 131.292 × 10⁻³) / 2( 8.314  )

T = 21984.14167 / 16.628

T = 1322 K

Therefore, required temperature of xenon is 1322 K

Answer:

Josiah Spiers in 1867 was when scripture union was founded

The reason why we are unlikely to see fluctuations in light output in extragalactic sources with a diameter of about one light day over timescales shorter than about one day is due to the size and distance of the source, as well as the speed of light.

When we observe an extragalactic source with a diameter of about one light day, we are essentially observing light that has traveled a very long distance through space to reach us. This light may have originated from a region of the source that is changing in brightness or emitting intense bursts of light, but by the time the light reaches us, these fluctuations are smeared out over a longer period of time due to the speed of light.

For example, if the source were emitting a burst of light that lasted for only a few hours, by the time that light travelled a distance of one light day (which is about 25 billion miles or 40 billion kilometres), the burst would be spread out over a longer period of time. This is because the light emitted at the beginning of the burst would have already traveled a significant distance away from the light emitted at the end of the burst by the time it reached us. As a result, we would observe the burst as a more gradual increase and decrease in light output over a period of several days, rather than a sharp increase and decrease over a few hours.

In addition, the turbulent interstellar and intergalactic media that the light passes through can also scatter and delay the light, further smearing out any short-term fluctuations in light output. This effect is known as interstellar scintillation and can make it even more difficult to observe short-term variations in the light output of extragalactic sources.

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