The valid digits in a measurement are called _____ digits. Question 10 options: insignificant significant uncertain non-zero

Complete question:

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:

a) the midpoint between the two rings?

b) the center of the left ring?

Answer:

a) the electric field strength at the midpoint between the two rings is 0

b) the electric field strength at the center of the left ring is 2712.44 N/C

Explanation:

Given;

distance between the two rings, d = 25 cm = 0.25 m

diameter of each ring, d = 10 cm = 0.1 m

radius of each ring, r = [tex]\frac{0.1}{2} = 0.05 \ m[/tex]

the charge on each ring, q = 20 nC

Electric field strength for a ring with radius r and distance x from the center of the ring is given as;

[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}}[/tex]

The electric field strength at the midpoint;

the distance from the left ring to the mid point , x = 0.25 m / 2 = 0.125 m

[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9}*0.125*20*10^{-9}}{(0.125^2 + 0.05^2)^{3/2}} \\\\E = 9210.5 \ N/C[/tex]

[tex]E_{left} = 9210.5 \ N/C[/tex]

The electric field strength due to right ring is equal in magnitude to left ring but opposite in direction;

[tex]E_{right} = -9210.5 \ N/C[/tex]

The electric field strength at the midpoint;

[tex]E_{mid} = E_{left} + E_{right}\\\\E_{mid} = 9210.5 \ N/C - 9210.5 \ N/C\\\\E_{mid} = 0[/tex]

(b)

The distance from the right ring to center of the left ring, x = 0.25 m.

[tex]E = \frac{KxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9} *0.25*20*10^{-9}}{(0.25^2 + 0.05^2)^{3/2}} \\\\E = 2712.44 \ N/C[/tex]

Answer:

The value of resistance of each resistor, R is 2.25 Ω

Explanation:

Given;

voltage across the three resistor, V = 1.5 V

power dissipated by the resistors, P = 3.00 W

the resistance of each resistor, = R

The effective resistance of the three resistors is given by;

R(effective) = R/3

Apply ohms law to determine the current delivered by the source;

V = IR

I = V/R

I = 3V/R

Also, power is calculated as;

P = IV

P = (3V/R) x V

P = 3V²/R

R = 3V² / P

R = (3 x 1.5²) / 3

R = 2.25 Ω

Therefore, the value of resistance of each resistor, R is 2.25 Ω

Answer:

M = F/3μ g - M₁/3

Explanation:

To solve this exercise we must use the equilibrium conditions translations

         ∑ F = 0

In the attachment we can see a free body diagram of each block

Block M (upper)

X axis

      fr₁ + F₂ -F = 0

      F = fr₁ + F₂              (1)

axis

     N₁-W = 0

     N₁ = Mg

the friction force has the formula

     fr₁ = μ N₁

     F = μ Mg + F₂

bottom block

X axis

     F₂ - fr₁ - fr₂ = 0

     F₂ = fr₁ + fr₂

Y axis

     N - W₁ -W = 0

     N = g (M + M₁)

we substitute

       F₂ = μ Mg + μ (M + M1) g

       F₂ = μ g (2M + M₁)

we substitute in 1

      F = μ M g + μ g (2M + M₁)

      F = μ g (3M + M₁)

we look for mass M    

      M = (F -  μ g M₁)/ 3μ g

      M = F/3μ g - M₁/3

the exercise does not have numerical data

Answer:

 I = 1.23 A

Explanation:

In an RL circuit current passing is described by

           I = E / R (1 - [tex]e^{-Rt/L}[/tex])

Let's reduce the magnitudes to the SI system

        L = 35 mH = 35 10⁻³ H

        t = 5.0 ms = 5.0 10⁻³ s

let's calculate

         I = 18/12 (1 - [tex]e^{-12 .. 5 {10}^{-3}/35 .. {10}^{-3} }[/tex]e (- 5 10-3 12/35 10-3))

         I = 1.5 (1- [tex]e^{-1.715}[/tex])

         I = 1.23 A

Answer:

d) (CATA + CBTB) / (CA + CB)

Explanation:

According to the given situation, the final temperature of both objects is shown below:-

We assume T be the final temperature

while m be the mass

So it will be represent

m CA (TA - T) = m CB (T - TB)

or we can say that

CATA - CA T = CB T - CBTB

or

(CA + CB) T = CATA + CBTB

or

T = (CA TA + CBTB) ÷ (CA + CB)

Therefore the right answer is d

The final temperature of both objects is [tex]T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex].

The given parameters;

The final temperature of both objects is calculated as follows;

heat lost by object A is equal to heat gained by object B

[tex]mC_A (T_A - T) = mC_B(T- T_B)\\\\C_AT_A-C_AT = C_BT - C_BT_B\\\\C_BT+C_AT = C_AT_A+ C_BT_B\\\\T(C_B + C_A) = C_AT_A+ C_BT_B \\\\T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex]

Thus, the final temperature of both objects is [tex]T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex].

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Answer:

If the older generation is lacking, the younger generation would likely have knowledge, skill, or a positive attitude in some combination, but it is relative to the culture.

The simple reason is the desirability for genetic variation using recessive genes.

In other words, if the older generation lacks something, it tends to be something they don’t need, but something that will look good on young people. But mostly relative to the culture and education system.

Hope this helps

Answer:

in an accident, when the body collides with the air bags, the collision time of impact between the two bodies will increase due to the presence of air bags in the car. Larger is the impact time smaller is the transformation of energy between the body and air bag. That is why air bags greatly reduce the chance of injury in a car accident.

Answer:

The working principle of a water pump mainly depends upon the positive displacement principle as well as kinetic energy to push the water.

Explanation:

it mainly depends upon the positive displacement principle and also kinetic energy to push water. hope this hepls!

Answer:

the two ice skater have the same momentum but the are in different directions.

Paula will have a greater speed than Ricardo after the push-off.

Explanation:

Given that:

Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.

A. Which skater, if either, has the greater momentum after the push-off? Explain.

The law of conservation of can be applied here in order to determine the skater that possess a greater momentum after the push -off

The law of conservation of momentum states that the total momentum of two  or more objects acting upon one another will not change, provided there are no external forces acting on them.

So if two objects in motion collide, their total momentum before the collision will be the same as the total momentum after the collision.

Momentum is the product of mass and velocity.

SO, from the information given:

Let represent the mass of Paula with [tex]m_{Pa}[/tex] and its initial velocity with [tex]u_{Pa}[/tex]

Let represent the mass of Ricardo with [tex]m_{Ri}[/tex] and its initial velocity with [tex]u_{Ri}[/tex]

At rest ;

their velocities will be zero, i.e

[tex]u_{Pa}[/tex] = [tex]u_{Ri}[/tex] = 0

The initial momentum for this process can be represented as :

[tex]m_{Pa}[/tex][tex]u_{Pa}[/tex] +  [tex]m_{Ri}[/tex][tex]u_{Ri}[/tex] = 0

after push off from each other then their final velocity will be [tex]v_{Pa}[/tex] and [tex]v_{Ri}[/tex]

The we can say their final momentum is:

[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] +   [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex] = 0

Using the law of conservation of momentum as states earlier.

Initial momentum = final momentum = 0

[tex]m_{Pa}[/tex][tex]u_{Pa}[/tex] +  [tex]m_{Ri}[/tex][tex]u_{Ri}[/tex] =  [tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] +   [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]

Since the initial velocities are stating at rest then ; u = 0

[tex]m_{Pa}[/tex](0) + [tex]m_{Pa}[/tex](0) = [tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] +   [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]

[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] +   [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]  = 0

[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] = - [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]

Hence, we can conclude that the two ice skater have the same momentum but the are in different directions.

 B. Which skater, if either, has the greater speed after the push-off? Explain.

Given that Ricardo weighs more than Paula

So [tex]m_{Ri} > m_{Pa}[/tex] ;

Then [tex]\mathsf{\dfrac{{m_{Ri}}}{m_{Pa} }= 1}[/tex]

The magnitude of their momentum which is a product of mass and velocity can now be expressed as:

[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] =  [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]

The ratio is

[tex]\dfrac{v_{Pa}}{v_{Ri}} =\dfrac{m_{Ri}}{m_{Pa}} = 1[/tex]

[tex]v_{Pa} >v_{Ri}[/tex]

Therefore, Paula will have a greater speed than Ricardo after the push-off.

(A) Both the skaters have the same magnitude of momentum.

(B) Paula has greater speed after push-off.

Given that two skaters Paula and Ricardo are initially at rest.

Ricardo weighs more than Paula.

Let us assume that the mass of Ricardo is M, and the mass of Paula is m.

Let their final velocities be V and v respectively.

(A) Initially, both are at rest.

So the initial momentum of Paula and Ricardo is zero.

According to the law of conservation of momentum, the final momentum of the system must be equal to the initial momentum of the system.

Initial momentum = final momentum

0 = MV + mv

MV = -mv

So, both of them have the same magnitude of momentum, but in opposite directions.

(B) If we compare the magnitude of the momentum of Paula and Ricardo, then:

MV = mv

M/m = v/V

Now, we know that M>m

so, M/m > 1

therefore:

v/V > 1

v > V

So, Paula has greater speed.

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Answer:

Check your router connections then restart your router.

Explanation:

Answer:

Check your router connections then restart your router.

Explanation:

Most internet access comes from routers so the problem is most likely the router.

Answer:

9.88 milivolt

Explanation:

Given: diameter d = 5.2 cm

magnetic field B_1 = 1.35 T, final magnetic field B_2 =0 T

t = 0.29 sec.

we know emf =  - dΦ/dt

and flux Φ = BA

A= area

therefore emf ε = -A(B_2-B_1)/Δt

[tex]=-\pi(d/2)^2\frac{B_2-B_1}{\Delta t} \\=-\pi(0.052/2)^2\frac{0-1.35}{0.29} \\=98.8\times10^4\\=9.88 mV[/tex]

Answer:

strong nuclear is the stronger force; it is related to mass

Explanation:

If we compare the force of gravity to strong nuclear force, we could conclude that strong nuclear is the stronger force; it is related to mass, therefore the correct answer is option C

The nuclear force is the interaction between the subatomic particles that make up a nucleus. There are two types of nuclear forces: the strong nuclear force and the weak nuclear force. Depending on the separation between the proton neutron and proton pairs, these nuclear forces can be both attracting and positive.

Both types of nuclear forces come under the four fundamental forces of nature. There are mainly four fundamental forces of nature electromagnetic force, gravitational force, strong nuclear force, and weak nuclear force.

Thus, Option C is the appropriate response since, when compared to the force of gravity, the strong nuclear force is the greater force because it is tied to mass.

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Answer:

True

Explanation:

Because the resistors in series is the sum of the two resistors given as

R= R1+R2

While that of resistors in parallel is the sum of the reciprocal of the resistance given as

1/R = 1/ R1+ 1/R2

So that of series connection will be greater

Answer:

The charge is  [tex]q = 1.50 *10^{-5} \ C[/tex]

Explanation:

From the question we are told that

   The electric field strength is  [tex]E = 1860 \ N/C[/tex]

    The force is  [tex]F = 0.02796 \ N[/tex]

Generally the charge on this particle is mathematically represented as

     [tex]q = \frac{F}{E}[/tex]

=>   [tex]q = \frac{0.02796}{ 1860}[/tex]

=>   [tex]q = 1.50 *10^{-5} \ C[/tex]

Answer:

Explanation:

Find:

Computation:

F = (9/5)C + 32

3 times that of the Celsius

If C = x

So F = 3x

So,

3x = (9/5)x + 32

15x = 9x +160

6x = 160

x = 26.67

So, C = 26.67° and F = 80°

1/5 times that of the Celsius

If C = x

So F = x/5

So,

x/5 = (9/5)x + 32

x = 9x + 160

x = -20

So, C = -20° and F = -4°

Answer:

They both have the same total K.E at the bottom

Explanation:

This Is because If assuming no work is done by non conservative forces, total mechanical energy must be conserved

So

K1 + U1 = K2 + U2

But If both hoops start from rest, and and at the bottom of the incline the level for gravitational potential energy is zero for reference

thus

K1 = 0 , U2 = 0

ΔK = ΔU = m g. h

But if the two inclines have the same height, and both hoops have the same mass m,

So difference in kinetic energy, must be the same for both hoops.

Explanations:

a) The physical properties of a mineral is as a result of the arrangement of the atoms in the minerals. The reason behind one mineral having a density of 2.1 g/ml which is lower than that of another mineral with density of 5.1 g/ml is the packing density of the minerals. Minerals with high density have their atoms more closely packed together, leaving less space between the atoms. This characteristics means that they have more atomic mass per unit volume for a given molecular space, when compared to another mineral with its atoms less closely packed.

b) The property of cleavage is due to the crystalline structure of a mineral species. Cleavage is used to describe the ease with which minerals cleaves. Cleavage is due to a weak bonding strength between the molecules of the mineral, or a due to a greater lattice spacing across the the cleavage plane than in other planes within the crystal. The greater the lattice spacing, the weaker the strength of the bond across a plane.

From these, we can clearly see that the property of cleavage is due to the physical properties of the crystalline structure of the mineral species.

Answer:

d)    α = 1693.5 rad / s² , a = 392.7 m / s² ,   a_total = α √(R² +1) ,

e)   tan θ = a / α

Explanation:

This is an exercise in linear and angular kinematics.

We initialize reduction of all the magnitudes to the SI system

   w₀ = 3000 rev / min (2π rad / 1rev) (1min / 60s) = 314.16 rad / s

   w = 6000 rev / mi = 628.32 rad / s

   θ = 12 rev = 12 rev (2π rad / 1 rev) = 75.398 rad

d) ask for centripetal, tangential and total acceleration.

Let's start by looking for centripetal acceleration, let's use the formula

          w² = w₀² + 2 α θ

          α = (w²- w₀²) / 2θ

we calculate

           α = (628.32²2 - 314.16²) / 2 75.398

           α = 1693.5 rad / s²

the quantity is linear and angular are related

the linear or tangential acceleration is

            a =    α  R

where R is the radius of the drum

            a = 1693.5 R

Unfortunately you do not give the radius of the drum for a complete calculation, but suppose it is a washing machine drum R = 20 cm = 0.20 m

           a = 1693.5 0.20

           a = 392.7 m / s²

the total acceleration is

           a_total = √(a² + α²)

           a_total = √ (α² R² + α²)

           a_total = α √(R² +1)

e) The centripetal acceleration is directed towards the center of the movement is radial and its magnitude is constant

Tangential acceleration is tangency to radius and its value varies proportionally radius

the total accelracicon is the result of the vector sum of the two accelerations and their directions given by trigonometry

            tan θ = a / α

the angular velocity increases linearly when with centripetal acceleration

Answer:

Explanation:

That Universe Consists of Matter

Answer:

a

   [tex]k = 11600000 N/m[/tex]

b

   [tex]\Delta L = 3.2323 *10^{-5} \ m[/tex]

c

  [tex]F = 3750.28 \ N[/tex]  

Explanation:

From the question we are told that

    The Young modulus is  [tex]E = 1.4 *10^{10} \ N/m^2[/tex]

     The length is  [tex]L = 0.35 \ m[/tex]

      The  area is  [tex]2.9 \ cm^2 = 2.9 *10^{-4} \ m ^2[/tex]

Generally the force acting on the tibia is mathematically represented as

       [tex]F = \frac{E * A * \Delta L }{L}[/tex]    derived from young modulus equation

Now this force can also be mathematically represented as

      [tex]F = k * \Delta L[/tex]    

So

     [tex]k = \frac{E * A }{L}[/tex]

substituting values

     [tex]k = \frac{1.4 *10^{10} * 2.9 *10^{-4} }{ 0.35}[/tex]

     [tex]k = 11600000 N/m[/tex]

    Since the tibia support half the weight then the force experienced by the tibia is  

        [tex]F_k = \frac{750 }{2} = 375 \ N[/tex]

 From the above equation the extension (compression) is mathematically represented as

          [tex]\Delta L = \frac{ F_k * L }{ A * E }[/tex]        

substituting values

           [tex]\Delta L = \frac{ 375 * 0.35 }{ (2.9 *10^{-4}) * 1.4*10^{10} }[/tex]

           [tex]\Delta L = 3.2323 *10^{-5} \ m[/tex]

From the above equation the maximum force is  

        [tex]F = \frac{1.4*10^{10} * (2.9*10^{-4}) * 3.233*10^{-5} }{ 0.35}[/tex]  

         [tex]F = 3750.28 \ N[/tex]  

Answer:

e. It is neither attracted nor repelled.

Explanation:

Electrostatic attraction or repulsion occurs between two or more charged particles or conductors. In this case, if the negatively charged ball is brought close to the neutral end A of the rod, there would be no attraction or repulsion between the rod end A and the negatively charged ball. This is because a charged particle or conductor has no attraction or repulsion to a neutral particle or conductor.

Answer:

In parallel

Explanation:

Ctotal = C1 + C2 + ... + Cn

Answer:

v = 6.28 m/s

Explanation:

It is given that,

A windmill on a farm rotates at a constant speed and completes one-half of a rotation in 0.5 seconds,

Number of revolution is half. It means angular velocity is 3.14 radians.

Let v is the angular speed. So,

[tex]v=\dfrac{\omega}{t}\\\\v=\dfrac{3.14}{0.5}\\\\v=6.28\ m/s[/tex]

So, the rotation speed is 6.28 m/s.

The angular velocity is the rotation speed, which is the angle of rotation

of the windmill per second, which is 2·π radians.

Response:

The given parameter are;

The angle of rotation the windmill rotates in 0.5 seconds = One-half a

rotation.

Required:

The rotational speed (angular velocity)

Solution:

The angle of one rotation = 2·π radians

Angle of one-half ration = [tex]\frac{1}{2}[/tex] × 2·π radians = π radians

[tex]Rotational \ speed = \mathbf{\dfrac{Angle \ of \ rotation}{Time}}[/tex]

Which gives;

[tex]Rotational \ speed, \omega = \dfrac{\pi}{0.5 \ s} = \mathbf{2 \cdot \pi \ rad/s}[/tex]

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Answer:

 t = 8.98 10⁻⁷ m

Explanation:

This is an exercise in interference by reflection, let's analyze what happens on each surface of the film.

* When the light ray shifts from a medium with a lower refractive index to a medium with a higher refractive index, the reflected ray has a reflection of 180

* The beam when passing to the middle its wavelength changes

           λ = λ₀ / n

if we take this into account, the constructive interference equation for normal incidence is

            2t = (m + ½) λ₀ / n

let's apply this equation to our case

for λ₀ = 479 nm = 479 10⁻⁹ m

             t = (m + ½) 479 10⁻⁹ / 1.33

             (m + ½) = 1.33 t / 479 10⁻⁹

for λ₀ = 798 nm = 798 10⁻⁹ m

             t = (m' + ½) 798 10⁻⁹ /1.33

            (m' + ½) = 1.33 t / 798 10⁻⁹

as they tell us that no other constructive interference occurs between the two wavelengths, the order of interference must be consecutive, let's write the two equat⁻ions

               (m + ½) = 1.33 t / 479 10⁻⁹

             ((m-1) + ½) = 1.33 t / 798 10⁻⁹

             (m + ½) = 1.33 t / 798 10⁻⁹ +1

resolve

             1.33 t / 479 10⁻⁹ = 1.33 t / 798 10⁻⁹ +1

             1.33 t / 479 10⁻⁹ = (1.33t + 798 10⁻⁹) / 798 10⁻⁹

             1.33t = (1 .33t + 798 10⁻⁹) 479/798

             1.33t = (1 .33t + 798 10⁻⁹) 0.6

             1.33 t = 0.7983 t + 477.6 10⁻⁹

             t (1.33 - 0.7983) = 477.6 10⁻⁹

             t = 477.6 10⁻⁹ /0.5315

             t = 8.98 10⁻⁷ m

Answer:

259 Hz or 269 Hz

Explanation:

Beat: This is the phenomenon obtained when two notes of nearly equal frequency are sounded together. The S.I unit of beat is Hertz (Hz).

From the question,

Beat = f₂-f₁................ Equation 1

Note: The frequency of the other instrument is either f₁ or f₂.

If the unknown instrument's frequency is f₁,

Then,

f₁ = f₂-beat............ equation 2

Given: f₂ = 264 Hz, Beat = 5 Hz

Substitute into equation 2

f₁ = 264-5

f₁ = 259 Hz.

But if the unknown frequency is f₂,

Then,

f₂ = f₁+Beat................. Equation 3

f₂ = 264+5

f₂ = 269 Hz.

Hence the beat could be 259 Hz or 269 Hz

Answer:

The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.

Explanation:

When a container is rigid, the process is supposed to be isochoric, that is, at constant volume. Then, the equation of state for ideal gases can be simplified into the following expression:

[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}[/tex]

Where:

[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Initial and final pressures, measured in pascals.

[tex]T_{1}[/tex], [tex]T_{2}[/tex] - Initial and final temperatures, measured in Kelvins.

In addtion, the specific heat at constant volume for monoatomic ideal gases, measured in joules per mole-Kelvin is given by:

[tex]\bar c_{v} = \frac{3}{2}\cdot R_{u}[/tex]

Where:

[tex]R_{u}[/tex] - Ideal gas constant, measured by pascal-cubic meters per mole-Kelvin.

If [tex]R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}[/tex], then:

[tex]\bar c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{2}}{mol\cdot K} \right)[/tex]

[tex]\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex]

And change in heat energy ([tex]Q[/tex]), measured by joules, by:

[tex]Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})[/tex]

Where:

[tex]n[/tex] - Molar quantity, measured in moles.

The final temperature of the monoatomic ideal gas is now cleared:

[tex]T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}[/tex]

Given that [tex]T_{1} = 300\,K[/tex], [tex]Q = 6000\,J[/tex], [tex]n = 4\,mol[/tex] and [tex]\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex], the final temperature is:

[tex]T_{2} = 300\,K + \frac{6000\,J}{(4\,mol)\cdot \left(12.471\,\frac{J}{mol\cdot K} \right)}[/tex]

[tex]T_{2} = 420.279\,K[/tex]

The final pressure of the system is calculated by the following relationship:

[tex]P_{2} = \left(\frac{T_{2}}{T_{1}}\right) \cdot P_{1}[/tex]

If [tex]T_{1} = 300\,K[/tex], [tex]T_{2} = 420.279\,K[/tex] and [tex]P_{1} = 6.00\times 10^{4}\,Pa[/tex], the final pressure is:

[tex]P_{2} = \left(\frac{420.279\,K}{300\,K} \right)\cdot (6.00\times 10^{4}\,Pa)[/tex]

[tex]P_{2} = 8.406\times 10^{4}\,Pa[/tex]

The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.

Answer: Ф = 17.2657 ≈ 17°

Explanation:

we simply apply ET =0 about the ending of the rod

so In.g.L/2sinФ - In.a.L/2cosФ = 0

g.sinФ - a.cosФ = 0

g.sinФ = a.cosФ

∴ tanФ = a/g

Ф =  tan⁻¹ a / g

Ф = tan⁻¹ ( 10 / 32.17405)

Ф = tan⁻¹ 0.31080948777

Ф = 17.2657 ≈ 17°

Therefore the angle of rotation of the rod during this acceleration is 17.2657 ≈ 17°

Answer:

At the moment contact is made with the battery, the voltage across the resistor is equal to the batteries terminal voltage

Explanation;

Because at series connection the battery and resistor have equal voltage

Answer:

The maximum wavelength of incident light that can produce photoelectrons from silver is 423.5 nm.

Explanation:

Given;

work function of silver, Φ = 2.93 eV = 2.93 x 1.602 x 10⁻¹⁹ J = 4.6939 x 10⁻¹⁹ J

Apply Einstein Photo electric effect;

E = K.E + Ф

Where;

E is the energy of the incident light

K.E is the kinetic of electron

Ф is the work function of silver surface

For the incident light to have maximum wavelength, the kinetic energy of the electron will be zero.

E = Ф

hf = Ф

[tex]h\frac{c}{\lambda} = \phi[/tex]

where;

c is speed of light = 3 x 10⁸ m/s

h is Planck's constant, = 6.626 x 10⁻³⁴ J/s

λ is the wavelength of the incident light

[tex]\lambda = \frac{hc}{\phi}\\\\\lambda =\frac{6.626*10^{-34} *3*10^8}{4.6939*10^{-19}} \\\\\lambda = 4.235 *10^{-7} \ m\\\\\lambda = 423.5 *10^{-9} \ m\\\\\lambda = 423.5 \ nm[/tex]

Therefore, the maximum wavelength of incident light that can produce photoelectrons from silver is 423.5 nm.

Answer:

The water pressure on the upper pipe is 92.5 kPa.

Explanation:

Given that,

Pressure in lower pipe= 120 kPa

Speed of water in lower pipe= 1 m/s

Acceleration due to gravity = 10 m/s²

Density of water = 1000 kg/m³

Radius of lower pipe = 12 m

Radius of uppes pipe = 6 m

Height of upper pipe = 2 m

We need to calculate the velocity in upper pipe

Using continuity equation

[tex]A_{1}v_{1}=A_{2}v_{1}[/tex]

[tex]\pi r_{1}^2\times v_{1}=\pi r_{2}^2\times v_{2}[/tex]

[tex]v_{2}=\dfrac{r_{1}^2\times v_{1}}{r_{2}^2}[/tex]

Put the value into the formula

[tex]v_{2}=\dfrac{12^2\times1}{6^2}[/tex]

[tex]v_{2}=4\ m/s[/tex]

We need to calculate the water pressure on the upper pipe

Using bernoulli equation

[tex]P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}[/tex]

Put the value into the formula

[tex]120\times10^{3}+\dfrac{1}{2}\times1000\times1^2+1000\times10\times0=P_{2}+\dfrac{1}{2}\times1000\times(4)^2+1000\times10\times2[/tex]

[tex]120500=P_{2}+28000[/tex]

[tex]P_{2}=120500-28000[/tex]

[tex]P_{2}=92500\ Pa[/tex]

[tex]P_{2}=92.5\ kPa[/tex]

Hence, The water pressure on the upper pipe is 92.5 kPa.