Figure (3) shows a car travelling along the route PQRST in 30 minutes. What is the average speed of the car in km/hour?

Figure (3) Shows A Car Travelling Along The Route PQRST In 30 Minutes. What Is The Average Speed Of The

Answers

Answer 1

Answer:

60 km/hour.

Explanation:

We'll begin by calculating the total distance traveled by the car. This is illustrated below:

Total distance traveled = sum of distance between PQRST

Total distance = 10 + 5 + 10 + 5

Total distance = 30 km

Next, we shall convert 30 mins to hour. This can obtained as follow:

Recall:

60 mins = 1 hour

Therefore,

30 mins = 30/60 = 0.5 hour.

Finally, we shall determine the average speed of the car as follow:

Distance = 30 km

Time = 0.5 hour

Speed =?

Speed = distance /time

Speed = 30/0.5

Speed = 60 km/hour

Therefore, the speed of the car is 60 km/hour.


Related Questions

Which is a “big idea” for space and time? Energy can be transferred but not destroyed. Forces describe the motion of the universe. The universe is very big and very old. The universe consists of matter.

Answers

Answer:

Explanation:

That Universe Consists of Matter

Two instruments produce a beat frequency of 5 Hz. If one has a frequency of 264 Hz, what could be the frequency of the other instrument

Answers

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

In the lab , you have an electric field with a strength of 1,860 N/C. If the force on a particle with an unknown charge is 0.02796 N, what is the value of the charge on this particle.

Answers

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]

Two hoops, staring from rest, roll down identical incline planes. The work done by nonconservative forces is zero. The hoops have the same mass, but the larger hoop has twice the radius. Which hoop will have the greater total kinetic energy at the bottom

Answers

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.

Air bags greatly reduces the chance og injury in a car accident.explain how they do si in terms of energy transfer

Answers

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.

Physical properties of a mineral are a result of the arrangement of the atoms in the mineral. Use this fact to explain the following:_________
A. One mineral has a density of 2.7 g/ml while another has a density of 5.1 g/ml
B. The mineral mica cleaves into thin flat sheets while olivine does not show cleavage

Answers

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.

Object A, with heat capacity CA and initially at temperature TA, is placed in thermal contact with object B, with heat capacity CB and initially at temperature TB. The combination is thermally isolated. If the heat capacities are independent of the temperature and no phase changes occur, the final temperature of both objects is

Answers

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;

heat capacity of object A = CAinitial temperature of object A = TAheat capacity of object B = CBinitial temperature of object B = TB

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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A block of mass M rests on a block of mass M1 which is on a tabletop. A light string passes over a frictionless peg and connects the blocks. The coefficient of kinetic friction between the blocks and between M1 and the tabletop is the same. A force F pulls the upper block to the left and the lower block to the right. The blocks are moving at a constant speed.
Determine the mass of the upper block. (Express your answer to three significant figures.)

Answers

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

At what temperature (degrees Fahrenheit) is the Fahrenheit scale reading equal to:_____
(a) 3 times that of the Celsius and
(b) 1/5 times that of the Celsius

Answers

Answer:

C = 26.67° and F = 80°C = -20° and F = -4°

Explanation:

Find:

3 times that of the Celsius and 1/5 times that of the Celsius

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°

A resistor and a capacitor are connected in series to an ideal battery of constant terminal voltage. At the moment contact is made with the battery, the voltage across the resistor is

Answers

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

Water is pumped with a 120 kPa compressor entering the lower pipe (1) and flows upward at a speed of 1 m/s. Acceleration due to gravity is 10 m/s and water density is1000 kg/m-3. What is the water pressure on the upper pipe (II).

Answers

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.

"A soap film is illuminated by white light normal to its surface. The index of refraction of the film is 1.33. Wavelengths of 479 nm and 798 nm and no wavelengths between are intensified in the reflected beam. The thickness of the film is"

Answers

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

on which principle does water pump work ?​

Answers

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!

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

Answers

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.

Conservation of momentum:

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.

Learn more about conservation of momentum:

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A 5.2 cm diameter circular loop of wire is in a 1.35 T magnetic field. The loop is removed from the field in 0.29 sec. Assume that the loop is perpendicular to the magnetic field. What is the average induced emf?

Answers

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]

3. What are the first steps that you should take if you are unable to get onto the Internet? (1 point)
O Check your router connections then restart your router.
O Plug the CPU to a power source and reboot the computer.
O Adjust the display properties and check the resolution.
Use the Control Panel to adjust the router settings.​

Answers

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.

This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating strings. Assume that the rod is initially electrically neutral. For convenience we will refer to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem, "strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two charged balls.A. A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?
What happens to end A of the rod when the ball approaches it closely this first time?a. It is strongly repelled.b. It is strongly attracted.c. It is weakly attracted.d. It is weakly repelled.e. It is neither attracted nor repelled.

Answers

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.

A rigid container holds 4.00 mol of a monatomic ideal gas that has temperature 300 K. The initial pressure of the gas is 6.00 * 104 Pa. What is the pressure after 6000 J of heat energy is added to the gas?

Answers

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.

Three resistors, each having a resistance, R, are connected in parallel to a 1.50 V battery. If the resistors dissipate a total power of 3.00 W, what is the value of R

Answers

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

A windmill on a farm rotates at a constant speed and completes one-half of a rotation in 0.5 seconds. What is its rotation speed

Answers

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 rotation speed is 2·π rad/s

How can the rotational speed of the windmill be calculated?

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]

The rotation speed is 2·π rad/s

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What is the maximum wavelength of incident light that can produce photoelectrons from silver? The work function for silver is Φ=2.93 eV.

Answers

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.

The tibia is a lower leg bone (shin bone) in a human. The maximum strain that the tibia can experience before fracturing corresponds to a 1 % change in length.
A. Young's modulus for bone is about Y = 1.4 x 10 N/m². The tibia (shin bone) of a human is 0.35 m long and has an average cross-sectional area of 2.9 cm. What is the effective spring constant of the tibia?
B. If a man weighs 750 N, how much is the tibia compressed if it supports half his weight?
C. What is the maximum force that can be applied to a tibia with a cross-sectional area, A = 2.90 cm?

Answers

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]  

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

Answers

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]

What would happen in a State if its citizens lack relevant knowledge, skills
and positive attitude?​

Answers

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

If we compare the force of gravity to strong nuclear force, we could conclude that
O gravity is the weaker force; it is related to mass
O gravity is the stronger force; it is related to distance
strong nuclear is the stronger force; it is related to mass
O strong nuclear is the weaker force; it is related to distance

Answers

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

What are nuclear forces?

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.

Learn more about nuclear forces here

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The equivalent resistance of two resistors connected in series is always greater than the equivalent resistance of the same two resistors connected in parallel. True False

Answers

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

An LR circuit consists of a 35-mH inductor, ac resistance of 12 ohms, an 18-V battery, and a switch. What is the current 5.0 ms after the switch is closed

Answers

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

A high school physics student claims her muscle car can achieve a constant acceleration of 10 ft/s/s. Her friend develops an accelerometer to confirm the feat. The accelerometer consists of a 1 ft long rod (mass=4 kg) with one end attached to the ceiling of the car, but free to rotate. During acceleration, the rod rotates. What will be the angle of rotation of the rod during this acceleration? Assume the road is flat and straight.

Answers

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°

You have two capacitors and want to connect them across a voltage source (battery) to store the maximum amount of energy. Should they be connected in series or in parallel?

Answers

Answer:

In parallel

Explanation:

Ctotal = C1 + C2 + ... + Cn

3. El tambor de una lavadora que gira a 3 000 revoluciones por minuto (rpm) se acelera uniformemente hasta que alcanza las 6 000 rpm, completando un total de 12 revoluciones.

d. Determina la aceleración tangencial, centrípeta y la total en m.s-2 cuando el tambor a alcanzado los 60000 rpm

e. Explica lo que ocurre con la magnitud y dirección de los vectores aceleración tangencial, aceleración centrípeta, aceleración total, aceleración angular, velocidad angular cuando la lavadora ha girado desde 3000 rpm hasta 6000 rpm.

Answers

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

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