The resolving power of a microscope is proportional to the wavelength used. A resolution of 1.0 10-11 m (0.010 nm) would be required in order to "see" an atom. (a) If electrons were used (electron microscope), what minimum kinetic energy would be required for the electrons

Answers

Answer 1

Answer:

K = 13448.64eV

Explanation:

(a) In order to calculate the kinetic energy of the electrons, to "see" the atom, you take into account that the wavelength of the electrons must be of the order of the resolution required (0.010nm).

Then, you first calculate, by using the Broglies' relation, the momentum of the electron associated to a wavelength of 0.010nm:

\(p=\frac{h}{\lambda}\)       (1)

p: momentum of the electron

h: Planck's constant = 6.626*10^-34 Js

λ: wavelength = 0.010nm

You replace the values of the parameters in the equation (1):

\(p=\frac{6.262*10^{-34}Js}{0.010*10^{-9}m}=6.262*10^{-23}kg\frac{m}{s}\)

With this values of the momentum of the electron you can calculate the kinetic energy of the electron by using the following formula:

\(K=\frac{p^2}{2m}\)    (2)

m: mass of the electron = 9.1*10^-31 kg

\(K=\frac{(6.262*10^{-23}kgm/s)^2}{2(9.1*10^{-31}kg)}=2.15*10^{-15 }J\)

In electron volts you obtain:

\(2.15*10^{-15}J*\frac{6.242*10^{18}eV}{1J}=13448.64eV\)

The kinetic energy required for the electrons must be, at least, of 13448.64 eV


Related Questions

Predict the magnitude of force applied on you if you push against a tree with a force of 50 and directed to the

Answers

Answer: -50N

Explanation:

Newton's 3rd law says that for every force applied, there is an opposite force with equal magnitude. The person pushing on the tree will experience a force of -50N

On a sunny spring day, Mr. Hane’s family went to a zoo in his town. His daughter looked at an elephant, and she found that the mass of the elephant is 2019 kg. What is the weight of the elephant in Newtons? Use the acceleration due to gravity (g) as 9.8 meters per second squared.

Answers

The weight of the elephant is approximately 19,780.2 Newtons, calculated by multiplying its mass (2019 kg) by the acceleration due to gravity (9.8 m/s²).

To calculate the weight of the elephant in Newtons, we need to use the formula:

Weight = Mass x Acceleration due to gravity

Given that the mass of the elephant is 2019 kg and the acceleration due to gravity is 9.8 m/s^2, we can substitute these values into the formula:

Weight = 2019 kg x 9.8 m/s^2

By multiplying these values, we find that the weight of the elephant is approximately 19,780.2 Newtons.

The weight of an object is the force with which it is attracted towards the center of the Earth due to gravity. The unit of weight is Newtons, and it represents the force required to support the object against gravity. In this case, the weight of the elephant tells us the force exerted by the Earth on the elephant, and it is approximately 19,780.2 Newtons.

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a horizontal 4-cm-diameter water jet with a velocity of 18 m/s. He impinges the jet normally upon a vertical plate of mass 750 kg. The plate rides on a nearly frictionless track and is initially stationary. When the jet strikes the plate, the plate begins to move in the direction of the jet. The water always splatters in the plane of the retreating plate. Determine (a)the acceleration of the plate when the jet first strikes it (time

Answers

Answer:

0.5429 m/s^2

Explanation:

velocity of waterjet = 18 m/s

diameter of water jet ( d ) = 4 cm = 0.04 m

mass of vertical plate(m) = 750 kg

Determine the acceleration of plate when the jet first strikes ( i.e. t = 0 )

first we will determine the impact force

F = β*A*V^2 ----- ( 1 )

where ; β = 1000 kg/m^3  ,  A = π/4 * d^2 , V = 18 m/s

input values into equation 1

F = 407.15 N

finally determine the acceleration at t = 0

F = m*a

a = F / m =  407.15 / 750 = 0.5429 m/s^2

7. A particle of mass 3 kg is held in equilibrium by two light unextensible strings. One string is horizontal, as shown in Figure 7.30. The tension in the horizontal string is PN and the tension in the other string is N. Find a) the value of 0 b) the value of P.​

Answers

The tension in the strings are 31.47 and 19.25 N respectively.

Mass of the block, m = 3 kg

From the figure, consider the vertical components,

T₁ sin45° + T₂ sin30° = mg

(T₁/√2) + (T₂/2) = 3 x 9.8 = 29.4

Also, consider the horizontal components,

T₁ cos45° = T₂ cos30°

T₁/√2 = T₂ x√3/2

T₁ = T₂ x √3/2 x √2

So,

T₁ = 0.612T₂

Applying in the first equation,

(T₁/√2) + (T₂/2) = 29.4

(0.612T₂/1.414) + 0.5T₂ = 29.4

0.434 T₂ + 0.5 T₂ = 29.4

0.934 T₂ = 29.4

Therefore, the tension,

T₂ = 29.4/0.934

T₂ = 31.47 N

So, the tension,

T₁ = 0.612 T₂

T₁ = 0.612 x 31.47

T₁ = 19.25 N

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A roller coaster car is released from rest as shown in the image below. If
friction is neglected, the car will oscillate back and forth across the "dip" in
the roller coaster. What is the approximate velocity of the roller coaster car
each time it reaches the bottom of the roller coaster in the image? (Recall
that g = 9.8 m/s2.)
TAS
81 m
O A. 40 m/s
B. 25 m/s
C. 30 m/s
D. 15 m/s

A roller coaster car is released from rest as shown in the image below. Iffriction is neglected, the

Answers

Answer:

40m/s

Explanation:

a=g

u=0

s=81

v²=u²+2as

v²=2(9.81)(81)

v=√1587.6=39.8446985181≈40m/s

The velocity of the roller coaster car each time it reaches the bottom is 40 ms⁻¹. The correct option is (A).

The rate at which the position of an object changes with respect to time is described by the physical quantity known as velocity. It has both magnitude and direction because it is a vector quantity.

Given:

Initial velocity, u = 0 m/s

Acceleration, a = -9.8 ms⁻²

Distance, d = 81 m

From the third equation of motion:

v² = u² - 2as

v² = 0 - 2×(-9.8)×81

v = 40 ms⁻¹

Hence, the velocity of the roller coaster car is 40 ms⁻¹. The correct option is (A).

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Consider a double-paned window consisting of two panes of glass, each with a thickness of 0.500 cm and an area of 0.760 m2 , separated by a layer of air with a thickness of 1.65 cm . The temperature on one side of the window is 0.00 ∘C; the temperature on the other side is 23.0 ∘C. In addition, note that the thermal conductivity of glass is roughly 36 times greater than that of air. Approximate the heat transfer through this window by ignoring the glass. That is, calculate the heat flow per second through 1.65 cm of air with a temperature difference of 23.0 ∘C . (The exact result for the complete window is 24.4 J/s .)

Answers

The approximate heat transfer through 1.65 cm of air with a temperature difference of 23.0 °C is approximately 24.4 J/s.

To approximate the heat transfer through the air layer in the double-paned window, we can assume that the glass layers have a negligible impact on the heat flow. The heat transfer can be calculated using Fourier's Law of Heat Conduction, which states that the heat flow (Q) is proportional to the temperature difference (ΔT) and inversely proportional to the thickness (L) and thermal conductivity (k) of the material.

First, we need to calculate the effective thermal conductivity of the air layer due to its thickness and the thermal conductivity ratio between air and glass. Let's denote the thermal conductivity of air as k_air and the thermal conductivity of glass as k_glass. Since glass has a thermal conductivity roughly 36 times greater than air, we have k_glass = 36 * k_air.

Next, we calculate the effective thermal conductivity of the air layer as:

k_eff = (k_air * L_air) / (L_air + k_glass)

Substituting the given values, we have:

k_eff = (k_air * 0.0165 m) / (0.0165 m + 0.005 m) = 0.01309 * k_air

Now, we can calculate the heat flow per second through the air layer using the formula:

Q = (k_eff * A * ΔT) / L_air

Substituting the given values, we get:

Q = (0.01309 * k_air * 0.760 m^2 * 23.0 K) / 0.0165 m = 24.4 J/s

Therefore, the approximate heat transfer through 1.65 cm of air with a temperature difference of 23.0 °C is approximately 24.4 J/s.

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(20) A rocket is launched vertically. At time t = 0 seconds, the rocket’s engine shuts down. At the time, the rocket has reached an altitude of 500m and is rising at a velocity of 125 m/s. Gravity then takes over. The height of the rocket as a function of time is h(t)=-9.8/2 t^2+125t+500,t>0. Using your function file from HW2A: Generate a plot of height (vertical axis) vs. time (horizontal axis) from 0 to 30 seconds. Include proper axis labels. Find the maximum height and the time at which it occurs: Analytically, showing your steps and equations. This part should be done entirely in the write-up: no coding Using the data cursor on the plot. Using the MAX function on your data from part (a) Using FMINSEARCH on your m file Comment on the differences between the methods. How closely does each method match the "true" (analytical) value? Find the time when the rocket hits the ground: Analytically, showing your equations. This part should be done entirely in the write-up: no coding Using the data cursor on the plot. Using FZERO on your m file Comment on the differences between the methods in each of part (B) and (C). How closely does each method match the "true" (analytical) value? Use a quantitative comparison to make your argument.

Answers

Answer:

Explanation:

Given that,

h(t) = -9.8t² / 2 + 125t + 500

for t > 0.

At t = 0, the rocket is at height h = 500m, at a velocity of Vo = 125m/s.

We want to find the maximum height reached by rocket

Using mathematics maxima and minima

let find the turning point when dh/dt = 0

dh/dt = -9.8t + 125

dh / dt = 0 = -9.8t + 125

9.8t = 125

t = 125 / 9.8

t = 12.76s

Let find the turning point to know if this time t = 12.76 is maximum or minimum point

Let find d²h / dt²

d²h / dt² = -9.8

Since, d²h/dt² < 0, then, at t = 12.76s is the maximum points.

Then, the maximum height reached is

h =  -9.8t² / 2 + 125t + 500

h =  -9.8(12.76)² / 2 + 125(12.76) + 500

h = -797.80 + 1595 + 500

h = 1297.2 m

The maximum height reached is 1297.2 m

From the attachment, the maximum height is 1297.2m at t = 12.76sec.

Comment, the result are the same for both the analysis aspect and the graphical aspect.

(20) A rocket is launched vertically. At time t = 0 seconds, the rockets engine shuts down. At the time,

What is the term used when a ball is hit and the batter reaches the following bases safely (without being called out)?
please help me
a) first base
b)second base
c)third base
d) homerun

Answers

First base is the answer
First base would be the term

Brainlist!! Help!! The nucleus contains protons and neutrons. Being positively charged, the protons repel each other. The nucleus should fly apart due to the repulsive force. Yet, the nuclei of most atoms are stable - explain.

Answers

Answer:

The reason that the nucleus of most atoms does not fall apart despite the oppositely charged protons exerting a repulsive force on each other is the strong nuclear force.

What is the strong nuclear force?

The strong nuclear force is one of the fundamental forces in nature that acts between protons and neutrons in the atomic nucleus. It is a short-range force that is much stronger than the electromagnetic force (which produces the repulsion between protons).

The strong nuclear force is responsible for holding the nucleus together.

Additionally, the ratio of protons to neutrons in a nucleus also affects its stability. Therefore, if there is an imbalance in this ratio, the repulsive force between the protons can become too strong, causing the nucleus to become unstable and undergo radioactive decay.

Overall, the nucleus remains stable due to the balance between the strong nuclear force and the repulsive force between the protons.

If the coefficient of kinetic friction between a crate and the floor is 0.20,
how much force is needed to slide a 92 kg crate uniformly across the floor?

Answers

Answer:

0.3 newtons

Explanation:

If the coefficient of kinetic friction between a crate and the floor is 0.20, then the fore needed to slide a 92 kg crate uniformly across the floor would be 180.504 N

What is friction?

Friction is a type of force that resists or prevents the relative motion of two physical objects when their surfaces come in contact.

The friction force prevents any two surfaces of objects from easily sliding over each other or slipping across one another. It depends upon the force applied to the object.

As given in the problem If the coefficient of kinetic friction between a crate and the floor is 0.20

The force needed = μN

where μ is the coefficient of the kinetic friction

μ =0.2

N is the normal force

N = mg

N = 92×9.81 N

Force needed to slide = 0.20× 92×9.81

                                     =180.504 N

Thus, The force needed to slide a 92 kg crate uniformly across the floor would be 180.504 N

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A cylindrical rod formed from silicon is 16.8 cm long and has a mass of 2.17 kg. The density of silicon is 2.33 g/cm3 . What is the diameter of the cylinder

Answers

Answer:

We are given the length, the mass and the density of the cylinder. First let us calculate for the volume by dividing the mass by the density.

volume = mass /density

where mass = 2.17 kg = 2170 g, therefore:

volume = 2170 g / (2.33 g/cm^3)

volume = 931.33 cm^3

 

We know that the volume of a cylinder has the formula:

volume = π r^2 h

since h = 16.8 cm, therefore calculating for radius:

931.33 cm^3 = π r^2 (16.8 cm)

r^2 = 17.646 cm^2

r = 4.2 cm

 

Hence the diameter (d) is:

d = 2 r

d = 8.4 cm

Explanation:

The diameter of the cylindrical rod is approximately 0.382 cm.

To find the diameter of the cylindrical rod, we can use the formula for the volume of a cylinder and then solve for the diameter.

The formula for the volume of a cylinder is:

V = π\(r^{2}\)h,

where V is the volume, r is the radius, and h is the height (or length) of the cylinder.

In this case, we know the length of the cylinder (h) is 16.8 cm. We need to find the radius (r) in order to calculate the diameter.

The mass of the cylinder can be related to its volume and density using the formula:

m = ρV,

where m is the mass, ρ is the density, and V is the volume.

Rearranging this formula, we can solve for V:

V = m / ρ.

Now we have two equations:

V = π\(r^{2}\)h,

V = m / ρ.

Setting these two equations equal to each other, we can solve for r:

π\(r^{2}\)h = m / ρ.

Substituting the given values:

π\(r^{2}\) * 16.8 cm = 2.17 kg / (2.33 g/\(cm^3\)).

Let's solve this equation for r:

\(r^{2}\) = (2.17 kg / (2.33 g/\(cm^3\))) / (π * 16.8 cm).

\(r^{2}\)  ≈ 0.036775 \(cm^2\).

Taking the square root of both sides:

r ≈ 0.191 cm.

Finally, we can find the diameter (d) by multiplying the radius by 2:

d ≈ 2 * 0.191 cm.

d ≈ 0.382 cm.

Therefore, the diameter of the cylindrical rod is approximately 0.382 cm.

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find the volume of the tire with dimensions
● Tire diameter: 26”
● Tube width diameter: 2.125”

Answers

The volume of the tire at the given diameter and thickness of tube is determined as  1,128.2 cubic inch.

What is volume?

Volume is a scalar quantity expressing the amount of three-dimensional space enclosed by a closed surface.

Volume of the tire

The volume of the tire is the measure of the product of area and thickness of the tire.

The volume of the tire is calculated as follows;

Radius of the tire = 0.5 x 26" = 13"

Volume of the tire = Area x thickness

Volume of the tire = πr² x h

where;

r is the radius of the tireh is the thickness of the tube

Volume of the tire = π(13)² x (2.125)

Volume of the tire =  1,128.2 cubic inch

Thus, the volume of the tire at the given diameter and thickness of tube is determined as  1,128.2 cubic inch.

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gap region between the diver's skin and her
wetsuit, forming a water layer about 0.5 mm thick.
Assume that the total surface area of the wetsuit
covering the diver is about 1.0 mº, and that the
water enters the suit at 10 °C and is warmed by
the diver to skin temperature of 35°C. The specific
heat of water is 1.00 kcal/kg - Cº.
Estimate how much energy (in
units of candy bars = 300 kcal)
is required by this heating
process.
Express your answer using
two significant figures.

Answers

The energy required by the heating process is 0.042 candy bars.

We'll begin by calculating the mass of the water.

Thickness = 0.5 mm = 0.5 / 1000 = 5×10¯⁴ mArea = 1 m²Volume = Area × Thickness = 1 × 5×10¯⁴ = 5×10¯⁴ m³Density of water = 1000 Kg/m³Mass of water =?

Mass = Density × Volume

Mass of water = 1000 × 5×10¯⁴

Mass of water = 0.5 Kg

Next, we shall determine the heat required

Mass of water (M) = 0.5 KgInitial temperature of water (T₁) = 10 °CFinal temperature (T₂) = 35 °CChange in temperature (ΔT) = T₂ – T₁ = 35 – 10 = 25 °CSpecific heat capacity of water (C) = 1 Kcal/KgºCHeat (Q) =?

Q = MCΔT

Q = 0.5 × 1 × 25

Q = 12.5 Kcal

Finally, we shall convert 12.5 Kcal to candy bar.

300 Kcal = 1 candy bar

Therefore,

12.5 Kcal = 12.5/300

12.5 Kcal = 0.042 candy bars

Therefore, the energy required in candy bar is 0.042 candy bars

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1. On each of your equipotential maps, draw some electric field lines with arrow heads indicating the direction of the field. (Hint: At what angle do field lines intersect equipotential lines?) Draw sufficient field lines that you can "see" the electric field.

Answers

Answer:

The angle between the electric field lines and the equipotential surface is 90 degree.

Explanation:

The equipotential surfaces are the surface on which the electric potential is same. The work done in moving a charge from one point to another on an equipotential surface is always zero.

The electric field lines are always perpendicular to the equipotential surface.

As

\(dV = \overrightarrow{E} . d\overrightarrow{r}\\\\\)

For equipotential surface, dV = 0 so

\(0 = \overrightarrow{E} . d\overrightarrow{r}\\\\\)

The dot product of two non zero vectors is zero, if they are perpendicular to each other.


Identify three main ideas about models

Answers

Answer:

Approaches mathematical learning through inquiry

-Explore real contexts, problems, situations, and models

-Learning through doing shifts the focus on the students

-Problems have multiple entry and exit points

-Links to other disciplines

Explanation:

quizlet

Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are tripled and their separation is also tripled, what happens to the magnitude of the electrostatic force that each charge exerts on the other

Answers

Answer:

The force remains the same.

Explanation:

Let the magnitude of the forces in each case be F1 and F2 respectively.

The charges are Q1 and Q2.

The distance of separation between them is R.

Hence, for F1;

F1 = KQ1Q2/R^2

For F2:

F2 = K * 3Q1 * 3Q2/(3R)^2

F2 = 9KQ1Q2/9R^2

F2 =KQ1Q2/R^2

Hence F1=F2

The force is the same in both cases!

Which tells us speed and direction?


1).speed


2).displacement


3).distance


4).velocity

Answers

Answer:

velocity

Explanation:

Answer:

Velocity

Velocity tells you the speed and the direction it's going. Speed only tells you the speed, with no direction.

Explanation:

You are driving your car at 35 mph .Your car has 55.7 cm radius tires . What is
the angular speed of the tires?

Answers

The angular speed of the car's tires is approximately 28.07 rad/s. To find the angular speed of the car's tires, we need to convert the linear speed (35 mph) to angular speed.

The linear speed of the car is given in miles per hour (mph), but we need to convert it to meters per second (m/s) for consistent units.

1 mile is approximately equal to 1.60934 kilometers, and 1 kilometer is equal to 1000 meters. Therefore, we can convert 35 miles per hour to meters per second as follows:

35 mph = \(35 * 1.60934 km/h * 1000 m/km / 3600 s/h\)

       ≈ 15.65 m/s

Now that we have the linear speed of the car (15.65 m/s), we can calculate the angular speed (ω) using the formula:

v = ω * r

where:

v is the linear speed,

ω is the angular speed, and

r is the radius of the tire.

Given that the radius of the tires (r) is 55.7 cm, we need to convert it to meters:

r = 55.7 cm * 0.01 m/cm

  = 0.557 m

Substituting the values into the formula, we have:

15.65 m/s = ω * 0.557 m

Solving for ω, we find:

ω = 15.65 m/s / 0.557 m

  ≈ 28.07 rad/s

Therefore, the angular speed of the car's tires is approximately 28.07 rad/s.

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A motorcycle, which has an initial linear speed of 5.0 m/s, decelerates to a speed of 3.5 m/s in 4.5 s. Each wheel has a radius of 0.60 m and is rotating in a counterclockwise (positive) directions. What is (a) the constant angular acceleration (in rad/s2) and (b) the angular displacement (in rad) of each wheel?

Answers

Explanation:

Given that,

A motorcycle, which has an initial linear speed of 5.0 m/s, decelerates to a speed of 3.5 m/s in 4.5 s. Each wheel has a radius of 0.60 m and is rotating in a counterclockwise (positive) directions.

Angular acceleration, \(\alpha =\dfrac{\omega_2-\omega_1}{t}\)

\(\alpha =\dfrac{\dfrac{v_2}{r}-\dfrac{v_1}{r}}{t}\)

Put all the values,

\(\alpha =\dfrac{\dfrac{3.5}{0.6}-\dfrac{5}{0.6}}{4.5}\\\\=-0.56\ rad/s^2\)

Angular displacement,

\(\theta=\dfrac{\omega_2^2-\omega_1^2}{2\alpha}\\\\\theta=\dfrac{(\dfrac{v_2}{r})^2-(\dfrac{v_1}{r})^2}{2\alpha}\\\\\theta=\dfrac{(\dfrac{3.5}{0.6})^2-(\dfrac{5}{0.6})^2}{2\times (-0.56)}\\\\=31.62\ rad\)

Hence

Answer:

(a) The angular acceleration is - 0.56 rad/s^2.

(b) The angular displacement is 31.6 rad.

Explanation:

initial velocity, u = 5 m/s

final velocity, v = 3.5 m/s

radius, r = 0.6 m

time, t = 4.5 s

initial angular velocity, wo = u/r = 5/0.6 = 8.33 rad/s

final angular velocity, w = v/r = 3.5 / 0.6 = 5.83 rad/s

(a) Use the first equation of motion to fine the angular acceleration.

\(w = w_o + \alpha t \\\\5.83 = 8.33 + \alpha \times 4.5\\\\\alpha = - 0.56 rad/s^2\)

(b) Use third equation of motion to find the angular displacement

\(w^2 = w_0^2 + 2\alpha \theta \\\\5.83^2 =8.833^2 - 2 \times 0.56\times \theta \\\\\theta =31.6 rad\)

If the frequency of a wave increases, the wavelength will
O decrease
O increase
O disappear
O remain unchanged

Answers

Answer:

the wavelength will decrease

Explanation:

If the frequency of a wave increases, the wavelength will decrease. This is because the speed of the wave is constant for a given medium, so if the frequency (the number of waves passing a fixed point per second) increases, then the distance between successive wave crests (i.e., the wavelength) must decrease to maintain a constant speed. This relationship is described by the wave equation:

v = f λ

where v is the speed of the wave, f is the frequency, and λ is the wavelength. If v is constant and f increases, then λ must decrease to keep the equation balanced.

A man on a road trip drives a car at different constant speeds over several legs of the trip. He drives for 55.0 min at 60.0 km/h, 18.0 min at 80.0 km/h, and 60.0 min at 60.0 km/h and spends
25.0 min eating lunch and buying gas.

Answers

Complete Question

A man on a road trip drives a car at different constant speeds over several legs of the trip. He drives for 55.0 min at 60.0 km/h, 18.0 min at 80.0 km/h, and 60.0 min at 60.0 km/h and spends

25.0 min eating lunch and buying gas

What is the total distance traveled over the entire trip (in km)

Answer:

The value is \(D = 139.02 \ km\)

Explanation:

From the question we are told that

For \( t_1 = 55 min = \frac{55}{60} = 0.917 \ h\) the speed is \(v_1 = 60 \ km/h\)

For \( t_2 = 18 min = \frac{18}{60} = 0.3 \ h\) the speed is \(v_2 = 80 \ km/h\)

For \( t_3 = 60 min = \frac{60}{60} = 1 \ h\) the speed is \(v_3 = 60\ km/h\)

The time taken to have lunch is \(t _l = 25 \ min = \frac{25}{60} = 0.42 \ h\)

Generally the total distance traveled over the entire trip (in km) is mathematically represented as

\(D = t_1 * v_1 + t_2 * v_2 + t_3 * v_3\)

=>   \(D = 0.917  *  60 +  0.3 * 80 +  1  * 60\)

=>   \(D = 55.02  + 24 +   60\)

=>   \(D = 139.02 \ km\)

At this rate, how long does it take to accelerate from 70 km/h to 110 km/h ?

Answers

Answer:

\({ \tt{acceleration = \frac{ \triangle \: v}{t} }} \\ \\ { \tt{a = \frac{110 - 70}{t} }} \\ \\ { \tt{t = \frac{40}{a} } \: s}\)

For a body under free fall; a = g = 9.8

\({ \tt{t = \frac{40}{9.8} }} \\ \\{ \tt{t = 4.08 \: s}}\)

The question is in the pictureplease answer problem 2Thank you

The question is in the pictureplease answer problem 2Thank you

Answers

We know that the average velocity for a moving object can be written as its displacement divided by the time of the movement. This can be formalized as:

\(v=\frac{\Delta s}{\Delta t}\)

In our case, as the velocity is constant, this will not only be an average velocity, but the exact value. Replace our data we get:

\(v=\frac{10}{5}=2\frac{m}{s}\)

Thus, v=2m/s

A hiker walks from
(x1, y1) = (−3.60 km, 3.25 km) to (x2, y2) = (3.05 km, 5.90 km). (Take east to lie along the positive x-axis and north to lie along the positive y-axis.)

(a) What distance has he traveled? (Assume the hiker walked in a straight line.)
(b) The hiker desires to return to his starting point. In what direction should he go? (Give the angle with respect to due east. Enter an angle between 0° and 360°.)

Answers

(a) The distance travelled by the hiker is 7.16 km

(b) If the hiker desires to return to his starting point, he should move at 201.71° with respect to east.

Form a right angled triangle with the distance between initial and final points as hypotenuse. The adjacent side will be the total distance travelled on x-direction and the opposite side will be the total distance travelled in y-direction.

Adjacent side = \(x_{2}\) - \(x_{1}\) = 3.05 - ( - 3.6 )

Adjacent side = 6.65 km

Opposite side = \(y_{2}\) - \(y_{1}\) = 5.9 - 3.25

Opposite side = 2.65 km

According to Pythagoras theorem,

\(Hypotenuse^{2}\) = \(Opposite side^{2}\) + \(Adjacent side^{2}\)

\(Hypotenuse^{2}\) = \(2.65^{2}\) + \(6.65^{2}\)

Hypotenuse = √ ( 7.02 + 44.22 )

Hypotenuse = 7.16 km

sin θ = Opposite side / Hypotenuse

sin θ = 2.65 / 7.16

θ = \(sin^{-1}\) ( 0.37 )

θ = 21.71°

This is the angle in which the hiker travelled from initial to final direction. The angle in which the hiker should travel to return to initial point with respect to east is

θ = 21.71° + 180°

θ = 201.71°

Therefore,

(a) The distance travelled by the hiker is 7.16 km

(b) If the hiker desires to return to his starting point, he should move at 201.71° with respect to east.

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What is the change in weight of a hollow cylinder of height 8 cm and radius 3 cm when the air is pumped out of it

Answers

This question involves the concepts of density, weight, and volume.

The change in weight of the cylinder will be "2.72 x 10⁻³ N".

The change in weight of the cylinder will be equal to the weight of the air inside the cylinder which is being removed from it:

\(\Delta W = W_a\)

where,

\(\Delta W\) = change in weight of cylinder = ?\(W_a\) = Weight of air inside cylinder = \(\rho Vg\) \(\rho\) = density of air = 1.225 kg/m³V = Volume of air inside the cylinder = πr²h = π(0.03 m)²(0.08 m) = 2.26 x 10⁻⁴ m³g = acceleration due to gravity = 9.81 m/s²

Therefore,

\(\Delta W = W_a = \rho Vg\\\\\Delta W = (1.225\ kg/m^3)(2.26\ x\ 10^{-4}\ m^3)(9.81\ m/s^2)\)

ΔW = 2.72 x 10⁻³ N

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Select all that apply.
A chemist is likely to do which of the following?
analyze the ingredients in ice cream
determine why planets orbit the Sun
determine how to separate gasoline from other substances in petroleum
determine the force necessary to push a button through a button hole

Answers

Answer:

Determine how to separate gasoline from other substances in petroleum

PLEASE PROVIDE EXPLANATION FOR D AND E

THANK YOU!

PLEASE PROVIDE EXPLANATION FOR D AND ETHANK YOU!

Answers

Answer:

d) 14.1 m

e) 8.57 m/s

Explanation:

(a) There are three forces on the ball:

Normal force Fn pushing up,

Weight force Fg pulling down,

Kinetic friction Fk pushing left (resisting motion).

FBD is the correct free body diagram.

(b) Sum of forces in the y direction:

∑F = ma

Fn − Fg = 0

Fn = Fg

Sum of forces in the x direction:

∑F = ma

-Fk = ma

-Fn μk = ma

-mg μk = ma

a = -g μk

a = -(10 m/s²) (0.25)

a = -2.5 m/s²

(c) Sum of torques about the ball's center:

∑τ = Iα

Fk r = ⅖ mr² α

mg μk r = ⅖ mr² α

g μk = ⅖ r α

α = 5g μk / (2r)

α = 5 (10 m/s²) (0.25) / (2 × 0.2 m)

α = 31.25 rad/s²

(d) The velocity and angular velocity at time t is:

v = at + v₀

v = -2.5t + 12.0

ω = αt + ω₀

ω = 31.25t + 0

When the ball begins to roll without slipping, v = ωr.

-2.5t + 12.0 = (31.25t) (0.2)

-2.5t + 12.0 = 6.25t

12.0 = 8.75t

t = 1.37

The distance the ball moves is:

Δx = v₀ t + ½ at²

Δx = (12) (1.37) + ½ (-2.5) (1.37)²

Δx = 14.1 m

(e) v = at + v₀

v = (-2.5 m/s²) (1.37 s) + 12.0 m/s

v = 8.57 m/s

Answer:

a

Explanation:

on egde 2020

Question 15 of 32
A bungee jumper jumps off a bridge and bounces up and down several times.
She finally comes to rest 30 m below the bridge from which she just jumped.
If her mass is 50 kg and the spring constant of the bungee cord is 10 N/m,
how much energy was lost due to air resistance while she was bouncing?
(Recall that g = 9.8 m/s²)
A. 7330 N
B. 9200 N
C. 10,200 N
D. 8605 N

Answers

C. 10,200 N is how much energy was lost due to air resistance while she was bouncing

How much energy was lost

The energy lost due to air resistance while the bungee jumper was bouncing can be calculated by finding the total mechanical energy of the system at the beginning of the jump and comparing it to the total mechanical energy at the end of the jump.

At the beginning of the jump, the total mechanical energy is given by:

Ei = mgh

where m is the mass of the bungee jumper, g is the acceleration due to gravity, and h is the height of the bridge. Therefore, at the beginning of the jump:

50 x 30 x 10 - 1/2 x 30^2 x 10

= 15000 - 4500

= 10,200 N

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A capacitor consists of two metal plates each 10 cm by 20 cm; they are separated by a 2.0 mm thick insulator with dielectric constant 4.1 and dielectric strength 6.0107 V/m. What is the capacitance in pF(10 −12
F)?

Answers

The equation gives the capacitance of a parallel-plate capacitor:C = εA/d. Where C is the capacitance, ε is the permittivity of the dielectric material between the plates, A is the area of each plate, and d is the distance between the plates.

In this case, the area of each plate is 10 cm × 20 cm = 200 cm^2 = 0.02 m^2. The distance between the plates is 2.0 mm = 0.002 m. The permittivity of the dielectric material is ε = ε0εr, where ε0 is the vacuum permittivity (8.85 × 10^-12 F/m), and εr is the relative permittivity or dielectric constant (4.1).

So, substituting these values into the equation, we get:

C = εA/d

= (ε0εr)(0.02)/(0.002)

= (8.85 × 10^-12)(4.1)(0.02)/(0.002)

= 7.26 × 10^-11 F

= 72.6 pF

Therefore, the capacitance parallel-plate capacitor is 72.6 pF.

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What is it called when data are arranged in rows and columns?
O A. Interpolation
O B. Correlation
O C. Table
OD. Bar graph

Answers

Answer:

C. Table

What is a Database table?

A organized collection of records is called a database table. The main key that uniquely identifies each row in the table. Since no two rows may have the same instantaneous value thanks to a unique key, it is simple to choose a row by its main key. A table can also contain several columns, each of which can hold a distinct kind of data. The information stored is verified using the column name and data type.  

The table name is the linguistic component of a database table. It can represent expressions or variable items, and it must be typed exactly. Square brackets denote a need for the item, whereas curly braces denote an optional element. An alternate choice is indicated by a vertical bar. A main key is a single field, and a secondary key might be a combination of many fields. This kind of information is crucial.

Any field in the data table can be referenced by the secondary key. A database's data field may be a string, a number, a date, or a time. A table's first field should always be the main key, followed by any foreign keys. Actual data should never be present in a primary key. Since the immediate key value is derived from external sources, it is crucial to use it properly. Otherwise, data input mistakes may occur.

The information kept in the database table is referred to as a secondary key. This kind is a primary key, which implies it is a secondary key. A primary key can be used as an alternate key to distinguish between different records in a table and to uniquely identify a record inside a database. Multiple fields that can be either main or alternative make up a composite key.

Depending on the type of data being stored and its intended use, a database table may have a single row or several. The main key determines the number of rows, which might be 50 or more. A secondary key can be used to establish connections between various tables and to distinguish between various data types in a database. A data table also includes a secondary key, which is essential for data retrieval and is separate from the primary key.

You can have local or global database tables. The two types of tables differ in terms of names, availability, and visibility. A global table is accessible to all users and may be utilized by any other user, while a local table is only visible to the present user. It is distinct from a typical table and requires more effort to build than a straightforward one does. Its main function is to maintain a database and store data.

Rows in a database table can be of any size. A table's rows can include any number of columns. A database table's rows can be arranged in any way. The number of rows on a data page might range from one to thousands. Each row may consist of a clustered index, a row, or both. A list of columns may also be present in a row. A clustered index can only contain information in a particular sequence.

A table can be set up in  two different ways. The information-containing row of data is its main key. The data column that contains the values for a row is known as a secondary key. A database that employs rows is the third form of table. Multiple primary keys are possible for a data row, and a secondary key is a secondary key that refers to a row column. A database will have a foreign key if it is formed with a single primary key.

There is no limit to the amount of rows and columns in a database table. Up to 1,024 rows are allowed in a typical user-defined table. The server's storage capacity determines how many rows can fit in a database table. A table can have characteristics added to it in addition to the column and row. A restriction can stop a row from having empty fields. The connection between two tables can be established via an important control.

Table
a bar graph is when data is arranged in bars, interpolation is data in the form of diagonal lines like on a x and y graph, and correlation is data in the form of dots in somewhat the shape of a interpolation
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