A professor of physics cannot see clearly at a distance shorter than 1 m. What should be the focal length of an eyeglass lens that would assist him in reading a newspaper while holding it at a desired distance of 25 cm

Answers

Answer 1

Answer:

Explanation:


Related Questions

If there is no slipping, a frictional force must exist between the wheels and the ground. In what direction does the frictional force from the ground on the wheels act

Answers

Answer:

tire advances to the right, the friction force must be directed to the left, that is, opposing the possible movement of the tire.

Explanation:

For the movement of the wheel to be composed of a rotating part and a translational part, it is necessary that there be a static friction force between the floor and the tire.

As the tire advances to the right, the friction force must be directed to the left, that is, opposing the possible movement of the tire.

An electron moves through a uniform electric field E = (2.60i + 5.90j) V/m and a uniform magnetic field B= 0.400k in m/s^2.) T.

Required:
a. Determine the acceleration of the electron when it has a velocity v= 8.0i m/s.
b. What If? For the electron moving along the x-axis in the fields in part (a), what speed (in m/s) would result in the electron also experiencing an acceleration directed along the x-axis?

Answers

A) The acceleration of the electron along the x -axis is ; 4.57 * 10⁻¹¹ m /s²

B) The speed that would result in the electron experiencing an acceleration  along the x-axis is 4.57 * 10⁻¹¹  * time  m/s

Given Data :

Electric field ( E ) = ( 2.60i + 5.90j ) V/m

Magnetic field ( B ) = 0.400 k T

Velocity ( v ) = 8.0i m/s

A) Determine the acceleration of the electron

Applying Lorentz force

F = q ( E + ( v * B ) )

  = 1.6 * 10⁻¹⁹ ( 2.60 i  +  5.90 j  + ( 8.0 i * 0.4 k ) ) N

  = 1.6 * 10⁻¹⁹ ( 2.60 i  +  5.90 j + ( 3.2 ( -j ) ) N

  = 1.6 * 10⁻¹⁹ (  2.60 i  + 2.70 j ) N

Ax = 4.57 * 10⁻¹¹ m /s²

B)  The speed of the electron moving along the x-axis

Ax = Fx / Mc

    = ( 1.6 * 10⁻¹⁹ * 2.60 ) / 9.1 * 10⁻³¹

    = ( 4.16 * 10⁻¹⁹ ) / 9.1 * 10⁻³¹

    = 0.457 * 10¹²

    = 4.57 * 10⁻¹¹ m /s²

Therefore The speed that would result in the electron experiencing an acceleration  along the x-axis is 4.57 * 10⁻¹¹  * time

Learn more about acceleration of an electron: https://brainly.com/question/15585270

Sammy is 5 feet and 5.3 inches tall. What is Sammy's height in inches? ​

Answers

Answer:

[tex]\boxed{\sf 65.3 \ inches}[/tex]

Explanation:

1 foot = 12 inches

Sammy is 5 feet tall.

5 feet = ? inches

Multiply the feet value by 12 to find in inches.

5 × 12

= 60

Add 5.3 inches to 60 inches.

60 + 5.3

= 65.3

65.3 Inches.
12 (1 Foot) X 5= 60 + 5.3 = 65.3

A charged particle moving through a magnetic field at right angles to the field with a speed of 25.7 m/s experiences a magnetic force of 2.98 10-4 N. Determine the magnetic force on an identical particle when it travels through the same magnetic field with a speed of 4.64 m/s at an angle of 29.2° relative to the magnetic field.

Answers

Answer:

The magnetic force would be:

[tex]F\approx 2.625\,\,10^{-5}\,\,N[/tex]

Explanation:

Recall that the magnetic force on a charged particle (of charge q) moving with velocity (v) in a magnetic field B, is given by the vector product:

F = q v x B

(where the bold represents vectors)

the vector product involves the sine of the angle ([tex]\theta[/tex]) between the vectors, so we can write the relationship between the magnitudes of these quantities as:

[tex]F=q\,v\,B\,sin(\theta)[/tex]

Therefore replacing the known quantities for the first case:

[tex]F=q\,v\,B\,sin(\theta)\\2.98\,\,10^{-4} \,\,N=q\,(25.7\,\,m/s)\,B\,sin(90^o)\\2.98\,\,10^{-4} \,\,N=q\,(25.7\,\,m/s)\,B\\q\,\,B=\frac{2.98\,\,10^{-4} }{25.7} \,\frac{N\,\,s}{m}[/tex]

Now, for the second case, we can find the force by using this expression for the product of the particle's charge times the magnetic field, and the new velocity and angle:

[tex]F=q\,v\,B\,sin(\theta)\\F=q\,(4.64\,\,m/s)\,B\,sin(29.2^o)\\F=q\,B(4.64\,\,m/s)\,\,sin(29.2^o)\\F=\frac{2.98\,\,10^{-4} }{25.7} \,(4.64\,\,m/s)\,\,sin(29.2^o)\\F\approx 2.625\,\,10^{-5}\,\,N[/tex]

A convex spherical mirror has a radius of curvature of magnitude36.0cm.

(a) Determine the position of the virtual image and the magnification for object distances of25.0cm. Indicate the location of the image with the sign of your answer.
image location =cm
magnification =
(b) Determine the position of the virtual image and the magnification for object distances of47.0cm. Indicate the location of the image with the sign of your answer.
image location =cm
magnification =
(c) Are the images in parts (a) and (b) upright or inverted?
The image in part (a) is---Select---uprightinverted
The image in part (b) is---Select---uprightinverted

Answers

Answer:

Explanation:

a )

focal length of convex spherical mirror

f = 36/2  cm = 18 cm

object distance u = - 25 cm

mirror formula

[tex]\frac{1}{v} + \frac{1}{u} = \frac{1}{f}[/tex]

[tex]\frac{1}{v} + \frac{1}{- 25} = \frac{1}{18}[/tex]

[tex]\frac{1}{v} = \frac{1}{25} + \frac{1}{18}[/tex]

v = 6.28  cm .

It is positive hence the image will be erect / upright  and formed on the back of the mirror.

For object distance of 47 cm

u = - 47 cm

Putting the values in the mirror formula

[tex]\frac{1}{v} + \frac{1}{- 47} = \frac{1}{18}[/tex]

[tex]\frac{1}{v} = \frac{1}{ 47} + \frac{1}{18}[/tex]

v = 13  cm

It is positive hence the image will be erect / upright and formed on the back of the mirror.

Which of the following exhibit the Tyndall Effect?


A flashlight beam through air in a room.

A laser beam through salt water.

Car headlights on a foggy night.

A laser beam through a soda drink.

Answers

Answer:

Probably all but (a)

The Tyndall Effect is caused by dispersion of the incident light by the individual molecules in the liquid.

Salt water and a foggy night will cause dispersion of the incident light.

A soda drink may also cause this dispersion, but and not sure.

An astronaut is traveling in a space vehicle that has a speed of 0.480c relative to Earth. The astronaut measures his pulse rate at 78.5 per minute. Signals generated by the astronaut's pulse are radioed to Earth when the vehicle is moving perpendicularly to a line that connects the vehicle with an Earth observer. (Due to vehicle's path there will be no Doppler shift in the signal.)
(a) What pulse rate does the Earth-based observer measure? beats/min
(b) What would be the pulse rate if the speed of the space vehicle were increased to 0.940c?
beats/min

Answers

Explanation:

The heart rate of the astronaut is 78.5 beats per minute, which means that the time between heart beats is 0.0127 min. This will be the time t measured by the moving observer. The time t' measured by the stationary Earth-based observer is given by

[tex]t' = \dfrac{t}{\sqrt{1 - \left(\dfrac{v^2}{c^2}\right)}}[/tex]

a) If the astronaut is moving at 0.480c, the time t' is

[tex]t' = \dfrac{0.0127\:\text{min}}{\sqrt{1 - \left(\dfrac{0.2304c^2}{c^2}\right)}}[/tex]

[tex]\:\:\:\:=0.0145\:\text{min}[/tex]

This means that time between his heart beats as measured by Earth-based observer is 0.0145 min, which is equivalent to 69.1 beats per minute.

b) At v = 0.940c, the time t' is

[tex]t' = \dfrac{0.0127\:\text{min}}{\sqrt{1 - \left(\dfrac{0.8836c^2}{c^2}\right)}}[/tex]

[tex]\:\:\:\:=0.0372\:\text{min}[/tex]

So at this speed, the astronaut's heart rate is 1/(0.0372 min) or 26.9 beats per minute.

What is temperature?
O A. The force exerted on an area
B. A measure of mass per unit volume
O C. The net energy transferred between two objects
OD. A measure of the movement of atoms or molecules within an
object​

Answers

Answer:

The net energy transferred between two objects

Explanation:

The physical property of matter that expresses hot or cold is called temperature. It demonstrates the thermal energy. A thermometer is used to measure temperature. It defines the rate to which the chemical reaction occurs. It tells about the thermal radiation emitted from an object.

The correct option that defines temperature is option C.

Answer:

A measure of the movement of atoms or molecules within an object

Explanation:

Process of elimination

Yeast are tiny fungi to make products like bread, other kinds of dough, beer, and wine. They do this by digesting sugar. A byproduct of this is the release of carbon dioxide gas which makes bread so light and fluffy. We can observe yeast through experimentation by putting them in a closed container with sugar with a ballon on top to trap any carbon dioxide released. What is a quantitative observation that could be made in this experiment?

Answers

Answer:

Compare the volume  (weight) of CO2  gas produced to the weights of sugar and yeast that were used to produce the CO2. Ideally, both the yeast and sugar would be entirely consumed.

explain why our sweat is salty?​

Answers

Answer:

Sweat also contains ammonia and urea, which are produced by the body when it breaks down proteins from the foods you eat.

Hope this helps..

Please help!
Much appreciated!​

Answers

Answer:

F = 2.7×10¯⁶ N.

Explanation:

From the question given:

F = (9×10⁹ Nm/C²) (3.2×10¯⁹ C × 9.6×10¯⁹ C) /(0.32)²

Thus we can obtain the value value of F by carrying the operation as follow:

F = (9×10⁹) (3.2×10¯⁹ × 9.6×10¯⁹) /(0.32)²

F = 2.7648×10¯⁷ / 0.1024

F = 2.7×10¯⁶ N.

Therefore, the value of F is 2.7×10¯⁶ N.

Match the following properties to the type of wave.

Answers

Answer:

hi there

Explanation:

1 - III

2- 1

3-1

HOPE IT HELP YOU

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Explanation:

1 . 3

2. 1

3. 2

I hope it is helpful to you.

Which one of the following lists gives the correct order of the electromagnetic spectrum from low to high frequencies?
A) radio waves, infrared, microwaves, ultraviolet, visible, x-rays, gamma rays
B) radio waves, ultraviolet, x-rays, microwaves, infrared, visible, gamma rays
C) radio waves, microwaves, infrared, visible, ultraviolet, x-rays, gamma rays
D) radio waves, microwaves, visible, x-rays, infrared, ultraviolet, gamma rays
E) radio waves, infrared, x-rays, microwaves, ultraviolet, visible, gamma rays

Answers

Answer:

C) radio waves, microwaves, infrared, visible, ultraviolet, x-rays, gamma rays

Explanation:

radio waves have lowest  energy , lowest  frequency and highest  wavelength

gamma rays  have highest  energy , highest  frequency and least  wavelength

Answer: C

Explanation:

IMPORTANT ANSWER ALL 3 PLEASE!

Answers

Answer:

4. Liters

5. Celsius

6. Grams

There are two cells, one with OER as 2.5 and other as 7. Which cell is more sensitive to radiation?
1)The cell with OER 2.5
2)The Cell with OER 7
3)Both the cells
4)Insufficient data

Answers

Answer:

2)The Cell with OER 7

Explanation:

OER is the acronym for Oxygen Enhancement Ratio. It is the measure of the  enhancement of the effect of ionizing radiation due to the presence of oxygen. The ionization effect can be detrimental or therapeutic (use in cancer treatment). OER is the ratio of radiation dose during the lack of oxygen (hypoxia), to the dosage in the presence of oxygen (air is used as a reference). From the definition, one can see that the higher the OER the higher the sensitivity of the cell.

What is the connection between speed, friction, and radius of the curve when turning when driving a car.

Answers

Answer:

hhhbbbbbbbbbbbbbbnnnnnbbhb

If the x-position of a particle is measured with an uncertainty of 1.00×10-10 m, then what is the uncertainty of the momentum in this same direction? (Useful constant: h-bar = 1.05×10-34 Js.)

Answers

Answer:

The uncertainty in momentum is 5.25x 10^25Jsm

Explanation:

We know that

h bar = h/2π

So

1.05x 10^34=h/2pπ

h=1.05x 10^ 34(2π)=6.597x 10^-34Js

dp=(6.597x10^-34/4pπ)/(1x10^-10)

=5.25x10^-25 Jsm

The pressure of sea water increases by 1.0atm for each 10m increase in the depth, by what percentage is the density of water increased in the deepest ocean of water of 12km. Compressibility is 5.0×10^-5 atm

Answers

The percentage by which the water density increased is 4.1[tex]\mathbf{\overline 6}[/tex] %

The known values are;

The increase in pressure per 10 meter increase in depth = 1.0 atm

The depth of the deepest ocean = 12 km = 12,000 m

The compressibility of the ocean = 5.0 × 10⁻⁵ 1/atm

The unknown

The percentage the density of water increased in the deepest ocean

Strategy;

Find the pressure at the deepest point of the deepest ocean and apply the compressibility

We have;

[tex]\mathbf{Compressibility = \dfrac{1}{V} \times \dfrac{\partial V}{\partial p}}[/tex]

The change in pressure, [tex]\partial p[/tex] = (12,000 m/(10 m)) × 1.0 atm = 1,200 atm

Therefore, we have for one cubic meter of water

[tex]\mathbf{5.0 \times 10^{-5} \ atm^{-1} = \dfrac{1}{1 \, m^3} \times \dfrac{\partial V}{1,200 \, atm}}[/tex]

Therefore;

[tex]\mathbf{\partial}[/tex]V = 5.0 × 10⁻⁵ atm⁻¹ × 1 m³ × 1,200 atm = 0.06 m³

The new volume = V - [tex]\mathbf{\partial}[/tex]V

∴ The new volume = 1 m³ - 0.06 m³ = 0.94 m³

The initial density = mass/(1 m³)

The new density = mass/(0.96 m³)

The percentage increase in density, [tex]\partial[/tex]ρ%, is given as follows;

[tex]\mathbf{\partial p \% = \dfrac{ \dfrac{Mass}{0.96 \ m^3} - \dfrac{Mass}{1 \ m^3} }{ \dfrac{Mass}{1 \ m^3}} \times 100 = \dfrac{25}{6} \% = 4.1 \overline 6 \%}[/tex]

∴  [tex]\mathbf{\partial}[/tex]ρ% =  4.1[tex]\mathbf {\overline 6}[/tex] %

The percentage by which the water density increased, [tex]\partial[/tex]ρ% = 4.1[tex]\mathbf{\overline 6}[/tex] %

Learn more about compressibility here;

https://brainly.com/question/18746977

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

A loop of wire is at the edge of a region of space containing a uniform magnetic field B. The plane of the loop is perpendicular to the magnetic field. Now the loop is pulled out of this region in such a way that the area A of the coil inside the magnetic field region is decreasing at the constant rate c. That is, dA/dt=−c, with c>0.Required:a. The induced emf in the loop is measuredto be V. What is the magnitude B of the magnetic field that the loop was in?b. For the case of a square loop of sidelength L being pulled out of the magneticfield with constant speed v, What is the rate of change of area c= -dA/dt

Answers

Answer:

The question is not clear enough. So i have attached a copy of the correct question.

A) B = V/c

B) c = Lv

Explanation:

A) we know that formula for magnetic flux is;

Φ = BA

Where B is magnetic field and A is area

Now,

Let's differentiate with B being a constant;

dΦ/dt = B•dA/dt

From faradays law, the EMF induced is given as;

E = -dΦ/dt

However, we want to express it in terms of V and E.M.F is also known as potential difference or Voltage.

Thus, V = -dΦ/dt

Thus, we can now say that;

-V = B•dA/dt

Now from the question, we are told that dA/dt = - c

Thus;

-V = B•-c

So, V = Bc

Thus, B = V/c

B) according to Faraday's Law or Lorentz Force Law, an electromotive force, emf, will be induced between the two ends of the sidelength:

Thus;

E =LvB or can be written as; V = LvB

Where;

V is EMF

L is length of bar

v is velocity

From the first solution, we saw that;

V = Bc

Thus, equating both of the equations, we have;

Bc = LvB

B will cancel out to give;

c = Lv

Explanation:

Reading glasses with a power of 1.50 diopters make reading a book comfortable for you when you wear them 1.8 cmcm from your eye. Part A If you hold the book 28.0 cmcm from your eye, what is your nearpoint distance

Answers

Answer:

The near point is  [tex]n =44.8 \ cm[/tex]

Explanation:

From the question we are told that

   The power is  [tex]P = 1.50[/tex]

   The  distance from the eye is  [tex]k = 1.8 \ cm[/tex]

    The distance of the book from the eye is [tex]z = -28 \ cm[/tex]

Generally the focal length of the glasses is  

       [tex]f = \frac{1}{P}[/tex]

=>   [tex]f = \frac{1}{1.50 }[/tex]

=>   [tex]f = 0.667 \ m[/tex]

=>   [tex]f = 66.7 \ cm[/tex]

The object distance is evaluated as

     [tex]u = z + k[/tex]

=>   [tex]u = -28 + 1.8[/tex]

=>  [tex]u = -26.2 \ cm[/tex]

The image distance is evaluated from lens formula as

       [tex]\frac{1}{v} = \frac{1}{f} + \frac{1}{u}[/tex]

=>   [tex]\frac{1}{v} = \frac{1}{66.7} + \frac{1}{-26.2}[/tex]

=>   [tex]v=- \frac{1}{0.0232}[/tex]

=>    [tex]v=- 43 \ cm[/tex]

The  near point is evaluated as

      [tex]n = -v + k[/tex]

=>    [tex]n =-(-43) + 1.8[/tex]

=>    [tex]n =44.8 \ cm[/tex]

An equation for the period of a planet is 4 pie² r³/Gm where T is in secs, r is in meters, G is in m³/kgs² m is in kg, show that the equation is dimensionally correct.​

Answers

Answer:

[tex]\displaystyle T = \sqrt{\frac{4\, \pi^{2} \, r^{3}}{G \cdot m}}[/tex].

The unit of both sides of this equation are [tex]\rm s[/tex].

Explanation:

The unit of the left-hand side is [tex]\rm s[/tex], same as the unit of [tex]T[/tex].

The following makes use of the fact that for any non-zero value [tex]x[/tex], the power [tex]x^{-1}[/tex] is equivalent to [tex]\displaystyle \frac{1}{x}[/tex].

On the right-hand side of this equation:

[tex]\pi[/tex] has no unit.The unit of [tex]r[/tex] is [tex]\rm m[/tex].The unit of [tex]G[/tex] is [tex]\displaystyle \rm \frac{m^{3}}{kg \cdot s^{2}}[/tex], which is equivalent to [tex]\rm m^{3} \cdot kg^{-1} \cdot s^{-2}[/tex].The unit of [tex]m[/tex] is [tex]\rm kg[/tex].

[tex]\begin{aligned}& \rm \sqrt{\frac{(m)^{3}}{(m^{3} \cdot kg^{-1} \cdot s^{-2}) \cdot (kg)}} \\ &= \rm \sqrt{\frac{m^{3}}{m^{3} \cdot s^{-2}}} = \sqrt{s^{2}} = s\end{aligned}[/tex].

Hence, the unit on the right-hand side of this equation is also [tex]\rm s[/tex].

An average sleeping person metabolizes at a rate of about 80 W by digesting food or burning fat. Typically, 20% of this energy goes into bodily functions, such as cell repair, pumping blood, and other uses of mechanical energy, while the rest goes to heat. Most people get rid of all this excess heat by transferring it (by conduction and the flow of blood) to the surface of the body, where it is radiated away. The normal internal temperature of the body (where the metabolism takes place) is 37∘C37 ∘ C, and the skin is typically 7C∘7C ∘ cooler. By how much does the person’s entropy change per second due to this heat transfer?

Answers

Answer:

-4.7 x 10^-3 J/K-s

Explanation:

The Power generated by metabolizing food = 80 W

The watt W is equivalent to the Joules per sec J/s

therefor power = 80 J/s

20% of this energy is not used for heating, amount available for heating is

==> H = 80% of 80 = 0.8 x 80 = 64 J/s

The inner body temperature = 37 °C = 273 + 37 = 310 K

The entropy of this inner body ΔS = ΔH/T

ΔS = 64/310 = 0.2065 J/K-s

The skin temperature is cooler than the inner body by 7 °C

Temperature of the skin =  37 - 7 = 30 °C = 273 + 30 = 303 K

The entropy of the skin = ΔS = ΔH/T

ΔS = 64/303 = 0.2112 J/K-s

change in entropy of the person's body = (entropy of hot region: inner body) - (entropy of cooler region: skin)

==> 0.2065 - 0.2112 = -4.7 x 10^-3 J/K-s



48. A patient presents with a thrombosis in
the popliteal vein. This thrombosis most likely
causes reduction of blood flow in which of the
following veins?

Answers

Answer:

the interation blood veins

Explanation:

Luz, who is skydiving, is traveling at terminal velocity with her body parallel to the ground. She then changes her body position to feet first toward the ground. What happens to her motion? She will continue to fall at the same terminal velocity because gravity has not changed. She will slow down because the air resistance will increase and be greater than gravity. She will speed up because air resistance will decrease and be less than gravity. She will begin to fall in free fall because she will have no air resistance acting on her.

Answers

Answer:

Option C - she will speed up because air resistance has reduced and be less than gravity

Explanation:

We are told that Luz is skydiving with terminal velocity and her body parallel to the ground. Now, at this point she will be experiencing a gravitational force acting downwards, and also air resistance as a result of the drag force on her body

Since the downward gravitational force on Luz is constant, she will fall with a net force of;

F_net = F_g - F_d

where;

F_net is the net force on Luz acting downwards

F_g is the gravitational force on Luz

F_d is the drag force on Luz

The drag force on her body is proportional to the surface area of attack.

We are now told that Luz changes her body position to feet first toward the ground. This means that the surface area of attack is reduced because the feet will consume less space than the frontal part of her body. Thus, the drag force will be lesser then before she changed her body position due to reduced air resistance on her body.

Now, from earlier, we saw that;

F_net = F_g - F_d

So, the lesser F_d is, the higher F_net becomes.

Thus, she will speed up because air resistance has reduced and be less than gravity.

Answer:

C

Explanation:

EDGE 2020

A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.5 cm before reaching its new equilibrium length. The block is then pulled down slightly and released. What is the frequency of oscillation?

Answers

Answer:

0.99Hz

Explanation:

Using F= -mx ( spring force)

At equilibrium the gravitational force will be balanced by the spring force so mg= kx

K= mg/ 0.25 N/m

But

Frequency f= 1/2pi √g/0.25

Frequency is 0.99Hz

The block is pulled down slightly and released so, Frequency of oscillation is 3.15 Hz

Frequency of oscillation based problem:

What information do we have?

Length starched = 2.5 cm

F = Kx

We know that

F = mg

So,

mg = Kx

K/m = g/x

[tex]f=\frac{1}{2\pi}\sqrt{\frac{g}{x} }\\f=\frac{1}{2\pi}\sqrt{\frac{9.8}{0.025} }[/tex]

Frequency of oscillation = 3.15 Hz

Find out more information about 'Oscillation'.

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We've seen that for thermal radiation, the energy is of the form AVT4, where A is a universal constant, V is volume, and T is temperature. 1) The heat capacity CV also is proportional to a power of T, Tx. What is x

Answers

Answer:

this raise the temperature is x = 3

Explanation:

Heat capacity is the relationship between heat and temperature change

          C = Q / ΔT

if the heat in the system is given by the change in energy and we carry this differential formulas

          [tex]c_{v}[/tex] = dE / dT

In this problem we are told that the energy of thermal radiation is

        E = A V T⁴

Let's look for the specific heat

        c_{v} = AV 4 T³

the power to which this raise the temperature is x = 3

If you stood on a planet having a mass four times higher than Earth's mass, and a radius two times 70) lon longer than Earth's radius, you would weigh:________
A) four times more than you do on Earth.
B) two times less than you do on Earth.
C) the same as you do on Earth
D) two times more than you do on Earth.

Answers

CHECK COMPLETE QUESTION BELOW

you stood on a planet having a mass four times that of earth mass and a radius two times of earth radius , you would weigh?

A) four times more than you do on Earth.

B) two times less than you do on Earth.

C) the same as you do on Earth

D) two times more than you do on Earth

Answer:

OPTION C is correct

The same as you do on Earth

Explanation :

According to law of gravitation :

F=GMm/R^2......(a)

F= mg.....(b)

M= mass of earth

m = mass of the person

R = radius of the earth

From law of motion

Put equation b into equation a

mg=GMm/R^2

g=GMm/R^2

g=GM/R^2

We know from question a planet having a mass four times that of earth mass and a radius two times of earth radius if we substitute we have

m= 4M

r=(2R)^2=4R^2

g= G4M/4R^2

Then, 4in the denominator will cancel out the numerator we have

g= GM/R^2

Therefore, g remain the same

Seismic attenuation and how spherical spreading affect amplitude, can anyone explain this please!

Answers

Answer:

Hey there!

This can be a confusing topic, so it's totally fine if you get confused...

First, Seismic Attenuation is how seismic waves lose energy as they expand and spread.

Secondly, when distance increases, amplitude decreases. This is because the distance (spherical spreading would mean radius) is inversely proportional to amplitude.

Let me know if this helps :)

The elastic limit of an alloy is 5.0×108 N/m2. What is the minimum radius rmin of a 4.0 m long wire made from the alloy if a single strand is designed to support a commercial sign that has a weight of 8000 N and hangs from a fixed point? To stay within safety codes, the wire cannot stretch more than 5.0 cm.

Answers

Answer:

4.5x 10^ -9m

Explanation:

See attached file

Answer:

The radius is  [tex]r_{min} = 0.00226 \ m[/tex]

Explanation:

   From the question we are told that

      The  elastic limit(stress) is [tex]\sigma = 5.0*10^{8} \ N /m^2[/tex]

      The length is  [tex]L = 4.0 \ m[/tex]

      The weight of the commercial sign is    [tex]F_s = 8000 \ N[/tex]

       The maximum extension of the wire is  [tex]\Delta L = 5.0 \ cm = 0.05 \ m[/tex]

Generally the elastic limit of an alloy (stress) is is mathematically represented as

            [tex]\sigma = \frac{ F_s }{ A }[/tex]

Where A is the cross-sectional area of the wire which is mathematically represented as

         [tex]A = \pi r^2[/tex]

here [tex]r = r_{min}[/tex] which is the minimum radius of the wire that support the commercial sign

So

          [tex]\sigma = \frac{ F_s }{ \pi r_{min}^2 }[/tex]

=>       [tex]r_{min} = \sqrt{\frac{F_s}{\sigma * \pi} }[/tex]

substituting values

             [tex]r_{min} = \sqrt{\frac{8000}{ 5.0* 10^8 * 3.142} }[/tex]

           [tex]r_{min} = 0.00226 \ m[/tex]

             

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