A cannon and a supply of cannonballs are inside a sealed railroad car of length L, as in Fig. 7-33. The cannon fires to the right; the car recoils to the left. The cannonballs remain in the car after hitting the far wall. (a) After all the cannonballs have been fired, what is the greatest distance the car can have moved from its original position

Answers

Answer 1

Answer:

Initially let n cannonballs with a total mass of m be to the left of the center of mass at L /2 and the mass of the car at L/2

x1 =  [-m / (m + M)] * L / 2   is the original position of the CM

x2 = (m (x + L/2) + M x) / (m + M) * L/2 final position of CM with all cannon balls to the right

[-m x - m L / 2 + m x - M x] / (M + m) * L/2

= - ( m L / 2 + M x) / (m + M) * L/2 = Xcm

Check the math, but maximum distance occurs when the cannonballs of mass m move from -L/2 to L/2 and the car of mass M moves from zero to -x


Related Questions

A car whose tire have radii 50cm travels at 20km/h. what is the angular velocity of the tires?

Answers

Radius=r=50cm=0.5mVelocity=20km/h=v

Convert to m/s

[tex]\\ \sf\longmapsto v=20\times 5/18=5.5m/s[/tex]

We know

[tex]\boxed{\sf \omega=\dfrac{rv}{|r|^2}}[/tex]

[tex]\\ \sf\longmapsto \omega=\dfrac{0.5(5.5)}{|0.5|^2}[/tex]

[tex]\\ \sf\longmapsto \omega=\dfrac{2.75}{0.25}[/tex]

[tex]\\ \sf\longmapsto \omega=11rad/s[/tex]

I REALLY NEED HELP WITH PHYSICS ASAP!!!
Vf^2 = v0^2 + 2a (xf - x0)


Solve for a

Answers

Answer:

a. solve for a

[tex]vf ^{2} = vo ^{2} + 2a(xf - xo) \\ 2a(xf - xo) = vf^{2} - vo ^{2} \\ a = \frac{vf^{2} - vo^{2} }{2(xf - xo)} \\ a = \frac{vf ^{2} - vo ^{2} }{2xf - 2xo} [/tex]

I hope I helped you ^_^

Describe the change in motion and kinetic energy of the particles as thermal energy is added to a liquid. Which change of state might happen?
please ill put brainliest!!!

Answers

If a liquid is heated the particles are given more energy and move faster and faster expanding the liquid. The most energetic particles at the surface escape from the surface of the liquid as a vapour as it gets warmer. Liquids evaporate faster as they heat up and more particles have enough energy to break away.

Answer:

If a liquid is heated the particles are given more energy and move faster and faster expanding the liquid. The most energetic particles at the surface escape from the surface of the liquid as a vapour as it gets warmer. Liquids evaporate faster as they heat up and more particles have enough energy to break away

PLEASE HELP ME WITH THIS PHYSICS QUESTION PLSSS!!!

Vf^2 = v0^2 + 2a (xf -x0)


Solve for v0

Answers

b. solve for Vo

[tex] vf ^{2} = vo^{2} + 2a(xf - xo) \\ vf ^{2} = vo ^{2} + 2axf - 2axo\\ vo ^{2} = vf ^{2} - 2axf + 2axo \\ vo = \sqrt{vf ^{2} - 2axf + 2axo } \\ vo = - \sqrt{vf^{2} - 2axf + 2axo } [/tex]

I hope I helped you ^_^

You are a detective investigating why someone was hit on the head by a falling flowerpot. One piece of evidence is a smartphone video taken in a 4th-floor apartment, which happened to capture the flowerpot as it fell past a window. In a span of 8 frames (captured at 30 frames per second), the flowerpot falls 0.84 of the height of the window. You visit the apartment and measure the window to be 1.27 m tall.

Required:
Assume the flowerpot was dropped from rest. How high above the window was the flowerpot when it was dropped?

Answers

Answer:

0.37 m

Explanation:

Given :

Window height, [tex]h_1[/tex] = 1.27 m

The flowerpot falls 0.84 m off the window height, i.e.

[tex]h_2[/tex] = (1.27 x 0.84 ) m in a time span of [tex]$t=\frac{8}{30}$[/tex]   seconds.

Assuming that the speed of the pot just above the window is v then,

[tex]h_2=ut+\frac{1}{2}gt^2[/tex]

[tex]$(1.27 \times 0.84) = v \times \left( \frac{8}{30} \right) + \frac{1}{2} \times 9.81 \times \left( \frac{8}{30} \right)^2$[/tex]

[tex]$v=\left(\frac{30}{8}\right) \left[ (1.27 \times 0.84) - \left( \frac{1}{2} \times 9.81 \times \left( \frac{8}{30 \right)^2 \right) \right]}$[/tex]

[tex]$v= 2.69$[/tex] m/s

Initially the pot was dropped from rest. So,  u = 0.

If it has fallen from a height of h above the window then,

[tex]$h = \frac{v^2}{2g}$[/tex]

[tex]$h = \frac{(2.69)^2}{2 \times 9.81}$[/tex]

h = 0.37 m

Một vật được ném lên trên theo phương thẳng đứng. Người quan sát
thấy vật đó đi qua vị trí có độ cao h hai lần và khoảng thời gian giữa hai lần đó là
t. Tìm vận tốc ban đầu và thời gian chuyển động của vật từ lúc ném đến khi vật
rơi về vị trí ban đầu.

Answers

Answer:

Language -English plz I cant understand

The double bond between two oxygen atoms (a molecule of oxygen air) has
two characteristics. What are they?
A. Four valence electrons are shared.
B. A metallic bond is formed.
C. Valence electrons are shared between oxygen atoms.
D. An ionic bond is formed.

Answers

Answer:

valance electrons are shared between oxygen atoms.. making them have eight in the outer most shells.

I hope this helps

The exponent of the exponential function contains RC for the given circuit, which is called the time constant. Use the units of R and C to find units of RC. Write ohms in terms of volts and amps and write farads in terms of volts and coulombs. Simplify until you get something simple. Show your work below.

Answers

Answer:

The unit of the time constant RC is the second

Explanation:

The unit of resistance, R is the Ohm, Ω and resistance, R = V/I where V = voltage and I = current. The unit of voltage is the volt, V while the unit of current is the ampere. A.

Since R = V/I

Unit of R = unit of V/unit of I

Unit of R = V/A

Ω = V/A

Also, The unit of capacitance, C is the Farad, F and capacitance, F = Q/V where Q = charge and V = voltage. the unit of charge is the coulomb, C while the unit of voltage is the volt, V

Since C = Q/V

Unit of C = unit of Q/unit of V

Unit of C = C/V

F = C/V

Now the time constant equals RC.

So, the unit of the time constant = unit of R × unit of C = Ω × F = V/A × C/V = C/A

Also. we know that the  1 Ampere = 1 Coulomb per second

1 A = 1 C/s

So, substituting 1 A in the denominator, we have

unit of RC =  C/A = C ÷ C/s = s

So, the unit of RC = s = second

So, the unit of the time constant RC is the second

A seesaw has an irregularly distributed mass of 30 kg, a length of 3.0 m, and a fulcrum beneath its midpoint. It is balanced when a 60-kg person sits on one end and a 78-kg person sits on the other end.

Required:
Find a displacement of the center of mass of the system relatively to the seesaw's midpoint.

Answers

Answer:

x = 0.9 m

Explanation:

For this exercise we must use the rotational equilibrium relation, we will assume that the counterclockwise rotations are positive

          ∑ τ = 0

          60 1.5 - 78 1.5 + 30 x = 0

where x is measured from the left side of the fulcrum

           90 - 117 + 30 x = 0

           x = 27/30

           x = 0.9 m       

In summary the center of mass is on the side of the lightest weight x = 0.9 m

The force of gravity is an inverse square law. This means that, if you double the distance between two large masses, the gravitational force between them Group of answer choices weakens by a factor of 4. strengthens by a factor of 4. weakens by a factor of 2. also doubles. is unaffected.

Answers

Answer:

the force decreases by a factor of 4

Explanation:

The expression for the law of universal gravitation is

          F = [tex]G \frac{m_1m_2}{r^2}[/tex]

let's call the force Fo for the distance r

          F₀ = [tex]G \frac{m_1m_2}{r^2}[/tex]

They indicate that the distance doubles

          r ’= 2 r

we substitute

          F = [tex]G \frac{m_1m_2}{(r')^2}[/tex]

          F = [tex]G \frac{m_1m_2}{r^2} \ \frac{1}{4}[/tex]

         

          F = ¼ F₀

consequently the correct answer is that the force decreases by a factor of 4

If the distance between two large masses are doubled, the gravitational force between them weakens by a factor of 4.

Let the initial force be F

Let the initial distance apart be r

Thus, we can obtain the final force as follow:

Initial force (F₁) = F

Initial distance apart (r₁) = r

Final distance apart (r₂) = 2r

Final force (F₂) =?

F = GM₁M₂ / r²

Fr² = GM₁M₂ (constant)

Thus,

F₁r₁² = F₂r₂²

Fr² = F₂(2r)²

Fr² = F₂4r²

Divide both side by 4r²

F₂ = Fr² /4r²

F₂ = F / 4

From the illustration above, we can see that when the distance (r) is doubled, the force (F) is decreased (i.e weakens) by a factor of 4

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Twin skaters approach each other with identical speeds. Then, the skaters lock hands and spin. Calculate their final angular velocity, given each had an initial speed of 2.50 m/s relative to the ice. Each has a mass of 70.0 kg, and each has a center of mass located 0.800 m from their locked hands. You may approximate

Answers

Answer:

[tex]\omega=3.135rad/s[/tex]

Explanation:

From the question we are told that:

initial Speed [tex]V_1=2.50[/tex]

Mass [tex]m=70.0kg[/tex]

Center of mass [tex]d=0.0.800m\[/tex]

Generally the equation for angular velocity is is mathematically given by

[tex]\omega=\frac{v}{r}\\\\\omega=\frac{2.50}{0.0800}[/tex]

[tex]\omega=3.135rad/s[/tex]

Tres personas, A, B y C jalan una caja con ayuda de cuerdas cuya masa es despreciable. Si la persona A aplica −3 en dirección horizontal y la persona B aplica a su vez 5 en dirección horizontal, ¿Cuál es el valor de la fuerza que debe ejercer la persona C, para que la caja esté en equilibrio físico?

Answers

Answer:

Un objeto se encuentra en equilibrio físico si la fuerza neta que se le aplica es igual a 0.

En este caso solo se aplican fuerzas en el eje horizontal, por lo que las podremos sumar directamente.

La persona A aplica una fuerza:

Fa = -3N

La persona B aplica una fuerza:

Fb = 5N

La persona C aplica una fuerza Fc, la cual aún no conocemos.

Pero sabemos que la caja está en equilibrio físico, por lo que:

Fa + Fb + Fc = 0N

reemplazando los valores que conocemos, obtenemos:

-3N + 5N + Fc = 0N

Ahora podemos resolver esto para Fc, la fuerza que aplica la persona C.

Fc = 0N + 3N - 5N

Fc = -2N

Podemos concluir que la persona C aplica una fuerza horizontal de -2N

A boat having stones floats in water. If stones are unloaded in water, what will happen to the level of water?​

Answers

Answer:

A boat having stones floats in water. If stones are unloaded in water, what will happen to the level of water?​

Explanation:

Which nucleus completes the following equation?
39 17 CI-> 0 -1 e+?

Answers

Answer:

[tex]_{18}^{39} } Ar[/tex]

Explanation:

The given equation shows the disintegration of an unstable isotope of chlorine to beta particle and Argon nucleus. The nucleus undergoes the emission of a beta particle to form a more stable nucleus of Argon.

[tex]_{17} ^{39} Cl[/tex] ⇒ [tex]_{-1}^{0} e[/tex] + [tex]_{18}^{39} } Ar[/tex]

Argon is a stable gas and is found in the group 8 on the periodic table of elements.

Answer:

Answer is below

Explanation:

39 18 Ar

Select the correct answer.
What is abstraction?
OA. the concept that software architecture can be separated into modules and that each module can be examined independently
OB. the process of containing information within a module, preventing any crossover or access to Irrelevant information
OC. the process of splitting a program both horizontally and vertically
OD. the process of cutting down irrelevant information so only the information that is useful for a particular purpose remains
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Answers

Answer:

OD. The process of cutting down irrelevant information so only the information that is useful for particular purpose remains

Abstraction is the process of cutting down irrelevant information so only the information that is useful for a particular purpose remains.

What is abstraction?

Abstraction is the practice of removing anything from a set of core features by eliminating or deleting attributes.

One of the three core ideas of object-oriented programming is abstraction order to decrease complexity and maximize efficiency, a programmer uses abstraction to conceal all but the important facts about an object.

Abstraction is the process of cutting down irrelevant information so only the information that is useful for a particular purpose remains.

Hence option D is correct.

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a model car moves round a circular path of radius 0.3m at 2 revolutions per secs what is its angular speed, the period of the car and the speed of the car

Answers

Answer:

a) T = 0.5 s

b) v = 1.2π m/s ≈ 3.77 m/s

Explanation:

It makes two revolutions in one second so makes one revolution in ½ second

circumference of the circle is

C = 2πr = 0.6π m

which it traverses in one time period

0.6π m / 0.5 s = 1.2π m/s

To solve this, we must be knowing each and every concept related to speed and its calculations. Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.

What is speed?

Speed may be defined as the distance traveled by an item in the amount of time it requires to travel that distance. In other words, it measures how rapidly an item travels but does not provide direction.

Speed may be calculated in Science. The speed equation is a scientific formula that is used to calculate various types of speed.

Mathematically, the formula for speed can be given as

speed= distance/time

Values that are given

Time period= 0.5 s

Circumference = 2πr = 0.6π m

substituting all the given values in the above equation, we get

speed     =0.6π m / 0.5 s

On calculations, we get

              = 1.2π m/s

              =3.77 m/s

Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.

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How to calculate voltage U1 ?

Please help!

Answers

Answer:

he is a baby art and design

Infrared and ultraviolet waves have different frequencies.
Both types of wave can have harmful effects on human beings.

Describe the harmful effects of infrared and ultraviolet waves, relating them to the frequencies of the waves.​​

Answers

Answer:

For infrared and ultraviolet waves have different frequencies. Both types of wave can have harmful effects on human beings. Describe the harmful effects of infrared and ultraviolet waves, relating them to the frequencies of the waves. Medical studies indicate that prolonged IR exposure can lead to lens, cornea and retina damage, including cataracts, corneal ulcers and retinal burns, respectively. To help protect against long-term IR exposure, workers can wear products with IR filters or reflective coatings.When you look at the EM spectrum, UV waves are quite a bit smaller in wavelength than infrared, and x-rays/gamma rays are even smaller. Therefore, UV waves are probably causing more harm than infrared waves, and x-rays/gamma rays are probably doing even more damage.

                                              Thank You  

                                     Please mark me Brainliest

Infrared and ultraviolet waves have different frequencies. Infrared waves have lower frequencies and longer wavelengths, while ultraviolet waves have higher frequencies and shorter wavelengths.

Harmful effects of Infrared waves:

Infrared waves have lower frequencies and are often associated with heat radiation. Prolonged exposure to intense infrared radiation can lead to thermal burns and damage to the skin and eyes. Infrared radiation can also cause dehydration and overheating of the body, especially in hot environments. While infrared radiation is not as harmful as ultraviolet radiation, excessive exposure can still lead to health issues.

Harmful effects of Ultraviolet waves:

Ultraviolet waves have higher frequencies and shorter wavelengths, making them more energetic than infrared waves. UV radiation from the sun is a well-known harmful agent. Short-term exposure to intense UV radiation can cause sunburn, skin redness, and eye irritation. Long-term exposure to UV radiation can lead to more serious health problems such as skin aging, cataracts, and an increased risk of skin cancer. UV radiation can also damage DNA in skin cells, leading to mutations and potential carcinogenesis.

It is essential to protect ourselves from both infrared and ultraviolet waves to prevent harmful effects. Using sunscreen and wearing protective clothing can help shield the skin from UV radiation. Limiting exposure to intense sources of infrared radiation, such as hot objects or infrared heaters, can help reduce the risk of thermal burns and overheating. Understanding the differences in the frequencies of these waves allows us to implement appropriate safety measures and protect our health.

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A brass road is 2cm long at instance to what is the lense for a temperature rise of 100k, If the expansivity of brass is 18x10^-6/k^-1

Answers

The length of the brass at a temperature rise of 100 K is 2.0036 m

From the question given above, the following data were obtained:

Original length (L₁) = 2 m

Temperature rise (ΔT) = 100 K

Coefficient of linear expansion (α) = 18×10¯⁶ K¯¹

Final length (L₂) =?

The final length of the brass can be obtained as follow:

α = L₂ – L₁ / L₁ΔT

18×10¯⁶ = L₂ – 2 / (2 × 100)

18×10¯⁶ = L₂ – 2 / 200

Cross multiply

L₂ – 2 = 18×10¯⁶ × 200

L₂ – 2 = 0.0036

Collect like terms

L₂ = 0.0036 + 2

L₂ = 2.0036 m

Thus, the length of the brass at a temperature rise of 100 K is 2.0036 m

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two bodies A and B with some asses 20 kg and 30 kg respectively above the ground which have greater potential​

Answers

Answer:

B has greater potential

Explanation:

We know;

Potential Energy (PE) = mgh

where, m=mass of body

g=acceleration due to gravity

h=height of body

From the formula,

PE is directly proportional to the mass of the body

so the body with greater mass has greater potential.

Để sử dụng nguồn điện xoay chiều 220V/50Hz thắp sáng bóng đèn 12V/3W, ta chọn điện trở giảm áp có giá trị:

Answers

Explanation:

Hi Linda,

How's it going?

Sorry I haven't been in touch for such a long time but I've had exams so I've been studying every free minute. Anyway, I'd love to hear all your news and I'm hoping we can get together soon to catch up. We just moved to a bigger flat so maybe you can come and visit one weekend?

How's the new job?

Looking forward to hearing from you!

Helga

Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 2 m3, is stirred until its temperature is 500 K. Assuming the ideal gas model, for the air, and ignoring kinetic and potential energy, determine

Answers

Answer:

The final pressure in bar will be "[tex]\frac{10}{3} \ Bar[/tex]".

Explanation:

As we know,

PV = nRT

[tex]\frac{P_1}{T_1} =\frac{P_2}{T_2} =CONST[/tex]

then,

⇒ [tex]\frac{2 \ bar}{300 \ K} = \frac{P_2}{500 \ K}[/tex]

⇒ [tex]P_2=(\frac{2}{300}\times 500 )Bar[/tex]

        [tex]=\frac{10}{3} \ Bar[/tex]

Thus the above is the correct answer.

1. A body moving with uniform acceleration of 10 m/s2 covers a distance of 320 m. if its initial velocity was 60 m/s. Calculate its final velocity.​

Answers

Answer:

v² = u² + 2as

v² = 3600 + 6400

v² = 10000

v = 100

Explanation:

final velocity is 100 m/s

initial velocity=u=60m/sAcceleration=10m/s^2=aDistance=s=320mFinal velocity=v

According to third equation of kinematics

[tex]\boxed{\sf v^2-u^2=2as}[/tex]

[tex]\\ \sf\longmapsto v^2=u^2+2as[/tex]

[tex]\\ \sf\longmapsto v^2=(60)^2+2(10)(320)[/tex]

[tex]\\ \sf\longmapsto v^2=3600+3400[/tex]

[tex]\\ \sf\longmapsto v^2=10000[/tex]

[tex]\\ \sf\longmapsto v=\sqrt{10000}[/tex]

[tex]\\ \sf\longmapsto v=100m/s[/tex]

4. A ball is thrown vertically upward from the ground with a velocity of 30m/s. (a) how long will it take to rise to the highest point? (b) How high does the ball rise? (c) How long after projection will the ball have a velocity of 10m/s upward? A velocity of 10m/s downward? (d) When is the displacement of the ball zero? (e) When is the magnitude of the ball’s velocity equal to half its velocity of projection? (f) When is the ball‘s displacement equal to half the maximum height to which it rises? (g) What is the magnitude and direction of the acceleration while the ball is moving upward? While moving downward? While at the highest point?​

Answers

All the answers are:

a) The time that will it take to rise to the highest point is 3.06 seconds.

b) The ball will rise to a height of 45.87 meters.

c) The time at which the ball will have a velocity of 10 m/s upward is 2.04 seconds.

The time when the ball has 10 m/s downward is 1.02 seconds.

d) The displacement of the ball will be zero at 6.12 seconds.

e) The time when the magnitude of the ball's velocity is equal to half its velocity of projection is 1.53 seconds.

f) The ball's displacement is equal to half the maximum height to which it rises after 0.90 seconds.

g) In each moment (upward and downward) the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Let's calculate the values for each case.

a) At the highest point, the final velocity is 0, so we can use the following equation.  

[tex]v_{f}=v_{i}-gt[/tex] (1)

Where:

v(i) is the initial velocityv(f) is the final velocityg is the acceleration due to gravity (9.81 m/s²)

We know that v(i) = 30 m/s.

[tex]0=30-9.81t[/tex]

Solve it for t:

[tex]t=3.06\: s[/tex]

Hence, the time is 3.06 s.

b) At the highest point, the final velocity is 0, so we can use the following equation.  

[tex]v_{f}^{2}=v_{i}^{2}-2gh[/tex] (2)

[tex]0=v_{i}^{2}-2gh[/tex]

We know that the initial velocity is 30 m/s.

[tex]0=30^{2}-2gh[/tex]

Solving it for h we have:  

[tex]h=\frac{30^{2}}{2*9.81}[/tex]

[tex]h=45.87 \: m[/tex]

Then, the height is 45.87 m.

c) Using equation (1) we can find the time (t).

[tex]10=30-(9.81t)[/tex]

So, the time elapsed to get 10 m/s is:

[tex]t_{upward}=2.04\: s[/tex]

We know the upward time is equal to the downward time. So the time from v=10 m/s to v=0 m/s will be.

[tex]t_{upward}=2.04+t[/tex]  

[tex]t=1.02\: s[/tex]

This is the time when the ball has 10 m/s downward.          

Therefore, the time upward is 2.04 s, and the time downward is 1.02 s.

d) It will be when the ball returns to the ground.

[tex]t=2t_{upward}[/tex]

[tex]t=2*3.06[/tex]      

[tex]t=6.12\: s[/tex]

The displacement will be zero after 6.12 s.  

e) Here we need to find the time when v(f) is 15 m/s

[tex]15=30-gt[/tex]

[tex]t=\frac{15}{9.81}[/tex]  

[tex]t=1.53\: s[/tex]

The time when the v(f) is 15 m/s is 1.53 s.

f) Here, we need to find t when h = 45.87/2 m = 22.94 m

We can use the next equation:

[tex]h=v_{i}t-0.5gt^{2}/tex]

[tex]22.94=30t-0.5*9.81*t^{2}/tex]

Solving this quadratic equation, t will be:

[tex]t=0.90\: s/tex]

Hence, the ball's displacement is equal to half the maximum h, at 0.90 s.

g) In each moment the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

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Puerto Rico is a U.S. territory that is southeast of the mainland United States. Which letter on the map represents the location of Puerto Rico?

Answers

The letter that represents the location of Puerto Rico on the map is A.

As the question noted, Puerto Rico is to the southeast of the United States. The first step therefore, would be to find the United States.

The United States is located in the center of the map and is painted orange. I have highlighted it in the attachment.

A place that is southeast of the United States would be both:

downward because it is south and,rightward because it is east.

The only place matching that description is labeled A so we can conclusively say that Puerto Rico is labeled the place labeled A.

For more information on Puerto Rico check https://brainly.com/question/23481183.

Answer:

its a im puerto rican so i know it is a

Explanation:

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What is a measure between the difference in start and end positions?

Answers

Answer:

Displacement

General Formulas and Concepts:

Kinematics

Displacement vs Total Distance

Explanation:

Displacement is the difference between the start position and end position.

Total Distance is the entire distance traveled between the start and end position.

Topic: AP Physics 1 Algebra-Based

Unit: Kinematics

A proton moves perpendicular to a uniform magnetic field at a speed of 1.75 107 m/s and experiences an acceleration of 2.25 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.

Answers

Answer:

B = 0.013(-j) T

Explanation:

Given that,

The speed of a proton, [tex]v=1.75\times 10^7\ m/s[/tex]

Acceleration experienced by the proton,[tex]a=2.25\times 10^3\ m/s[/tex]

We need to find the magnitude and the direction of the magnetic field. At equilibrium,

[tex]ma=qvB\\\\B=\dfrac{ma}{qv}\\\\B=\dfrac{1.67\times 10^{-27}\times 2.25\times 10^{13}}{1.6\times 10^{-19}\times 1.75\times 10^{7}}\\\\B=0.013\ T[/tex]

The velocity is in +z direction, force in +x direction, then the field must be in -y direction.

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms. At what depth did this reflection occur? (The average propagation speed for sound in body tissue is 1540 m/s)

Answers

Answer:

10.01 cm

Explanation:

Given that,

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.

The average propagation speed for sound in body tissue is 1540 m/s.

We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,

[tex]v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m[/tex]

or

d = 10.01 cm

So, the reflection will occur at 10.01 cm.

which one of the following is a product of an acid base reaction? A. Base B. Acid C. Salt D. Fire

Answers

Answer:

salt

Explanation:

salt is a component for many acid base reactions

A bullet 2cm log is fired at 420m/s and passes straight a 10cm thick board exiting at 280m/s
a) what is the average acceleration of the bullet through the board?
b)what is the total time the bullet is in contact with the board?
c)what minimum thickness could the board have if it was supposed to bring the bullet to a stop?

Answers

Solving for the acceleration of the bullet

acceleration = (vf^2 – vi^2) / 2d

acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)

acceleration = (78400 - 176400) / 0.24 m

acceleration = -98000 / 0.24

acceleration = -408333 m/s^2

Solving for contact time with board

t^2 = 2d/a

t^2 = 2 * 0.12 m / 408333 m/s^2

t^2 = 0.24 m / 408333 m/s^2

t^2 = 5.8775558 x 10^-7

t = 0.0007666 s or 767 microseconds


(I was only able to do A and B)

Answer:

Explanation:

(a)Solving for the acceleration of the bullet

acceleration = (vf^2 – vi^2) / 2d

acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)

acceleration = (78400 - 176400) / 0.24 m

acceleration = -98000 / 0.24

acceleration = -408333 m/s^2

(a)Solving for contact time with board

t^2 = 2d/a

t^2 = 2 * 0.12 m / 408333 m/s^2

t^2 = 0.24 m / 408333 m/s^2

t^2 = 5.8775558 x 10^-7

t = 0.0007666 s or 767 microseconds

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