Your teacher placed a 3.5 kg block at the position marked with a “ + ” (horizontally, 0.5 m from the origin) on a large incline outlined on the graph below and let it slide, starting from rest. ***There are two images included!***

Your Teacher Placed A 3.5 Kg Block At The Position Marked With A + (horizontally, 0.5 M From The Origin)
Your Teacher Placed A 3.5 Kg Block At The Position Marked With A + (horizontally, 0.5 M From The Origin)

Answers

Answer 1

Answer:

 x = 10.75 m

Explanation:

For this problem we will solve it in two parts, the first using energy and the second with kinematics

Let's use the energy work relationship to find the velocity of the block as it exits the ramp

       W = [tex]Em_{f}[/tex] - Em₀

Starting point. Higher

       Em₀ = U = m g h

the height from the edge of the ramp of the graph has a value

        h = 9-3 = 6 m

Final point. At the bottom of the ramp

       Em_{f} = K = ½ m v²

Friction force work

      W = - fr  d

The friction force has the formula

      fr = μ N

 

On the ramp, we can use Newton's second law

         N - W cos θ = 0

         N = W cos θ

where the angle is obtained from the graph

         tan θ = (9-3) / (0.5-4) = -6 / 3.5

         θ = tan⁻¹ (-1,714)

         θ = -59.7º

the distance d is

         d = √ (Δx² + Δy²)

         d = √ [(0.5-4)² + (9-3)²]

         d = 6.95 m

for which the work is

       W = - μ mg cos 59.7 d

we substitute

        W = Em_{f} -Em₀

        - μ mg cos 59.7 d = ½ m v² - m g h

In the graph o text the value of the friction coefficient is not observed, suppose that it is μvery = 0.2

        - μ g cos 59.7 d = ½ v² - g h

         v² = 2g (h - very d coss 59.7)

let's calculate

         v² = 2 9.8 (6 - 0.2 6.95 cos 59.7)

         v = √ 103.8546

         v = 10.19 m / s

in the same direction as the ramp

in the second part we use projectile launch kinematics

       

let's look for the components of velocity

         v₀ₓ = vo cos -59.7

         [tex]v_{oy}[/tex] = vo sin (-59,7)

         v₀ₓ = 10.19 cos (-59.7) = 5.14 m / s

         v_{oy} = 10.19 if (-59.7) = -8.798 m / s

Let's find the time to get to the floor (y = o)

          y = y₀ + v_{oy} t - ½ g t²

to de groph y₀=3 m

          0 = 3 - 8.798 t - ½ 9.8 t²

          t² - 1.796 t - 0.612 = 0

we solve the quadratic equation

          t = [1.796 ±√(1.796² + 4 0.612)] / 2

          t = [1,795 ± 2,382] / 2

          t₁ = 2.09 s

          t₂ = -0.29 s

since time must be a positive quantity the correct value is t = 2.09 s

we calculate the horizontal displacement

          x = v₀ₓ t

          x = 5.14 2.09

          x = 10.75 m

Answer 2

The motion of the box, after it exits the incline is the motion and trajectory

of a projectile.

Horizontal distance from the right-hand edge of the incline to the point of

contact with the floor is approximately 1.24613 m.

Reasons:

Mass of the block,  m = 3.5 kg

Coefficient of kinetic friction, μ = 1.2

Location of the = 0.5 m from the origin

Required:

Horizontal distance between the block's point of contact with the floor and

the bottom right-hand edge of the incline.

Solution:

Let θ represent the angle the incline make with the horizontal.

The normal reaction of the incline on the block, [tex]F_N[/tex] = m·g·cos(θ)

Work done on friction = [tex]F_N[/tex]×μ×Length of the incline, L

Rise of the incline = 10 - 3 = 7

Run of the incline = 4

L = √(6.125² + 3.5²) = [tex]\dfrac{7 \times \sqrt{65} }{8}[/tex]

Let ΔP.E.₁  represent the potential energy transferred to kinetic energy

and work along the incline, we have;

Energy of the block at the bottom of the incline, M.E.₂, is found as follows;

K.E.₂ = mgh - m·g·μ·cos(θ)·L

[tex]K.E. =\frac{1}{2} \times 3.5 \times v^2 = 3.5 \times 9.81 \times 6.125 - 3.5 \times 9.81 \times 1.2 \times \dfrac{4}{\sqrt{65} } \times \dfrac{7 \times \sqrt{65} }{8}[/tex]

v ≈ 6.1456 m/s

The vertical component of the velocity is therefore;

[tex]v_y = v \cdot sin(\theta)[/tex]

[tex]v_y = 6.1456 \times \dfrac{7}{\sqrt{65} } \approx 5.33588[/tex]

From the equation, h = u·t + 0.5·g·t² derived from Newton's Laws of motion, we have;

ΔP.E.₁ = 3.5×9.81×7

3 = 5.33588·t + 0.5×9.81·t²

Factorizing, the above quadratic equation, we get;

The time it takes the block to reach the floor, t ≈ 0.40869 seconds

Horizontal component of the velocity is [tex]v_x \approx 6.1456 \times \dfrac{4}{\sqrt{65} } \approx 3.04908[/tex]

The horizontal distance, x = vₓ × t

∴ x = 3.04908 × 0.40869 ≈ 1.08194

Horizontal distance from the right-hand edge of the incline to the point of

contact with the floor, x ≈ 1.24613 m.

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Your Teacher Placed A 3.5 Kg Block At The Position Marked With A + (horizontally, 0.5 M From The Origin)

Related Questions

an object falls from a hovering helicopter and hits the ground at a speed of 30m per seconds. how long does it take the object to reach the ground and how far does it fall? sketch a velocity-time graph for the object ( ignore air resistance

Answers

Answer:

45.9m

Explanation:

Given parameters:

Final velocity  = 30m/s

Initial velocity  = 0m/s

Unknown:

Time it takes for the object of fall  = ?

Height of fall = ?

Solution:

For the first problem, we use the equation below to solve for t;

      V = U + gt

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

t is the time taken

          30  = 0 + 9.8 x t

          30  = 9.8t

            t  = [tex]\frac{30}{9.8}[/tex]  = 3.1s

Now, height of fall;

  V² = U²  + 2gH

   30² = 0² + 2 x 9.8 x H

   900  = 19.6H

      H  = 45.9m

1
Betty is sitting on of her surfboard out in the ocean. She is waiting for the
perfect wave to come along so she can ride it in to shore. As she waits, she
notices that the waves roll by in patterns, or sets. As the top of each wave
passes by Betty, it pushes her up. Which part of the wave does this? *

Answers

Explanation:

yooooooooo

A footballer kicks a ball from rest. The foot is in contact with the ball for 0.30s and the final velocity of the ball is 15ms-1 .What is the average acceleration of the ball?​

Answers

Answer:

50m/s^2

Explanation:

Step one:

given data

initial velocity u= 0m/s since the ball is at rest

time of contact t= 0.3s

final velocity v=15m/s

Required

acceleration a

from the first law of motion

v=u+at

substitute our given data

15=0+a*0.3

15=0.3a

divide both sides by 0.3

a=15/0.3

a=50m/s

The average acceleration is 50m/s^2

STATION 1
Jane moved a 800kg piano to the right across the
carpet with a coefficient of friction of 0.4. What is the
magnitude of the force of friction acting on the
piano?
If she moved it at a constant velocity what is the
applied force acting on the piano?

Answers

Chedda chease snadwhik

5. How does the existence of humans compare with Earth's age?

Answers

Answer:

Human activity has fundamentally changed our planet. We live on every continent and have directly affected at least 83% of the planet’s viable land surface. Our influence has impacted everything from the makeup of ecosystems to the geochemistry of Earth, from the atmosphere to the ocean. Many scientists define this time in the planet’s history by the scale of human influence, and label it as a new geological epoch called the Anthropocene.

Explanation:

Describe and give an example of mutualism.


Describe and give an example of commensalism.


Describe and give an example of parasitism.


Describe and give an example of competition.


Describe and give an example of predation.

Answers

Answer:

Mutualism - Bee to flower. Bee eats - flower reproduces

Commensalism - Tree Frog to plant or tree. Frog uses plant for protection.

Parasitism - Flea or tick to host. Parasite feeds off host.

Explanation:

Competition - relationship between organisms that strive for same resources. intraspecific and interspecific. ex) two males competing for mates.

predation - one organism kills and consumes another. wolf hunting moose, cat hunting mouse. venus fly trap killing insect

A car is traveling west for 12 s its speed is 36.12 m/s in the same direction find the total distance the car traveled

Answers

We are given:

constant speed of the car (u) = 36.12 m/s

time in question (t) = 12 seconds

Solving for the Distance and Displacement:

from the second equation of motion:

s = ut + 1/2 at^2

since we have 0 acceleration:

s = ut

replacing the variables

s = 36.12 * 12

s = 433.44 m

Since the car is travelling in a straight line towards the same direction, it's Distance will be equal to its Displacement

Hence, both the Displacement and Distance covered by the car is

433.44 m

but since Displacement also has a direction vector along with it,

the Displacement will be  433.44 m due west

Using a maximum allowable shear stress of 70 MPa, find the shaft diameter needed to transmit 40 kW when (a) The shaft speed is 2500 rev/min. (b) The shaft speed is 250 rev/min

Answers

Answer:

a

 [tex]d = 0.0223 \ m[/tex]

b

[tex]d = 0.0481 \ m[/tex]  

Explanation:

From the question we are told that

   The maximum allowable shear stress is  [tex]\sigma = 70 MPa = 70 *10^{6} \ Pa[/tex]

    The power is  [tex]P = 40 \ kW = 40 *10^{3} \ W[/tex]

considering question a

   The shaft speed is given as [tex]v = 2500\ rev/min[/tex]

Generally the torque experienced by the shaft is mathematically represented as

       [tex]\tau = \frac{ 9.55 * P}{v}[/tex]

=>    [tex]\tau = \frac{ 9.55 * 40 *10^{3}}{ 2500}[/tex]

=>    [tex]\tau = 152.8 \ N \cdot m[/tex]

Generally the maximum torque experienced by the shaft is mathematically represented as

          [tex]\tau_m = \frac{2 \tau }{ \pi r^2 }[/tex]

Generally diameter  =  2 *  radius (r)

So

        [tex]\tau_m = \frac{2 \tau }{ \pi 4 d^2 }[/tex]

Generally the maximum allowable shear stress is mathematically represented as

                [tex]\sigma = \frac{2 \tau }{ \pi 4 d^2 } * \frac{32}{d}[/tex]

=>             [tex]\sigma = \frac{16 \tau }{ \pi d^3}[/tex]

=>            [tex]d = \sqrt[3]{\frac{16 * \tau }{ \pi \sigma } }[/tex]

=>          [tex]d = \sqrt[3]{\frac{16 * 152.8 }{ \pi * 70 *10^{6} } }[/tex]

=>          [tex]d = 0.0223 \ m[/tex]

considering question b

   The shaft speed is given as [tex]v = 250\ rev/min[/tex]

Generally the torque experienced by the shaft is mathematically represented as

       [tex]\tau = \frac{ 9.55 * 40 *10^{3}}{250 }[/tex]

=>    [tex]\tau = 1528 \ N \cdot m[/tex]

Generally the shaft diameter is mathematically represented as

   [tex]d = \sqrt[3]{\frac{16 * \tau }{ \pi \sigma } }[/tex]

=>[tex]d = \sqrt[3]{\frac{16 * 1528 }{ 3.142 * 70 *10^{6} } }[/tex]  

=>[tex]d = 0.0481 \ m[/tex]  

A boat travels 28 m while it reduces its velocity from 27.5 m/s to 14.5m/s. What is the boat’s acceleration while it travels that distance?

Answers

Answer:

9.75 m/s²

Explanation:

Given that,

v= 14.5 m/s

u = 27.5 m/s

s = 28 m

a = ?

v² = u² -2as [ minus sign due to deceleration]

14.5² = 27.5² - 2 × a × 28

210.25 - 756.25 = -56a

-56a = -546

a = 9.75 m/s²

A car has a mass of 850 kg. By pushing on the car, Evan increases its speed
from 3.5 m/s to 5 m/s. What impulse did Evan apply to the car?

A. 4250 kg•m/s
B. 1275 kg•m/s
C. 850 kg•m/s
D. 2975 kg•m/s

Answers

Answer:

B. 1275 kg*m/s

Explanation:

I = F(deltaT) = (deltaP) = mv2- mv1

Therefore,

I = mv2-mv1

m = 850 kg

v2 = 5 m/s

v1 = 3.5 m/s

I = (850)(5)-(850)(3.5)

I = 1275 kg* m/s

A circular coil of wire 8.6 cm in diameter has 15 turns and carries a current of 2.7 A. The coil is in a region where the magnetic field is 0.56 T. (a) What orientation of the coil gives the maximum torque on the coil, and what is this maximum torque? (b) For what orientation of the coil is the magnitude of the torque 71% of the maximum found in part (a)?

Answers

Answer:

(a) (i) The orientation of the coil which gives maximum torque on the coil is 90⁰

(a)(ii) The maximum torque is 0.132 Nm

(b) The orientation of the coil is 45⁰

Explanation:

Given;

diameter of the circular wire, d = 8.6 cm = 0.086 m

radius of the wire, r = d /2 = 0.043 m

number of turns, N = 15 turns

magnetic field, B = 0.56 T

The torque on the wire is given by;

τ = NIABsinθ

where;

θ is the orientation of the wire

(a) maximum torque occurs when the orientation of the wire is at 90⁰

The maximum torque is given by;

τ = NIABsin(90⁰)

τ = NIAB

τ = (15)(2.7)(π x 0.043²)(0.56)

τ = 0.132 Nm

(b)

71% of 0.132 = 0.71 x 0.132 = 0.0937 Nm

[tex]\tau = NIAB sin\theta\\\\sin\theta = \frac{\tau}{NIAB }\\\\ sin\theta = \frac{0.0937}{(15)(2.7)(\pi *0.043^2)(0.56)} \\\\sin\theta = 0.7111\\\\\theta = sin^{-1}(0.7111)\\\\\theta =45.32\\\\\theta = 45^0[/tex]

How long does it take a vehicle to reach a velocity of 32 m/s if it accelerates from rest at a rate of 4.2 m/s^2?
What is the initial velocity of the vehicle?
What is the final velocity of the vehicle?
What is the acceleration of the vehicle?
Write the equation you will use to solve the problem.
How long does it take the vehicle to reach its final velocity?
0.13 seconds
18.1 seconds
7.62 seconds
134.4 seconds

Answers

Answer:

7.62

Explanation:

because you have to divide 32/4.2

and can you do a friend request so i can accept it

In your new job, you are the technical advisor for the writers of a gangster movie about Bonnie and Clyde. In one scene Bonnie and Clyde are being pursued by a police car. They are 750m from a level railroad crossing travelling at 100 km/hr. A train is 500 meters from the crossing travelling at 130 km/hr. The level crossing is on the state line, and so if they can beat the train, they could evade capture, at least for a while until they become Federal fugitives. They accelerate at a constant rate of 4 m/s2 toward the crossing. The writers want to know if Bonnie and Clyde make it across the crossing before the train.

Answers

Answer:

Explanation:

Velocity of train 130 km /hr

distance of crossing s = 500 m = .5 km

Time taken by train to reach crossing = .5 / 130 = .003846 hr = 13.85 s

Time taken by gangster to reach the crossing = t

initial velocity u = 100km/h = 27.77 m /s

distance of crossing s = 750 m

acceleration a = 4 m /s

s = ut + .5 a t²

750 = 27.77 t + .5 x 4 x t²

2 t² + 27.77 t - 750 = 0

t = - 27.77 ± √(27.77² + 4 x 2 x 750) / 2 x 2

= - 27.77 ±√ ( 771.17 + 6000) / 4

= - 27.77 ±82.28 / 4

= 13.62 s

So this time is less than time taken by train so it will be able to cross the crossing before train arrives .

(1.5 pts) A woman pushes on a box to the left. If the box is accelerating, what forces are working on the
Question 2:
box? (Draw both y and x forces)

Answers

Answer:

Nope

Explanation:

how does the uneaven heating of earths surface affects earths weather patterns

Answers

Answer: it causes some parts of the earth to get more radiation than others.

Explanation: earth rotates around the sun on a tilted axis so the Rays of the sun cause earth to have more radiation than it needs.

Ruby has an index of refraction of 1.7. Calculate the maximum angle with respect to the normal at which a light ray escapes a ruby if the ruby is in air.

a. 17°
b. 36°
c. 53°
d. 90°
e. Light escapes at all incident angles.

Answers

Answer:

Umm sorry just came for the points.

Try google! I’m trying to find you an answer now and will put it in the comments. One moment!

At a young age, who has the greatest influence on your choices?
Why do stretching exercises increase flexibility more than cardio exercises?

Answers

An Adult like a parent or maybe and older sibling. They are older and are a role model. They influence you to make choices everyday. For example, if you see your parents work out, that might increase your thoughts about working out because they started working out and they are older and wiser so it influences you that working out would be good for your body.

Stretching exercises increase flexibility more than cardio exercises because being flexible requires you to stretch parts of your body. For instance, when you touch your toes your being flexible by stretching your arms to touch your toes.

Answer:

Explanation:

No da esta aplicación y si para que usted vella

Sam heats an 8kg sample of sand, with a specific heat of 664 J/kg·C°, from 20° to 40°. What is the change in thermal energy?

Answers

Answer:

106.24 kJ.

Explanation:

Given that,

Mass of sample of sand, m = 8 kg

Specific heat of sand, c = 664 J/kg-°C

The temperature changes from 20° C to 40° C. We need to find the change in thermal energy. It is given by :

[tex]Q=mc\Delta T\\\\Q=8\times 664(40-20)\\\\=106240\ J\\\\=106.24\ kJ[/tex]

So, the change in thermal energy is 106.24 kJ.

What is the difference between the two graphs

Answers

Answer:

one of the graph is postion-time graph while the other one is velocity-time graph


If I ride my bike at 10 mph and traveled 5 miles, how long did I ride in both hours and
minutes?

Answers

1/2 hour or 30 minutes is your answer!

5. Peter had an inflated balloon that he released and flies across the room. The balloon slows down and then stops on top of the dinning table, As the balloon slows downs the force becomes




A.balanced
B.frictional
C.restricted
D.unbalanced​

Answers

I think it is c because it was slowing down I don’t know for sure
balanced as it reaches terminal velocity

What horizontal speed must a pumpkin be thrown to hit a car 13.4 meters away from a building which stands 10.4 meters tall?
A) 1.5 m/s
B) 2.1 m/s
C)6.1 m/s
D) 8.9 m/s​

Answers

Answer:

V₀ₓ = 9.2 m/s

Nearest answer:

D) 8.9 m/s

Explanation:

First we find the time taken by the pumpkin to hit the car. For that purpose we apply 2nd equation of motion to the pumpkin:

h = V₀y t + (1/2)gt²

where,

h = height of building = 10.4 m

V₀y = vertical component of initial speed = 0 m/s

t = time = ?

g = 9.8 m/s²

Therefore,

10.4 m = (0 m/s)(t) + (1/2)(9.8 m/s²)t²

t² = (10.4 m)(2)/(9.8 m/s²)

t = √[2.122 s²]

t = 1.45 s

Now, we analyze horizontal motion for horizontal component of initial velocity. We assume air friction to be zero so that the horizontal motion is uniform. Therefore,

s = V₀ₓ t

where,

s = horizontal distance between building and car = 13.4 m

V₀ₓ = Horizontal Component of Initial Velocity = ?

Therefore,

13.4 m = V₀ₓ(1.45 s)

V₀ₓ = 13.4 m/1.45 s

V₀ₓ = 9.2 m/s

if a toy car has a centripetal acceleration of 50 m/s2 and was making the turn at 10 m/s. what was his radius

a 2m
ь 500m
C 5m
d.25m​

Answers

Answer:

Explanation:

The centripetal acceleration is expressed as;

a = v²/r

a is the acceleration = 50m/s²

v is the velocity = 10m/s

r is the radius

To get the radius

r = v²/a

r = 10²/50

r = 100/50

r = 2m

Hence its radius is 2m

What is potential energy? Use in your own words.

Answers

Answer:

Potential energy is the latent energy in an object at rest.

Explanation:

Hope this helps- this is how I would answer :)

A 27-g steel-jacketed bullet is fired with a velocity of 640 m/s toward a steel plate and ricochets along path CD with a velocity 500 m/s. Knowing that the bullet leaves a 50-mm scratch on the surface of the plate and assuming that it has an average speed of 600 m/s while in contact with the plate, determine the magnitude and direction of the impulsive force exerted by the plate on the bullet.

Answers

Answer:

F = - 3.56*10⁵ N

Explanation:

To attempt this question, we use the formula for the relationship between momentum and the amount of movement.

I = F t = Δp

Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say

v = d / t

t = d / v

Given that

m = 26 g = 26 10⁻³ kg

d = 50 mm = 50 10⁻³ m

t = d/v

t = 50 10⁻³ / 600

t = 8.33 10⁻⁵ s

F t = m v - m v₀

This is so, because the bullet bounces the speed sign after the crash is negative

F = m (v-vo) / t

F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵

F = - 3.56*10⁵ N

The negative sign is as a result of the force exerted against the bullet

Staying with the dragster problem...two more... If the dragster did the amount of work you calculated in the previous problem in 4.38 seconds, how much power was generated in Watts?

Answers

Answer:

1,439,283.10Watts

Explanation:

The question looks incomplete. We need the workdone to get the power generated.

Power = Workdone/Time

Power = Force * Distance/Time

Let

Force = 24340 Newtons

Distance covered = 259meters

Given

Time = 4.38secs

Required

Power generated

Substitute the given parameters into the formula;

Power generated = 24340*259/4.38

Power generated = 6,304,060/4.38

Power generated = 1,439,283.10

Hence the amount of power generated is 1,439,283.10Watts

In a certain time, light travels 3.50 km in a vacuum. During the same time, light travels only 2.35 km in a liquid. What is the refractive index of the liquid?

Answers

Answer:

1.45

Explanation:

Refractive index of the liquid is given as;

Refractive index = [tex]\frac{speed of light in vacuum}{speed of light in liquid}[/tex]

But,

speed = [tex]\frac{distance}{time}[/tex]

Since a certain light of specific wavelength was used during the same time, let the time be represented by t.

So that;

speed of light in vacuum = [tex]\frac{3500}{t}[/tex]

speed of light in the liquid = [tex]\frac{2350}{t}[/tex]

Refractive index = [tex]\frac{3500}{t}[/tex] ÷ [tex]\frac{2350}{t}[/tex]

                            = [tex]\frac{3500}{t}[/tex] x [tex]\frac{t}{2350}[/tex]

Refractive index = [tex]\frac{3500}{2350}[/tex]

                          = 1.4536

                          = 1.45

The refractive index of the liquid is 1.45.

Which element is used in the manufacture of mirrors and bronze?

Answers

Answer:

Silver

Explanation:

Silver is an important element in the manufacturing process of mirrors. Silver is used to make mirrors through the process we call "silvering". Silvering is a process in which a glass is coated with reflective substances so as to produce reflections, and then mirrors.

In Silvering process, Chlorine is also used. Stannous Chloride is the particular compound used to carry out the silvering, it has the chemical formula, SnCl₂

Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton

Answers

Answer:

a

[tex]\lambda = 3.68 *10^{-36} \ m[/tex]

b

[tex]\lambda_p = 1.28*10^{-14} \ m[/tex]

Explanation:

From the question we are told that

   The mass of the person is  [tex]m = 180 \ kg[/tex]

    The speed of the person is  [tex]v = 1 \ m/s[/tex]

    The energy of the proton is  [tex]E_ p = 5 MeV = 5 *10^{6} eV = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \ J[/tex]

Generally the de Broglie wavelength is mathematically represented as

      [tex]\lambda = \frac{h}{m * v }[/tex]

Here  h is the Planck constant with the value

      [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]

So  

     [tex]\lambda = \frac{6.62607015 * 10^{-34}}{ 180 * 1 }[/tex]

=> [tex]\lambda = 3.68 *10^{-36} \ m[/tex]

Generally the energy of the proton is mathematically represented as

         [tex]E_p = \frac{1}{2} * m_p * v^2_p[/tex]

Here [tex]m_p[/tex]  is the mass of proton with value  [tex]m_p = 1.67 *10^{-27} \ kg[/tex]

=>     [tex]8.0*10^{-13} = \frac{1}{2} * 1.67 *10^{-27} * v^2[/tex]

=>   [tex]v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }[/tex]

=>   [tex]v = 3.09529 *10^{7} \ m/s[/tex]

So

        [tex]\lambda_p = \frac{h}{m_p * v_p }[/tex]

so    [tex]\lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }[/tex]

=>     [tex]\lambda_p = 1.28*10^{-14} \ m[/tex]

     

A pressure antinode in a sound wave is a region of high pressure, while a pressure node is a region of low pressure.
True
False

Answers

A pressure antinode in a sound wave is not a region of high pressure, while a pressure node is not a region of low pressure.

The answer is false

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