Catching a wave, a 77-kg surfer starts with a speed of 1.3 m>s, drops through a height of 1.65 m, and ends with a speed of 8.2 m>s. How much nonconservative work was done on the surfer

Answers

Answer 1

Answer:

W = 2523.67 J

Explanation:

Given that,

The mass of surfer, m = 77 kg

He starts with a speed of 1.3 m/s

It drops through a height of 1.65 m and ends with a speed of 8.2 m/s.

We know that, the work done is equal to the change in kinetic energy. So,

[tex]W=\dfrac{1}{2}m(v_2^2-v_1^2)\\\\=\dfrac{1}{2}\times 77\times (8.2^2-1.3^2)\\W=2523.67\ J[/tex]

So, the required work done is equal to 2523.67 J.


Related Questions

An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X is stationary. An observer on Planet X sees rocket 2 travel toward Earth at 0.4c. What is the speed of rocket 1 according to an observer on rocket 2?

Answers

Answer:

0.625 c

Explanation:

Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.

In the context,

The relative speed of body 2 with respect to body 1 can be expressed as :

[tex]$u'=\frac{u-v}{1-\frac{uv}{c^2}}$[/tex]

Speed of rocket 1 with respect to rocket 2 :

[tex]$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$[/tex]

[tex]$u' = \frac{0.7 c}{1.12}$[/tex]

[tex]u'=0.625 c[/tex]

Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c

Snell's Law: Light goes from material having a refractive index of n 1 into a material with refractive index n 2. If the refracted light is bent away from the normal, what can you conclude about the indices of refraction

Answers

Answer:

a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.

b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2

Explanation:

The expression for the angle of refraction is

          n₁ sin θ₁ = n₂ sin θ₂

refractive index n₁ is for incident light and n₂ is for transmitted light.

We have two cases

a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.

b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2

Our system is a block attached to a horizontal spring on a frictionless table. The spring has a spring constant of 4.0 N/m and a rest length of 1.0 m, and the block has a mass of 0.25 kg.

Compute the PE when the spring is compressed by 0.50 m.

Answers

Answer

E - 1/2 K x^2      potential energy of compressed spring

E = 1/2 * 4 N / m * (.5 m)^2 = 2 * .5^2 N-m = .5 N-m

Condensation is the process of ____________________.
a. planetesimals accumulating to form protoplanets.
b. planets gaining atmospheres from the collisions of comets.
c. clumps of matter adding material a small bit at a time.
d. clumps of matter sticking to other clumps.
e. clouds formed from volcanic eruptions.

Answers

D. I hope my answer helps you!

A student graphs power (p) on the vertical axis and time (t) on the horizontal axis. The graph appears to be a hyperbola.

a) What should the student graph on each axis to test whether the relationship is actually
hyperbolic?

b) If the relationship is actually hyperbolic, what is the general equation for the relationship between power and time?

Answers

Answer: it would be daddy

Explanation:

Because I’m daddy

a person lifts 60kg on the surface of the earth, how much mass can he lift on the surface of the moon if he applies same magnitude of force​

Answers

Explanation:

Hey there!

According to the question;

A person can lift mass of 60 kg on earth.

mass(m1) = 60kg

acceleration due to gravity on earth (a) = 9.8m/s²

Now;

force (f) = m.a

= 60*9.8

= 588 N

Since, there is application of same magnitude of force on moon,

mass(m) =?

acceleration due to gravity on moon (a) = 1.67m/s²

Now;

force (f) = m.a

588 = m*1.67

m = 352.09 kg

Therefore, the person who can lift the mass of 60 kg on earth can lift mass of 352 kg on moon.

Hope it helps!

Which of the following groups is the largest ?

population
community
ecosystem
biome

Answers

Answer:

B. Community

Took science classes for 6 years now

Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B​

Answers

[tex]\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

Explanation:

Given:

[tex]\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}[/tex]

[tex]\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

The cross product [tex]\textbf{A}×\textbf{B}[/tex] is given by

[tex]\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|[/tex]

[tex]= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}[/tex]

[tex]= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

a concave mirror has a radius of curvature of 60cm. How close to the mirror should an object be placed so that the rays travel parallel to each other after reflection

Answers

Answer:

Answer:30 cm

Answer:30 cmExplanation:

Answer:30 cmExplanation:Given=ROC= 60cm

Answer:30 cmExplanation:Given=ROC= 60cmObject be placed so that the rays that came from the object to them mirror are reflected from the mirror, and, then travel parallel to each other= 30cm at focus.

Large cockroaches can run as fast as 1.50 m/s in short bursts. Suppose you turn on the light in a cheap motel and
see one scurrying directly away from you at a constant 1.50 m/s. If you start 0.90 m behind the cockroach with
an initial speed of 0.80 m/s toward it, what minimum constant acceleration would you need to catch up with it
when it has traveled 1.20 m, just short of safety onder a counter?

Answers

Answer:

The time that you need to use 1.2/1.5 because this is how long it took the cockroach to travel the 1.2 meters to the counter. That is therefore how long you have to catch up to it.

Explanation:

Consider newtonian mechanics here.

Dynamic equation is

The time we have to use 1.2/1.5 this how long it took the cockroach to travel the 1.2 meters to the counter.

we'll consider newtonian mechanics here.

so the dynamic equations is S = ut + 0.5at^2

we know u=0.8

S=1.2+0.9

t=1.2/1.5

find a.

A 2.0 kg puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. How far does the puck move from rest in 2.25 s?

Answers

Answer:

d = 6.32 m

Explanation:

Given that,

The mass of a puck, m = 2 kg

It is pushed straight north with a constant force of 5N for 1.50 s and then let go.

We need to find the distance covered by the puck when move from rest in 2.25 s.

We know that,

F = ma

[tex]a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2[/tex]

Let d is the distance moved in 2.25 s. Using second equation of motion,

[tex]d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m[/tex]

So, it will move 6.32 m from rest in 2.25 seconds.

If the mass of an object is 15 kg and the velocity is -4 m/s, what is the momentum?

Answers

momentum p= m x v = 15 x -4 = -60 N.s

a student standing between two walls shouts once.he hears the first echo after 3 seconds and the next after 5 seconds. calculate the distance between the walls.​

Answers

Explanation:

It took [tex]t_1 =1.5\:\text{s}[/tex] for the sound to reach the 1st wall and at the same time time, the same sound took [tex]t_2 = 2.5\:\text{s}[/tex] to reach the 2nd wall. Assuming that the sound travels at 343 m/s, then let [tex]x_1[/tex] be the distance of the person to the 1st wall and [tex]x_2[/tex] be the distance to the 2nd wall. So the distance between the walls X is

[tex]X = x_1 + x_2 = v_st_1 + v_st_2 = v_s(t_1 + t_2)[/tex]

[tex]\:\:\:\:\:= (343\:\text{m/s})(4.0\:\text{s}) = 1372\:\text{m}[/tex]

A 30-year-old astronaut goes off on a long-term mission in a spacecraft that travels at speeds close to that of light. The mission lasts exactly 20 years as measured on Earth. Biologically speaking, at the end of the mission, the astronaut's age would be:_______.
a) exactly 50 years.
b) exactly 25 years.
c) exactly 30 years.
d) less than 50 years.
e) more than 50 years.

Answers

Answer:

I think D) less than 50 years

Biologically speaking, at the end of the mission, the astronaut's age would be less than 50 years. The correct option is d.

Who is an astronaut?

An astronaut observes and performs the experiments based on the universe.

A 30-year-old astronaut goes off on a long-term mission in a spacecraft that travels at speeds close to that of light. The mission lasts exactly 20 years as measured on Earth.

Due to special relativity, between space and Earth, both moving with different speeds.

The total age will be less than 30 +20 =50 years. In space, he is moving with speed of light. So, time will move slowly. As measured with respect to Earth, exact time spent in space 20 years will be less on Earth.

So, biologically speaking, at the end of the mission, the astronaut's age would be less than 50 years.

Thus, the correct option is d.

Learn more about astronaut.

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What is the energy of a photon with a frequency of 3.6 × 1015 Hz? Planck’s constant is 6.63 × 10–34 J•s.

1.8 × 10–49 J
2.4 × 10–19 J
1.8 × 10–18 J
2.4 × 10–18 J

Answers

Frequency =v=3.6×10^15Hz

We know

[tex]\boxed{\sf E=hv}[/tex]

[tex]\\ \sf\longmapsto E=6.63\times 10^{-34}J\times 3.6\times 10^{15}s^{-1}[/tex]

[tex]\\ \sf\longmapsto E=23.86\times 10^{-19}J[/tex]

[tex]\\ \sf\longmapsto E=2.38\times 10^{-18}J[/tex]

[tex]\\ \sf\longmapsto E=2.4\times 10^{-18}J[/tex]

Answer:

D!!!!!

Explanation:

Flag question
Consider the pressure and force acting on the
dam retaining a reservoir of water. Suppose the
dam is 500-m wide
and the water is 80.0-m
deep at the dam, as illustrated below. What is
the average pressure on the dam due to the
water?​

Answers

Answer:

P = density (p) * g * h

P = 1000  kg/m^3 * 9.8 m/s^2 * 40 m = 392,000 N/m^2

since kg m / s^2 = Newtons

The average pressure is 1/2 (pressure at 0m +  pressure 80 m) for liquid of uniform density

Keisha writes that if an object has any external forces acting on it, then the object can be in dynamic equilibrium but not
static equilibrium
Which statement best describes Keisha's error?
An object that is not moving is always in static equilibrium.
O An object that is moving must be in dynamic equilibrium.
An object in either state of equilibrium must have no forces acting on it.
An object in either state of equilibrium must have no net force acting on it.

Answers

Answer:

An object in either state of equilibrium must have no net force acting on it.

Explanation:

Answer: An object in either state of equilibrium must have no net force acting on it.

Explanation:

During normal beating, the heart creates a maximum 4.10-mV potential across 0.350 m of a person's chest, creating a 1.00-Hz electromagnetic wave. (a) What is the maximum electric field strength created? V/m (b) What is the corresponding maximum magnetic field strength in the electromagnetic wave? T (c) What is the wavelength of the electromagnetic wave?

Answers

Explanation:

Given that,

Maximum potential, V = 4. mV

Distance, d = 0.350 m

Frequency of the wave, f = 100 Hz

(a) The maximum electric field strength created is given by:

[tex]E=\dfrac{V}{d}\\\\E=\dfrac{4.1\times 10^{-3}}{0.350 }\\\\E=0.0117\ V/m[/tex]

(b) The corresponding maximum magnetic field strength in the electromagnetic wave is given by :

[tex]B=\dfrac{E}{c}\\\\B=\dfrac{0.0117}{3\times 10^8}\\\\B=3.9\times 10^{-11}\ T[/tex]

(c) The wavelength of the electromagnetic wave can be calculated as :

[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{100}\\\\=3\times 10^6\ m[/tex]

So, the wavelength of the electromagnetic wave is [tex]3\times 10^6\ m[/tex].

In which situation should a parent be proactive and act to assume responsibility?

Answers

Answer: Patsy is eager to learn how to bake a cake but does not know how to do it.

Explanation: i picked this and it is correct, you’re welcome:)

Outside a spherically symmetric charge distribution of net charge Q, Gauss's law can be used to show that the electric field at a given distance:___________.
A) must be directed inward.
B) acts like it originated in a point charge Q at the center of the distribution.
C) must be directed outward.
D) must be greater than zero.
E) must be zero.

Answers

Answer:

Q at the center of the distribution.

Explanation:

The Gauss's law is the law that relates to the distribution of electrical charges to the resulting electrical field. It states that a flux of electricity outside the arabatory closed surface is proportional to the electricitical harg enclosed by the surface.

An object is acted upon by two and only two forces that are equal magnitude and oppositely directed. Is the objected necessarily in static equilibrium? Explain. You can draw a picture if that helps explain.

Answers

Answer:

the body is subjected to a continuous rotation and the body is not in rotational equilibrium

Explanation:

For an object to have a static equilibrium, it must meet two relationships

             ∑ F = 0

             ∑ τ =0              

force acting on a body fulfills the relation of

         sum F = F - F = 0

when two forces do not move from position.

To find the torque we assume that the counterclockwise rotations are positive

        Σ τ = - F r - F r

        Στ = -2 Fr <> 0

consequently the body is subjected to a continuous rotation and the body is not in rotational equilibrium

What is not one of the main uses of springs?
A. Car suspension
B. Bike suspension
C. The seasons
D. Clock making

Answers

Hi! I believe the answer is C. or The seasons because all the other options use springs when making cars, bikes, and clocks. I hope this helped. Goodluck :)

1 of 3 : please help got an extra day for a test and i don’t get this (must show work) points and brainliest!

Answers

Answer:

y = 1/2at^2

we could also write it as-

y = (at^2)/2

2y = at^2

2y/a = t^2

√2y/a = t

hope it helps

which of the following cannot be increased by using a machine of some kind? work, force, speed, torque

Answers

Explanation:

Work cannot be increased by using a machine of some kind.

Work cannot be increased by using a machine of some kind.

A machine is any device in which the effort applied at one end overcomes a load at the other end.

Machines are generally used to perform different tasks faster.

However, a simple machine can not be used to increase the amount of work done at any time.

Force, speed and torque can all be increased using machines.

Learn more: https://brainly.com/question/15365822

The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 6 months. The lines of both stars shift by equal amounts, and the amount of the Doppler shift indicates that each star has an orbital speed of 64,000 m/s. What are the masses of the two stars

Answers

Answer:

the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg

Explanation:

Given the data in the question;

Time period = 6 months = 1.577 × 10⁷ s

orbital speed v = 64000 m/s

since its a circular orbit,

v = 2πr / T

we solve for r

r = vT/ 2π

r = ( 64000 × 1.577 × 10⁷ ) / 2π

r = 1.6063 × 10¹¹ m = ( (1.6063 × 10¹¹) / (1.496 × 10¹¹) )AU = 1.0737 AU

Now, from Kepler's law

T² = r³ / ( m₁ + m₂ )

T = 6 months = 0.5 years

we substitute

(0.5)² = (1.0737)³ / ( m₁ + m₂ )

0.25 = 1.2378 / ( m₁ + m₂ )

( m₁ + m₂ ) = 1.2378 / 0.25

( m₁ + m₂ ) = 4.9512

m₁ = m₂  = 4.9512 / 2 = 2.4756 solar mass

we know that solar mass = 1.989 × 10³⁰ kg

so

m₁ = m₂ = 2.4756 × 1.989 × 10³⁰ kg

m₁ = m₂ = 4.92 × 10³⁰ kg

Therefore, the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg

State the law of conservation of momentum

Answers

Explanation:

Conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant

The potential difference between the plates of a capacitor is 234 V. Midway between the plates, a proton and an electron are released. The electron is released from rest. The proton is projected perpendicularly toward the negative plate with an initial speed. The proton strikes the negative plate at the same instant the electron strikes the positive plate. Ignore the attraction between the two particles, and find the initial speed of the proton.
I have tried looking at the cramster.com solution manual and do not like the way it is explained. Simply put, I cannot follow what is going on and I am looking for someone who can explain it in plain man's terms and help me understand and get the correct answer. I am willing to give MAX karma points to anyone who can help me through this. Thank you kindly.

Answers

Answer:

The speed of proton is 2.1 x 10^5 m/s .

Explanation:

potential difference, V = 234 V

let the initial speed of the proton is v.

The kinetic energy of proton is

KE = q V

[tex]0.5 mv^2 = e V \\\\0.5\times 1.67\times 10^{-27} v^2 = 1.6\times 10^{-19} \times 234\\\\v=2.1\times 10^5 m/s[/tex]

What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from its siren? The speed of sound on this day is 345 m/s. Group of answer choices

Answers

Answer:

check photo for solve

Explanation:

Present the ways to observe interference patterns on thin films. Why the thickness of thin films should be in scale of wavelength?

Answers

the president of The proposal ahr stay Sheba SBR Abba and and I send svrvs you svrvs

A magnifying glass produces a maximum angular magnification of 5.4 when used by a young person with a near point of 20 cm. What is the maximum angular magnification obtained by an older person with a near point of 65 cm

Answers

67 points to the 55th power add it then you get exactly 1500
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