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3 years ago

PLEASE HELP ME WITH THIS PROBLEM

Physics

1 answer:

valentinak56 [21]3 years ago

3 0

1) The mass of the continent is $2.13\cdot 10^{21} kg$

2) The kinetic energy of the continent is 274.8 J

3) The speed of the jogger must be 2.76 m/s

Explanation:

The continent is a slab of side 5900 km (so the surface is 5900 x 5900, assuming it is a square) and depth 26 km, therefore its volume is:

$V=(36)(4600)^2=7.62\cdot 10^8 km^3 = 7.62\cdot 10^{17} m^3$

The mass of the continent is given by

$m=\rho V$

where:

$\rho = 2790 kg/m^3$ is its density

$V=7.62\cdot 10^{17} m^3$ is its volume

Substituting, we find the mass:

$m=(2790)(7.62\cdot 10^{17})=2.13\cdot 10^{21} kg$

To find the kinetic energy, we need to convert the speed of the continent into m/s first.

The speed is

v = 1.6 cm/year

And we have:

1.6 cm = 0.016 m

$1 year = (365)(24)(60)(60)=3.15\cdot 10^7 s$

So, the speed is

$v=\frac{0.016 m}{3.15 \cdot 10^7 s}=5.08\cdot 10^{-10}m/s$

Now we can find the kinetic energy of the continent, which is given by

$K=\frac{1}{2}mv^2$

where

$m=2.13\cdot 10^{21} kg$ is the mass

$v=5.08\cdot 10^{-10}m/s$ is the speed

Substituting,

$K=\frac{1}{2}(2.13\cdot 10^{21})(5.08\cdot 10^{-10})^2=274.8 J$

The jogger in this part has the same kinetic energy of the continent, so

K = 274.8 J

And its mass is

m = 72 kg

We can write his kinetic energy as

$K=\frac{1}{2}mv^2$

where

v is the speed of the man

And solving the equation for v, we find his speed:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(274.8)}{72}}=2.76 m/s$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

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A Hooke's law spring is compressed 12.0 cm from equilibrium, and the potential energy stored is 72.0 J. What compression (as mea

Anuta_ua [19.1K]

Answer:14.14 cm

Explanation:

Given

Spring Compression $x=12 cm$

Potential energy Stored in spring $=72 J$

Suppose k is the spring constant of spring

Potential Energy of spring is given by $=\frac{kx^2}{2}$

$\frac{k(0.12)^2}{2}=72$

$k(0.12)^2=144$

$k=10,000 N/m$

$k=10 kN/m$

for 100 J energy

$\frac{k(x_0)^2}{2}=100$

$10\times 10^3\cdot (x_0)^2=200$

$(x_0)^2=2\times 10^{-2}$

$x_0=0.1414$

$x_0=14.14 cm$

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3 years ago

A ball is falling after rolling off a tall roof<br> The ball has what type of energy.

CaHeK987 [17]

Answer:

Linear and rotational Kinetic Energy + Gravitational potential energy

Explanation:

The ball rolls off a tall roof and starts falling.

Let us first consider the potential energy or more specifically gravitational potential energy ( $mgh$ ; $m$ = mass of the ball, $g$ = acceleration due to gravity, $h$ = height of the roof). This energy comes because someone or something had to do work to take the ball to the top of the roof against the force of gravity. The potential energy is naturally maximum at the top and minimum when the ball finally reaches the ground.

Now, the ball starts to roll and falls off the roof. It shall continue rotating because of inertia (Newton's first law). This contributes to the rotational kinetic energy ( $\frac{1}{2}I\omega^2$ ; $I$ =moment of inertia of the ball & $\omega$ = angular velocity).

Finally comes the linear kinetic energy or simply, kinetic energy ( $\frac{1}{2}mv^2$ ) which is caused due to the velocity $v$ of the ball.

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3 years ago

An electron moving to the left at 0.8c collides with a photon moving to the right. After the collision, the electron is moving t

SVETLANKA909090 [29]

Answer:

Wavelength = 2.91 x 10⁻¹² m, Energy = 6.8 x 10⁻¹⁴

Explanation:

In order to show that a free electron can’t completely absorb a photon, the equation for relativistic energy and momentum will be needed, along the equation for the energy and momentum of a photon. The conservation of energy and momentum will also be used.

E = y(u) mc²

Here c is the speed of light in vacuum and y(u) is the Lorentz factor

y(u) = 1/√[1-(u/c)²], where u is the velocity of the particle

The relativistic momentum p of an object of mass m and velocity u is given by

p = y(u)mu

Here y(u) being the Lorentz factor

The energy E of a photon of wavelength λ is

E = hc/λ, where h is the Planck’s constant 6.6 x 10⁻³⁴ J.s and c being the speed of light in vacuum 3 x 108m/s

The momentum p of a photon of wavelenght λ is,

P = h/λ

If the electron is moving, it will start the interaction with some momentum and energy already. Momentum of the electron and photon in the initial and final state is

p(pi) + p(ei) = p(pf) + p(ef), equation 1, where p refers to momentum and the e and p in the brackets refer to proton and electron respectively

The momentum of the photon in the initial state is,

p(pi) = h/λ(i)

The momentum of the electron in the initial state is,

p(ei) = y(i)mu(i)

The momentum of the electron in the final state is

p(ef) = y(f)mu(f)

Since the electron starts off going in the negative direction, that momentum will be negative, along with the photon’s momentum after the collision

Rearranging the equation 1 , we get

p(pi) – p(ei) = -p(pf) +p(ef)

Substitute h/λ(i) for p(pi) , h/λ(f) for p(pf) , y(i)mu(i) for p(ei), y(f)mu(f) for p(ef) in the equation 1 and solve

h/λ(i) – y(i)mu(i) = -h/λ(f) – y(f)mu(f), equation 2

Next write out the energy conservation equation and expand it

E(pi) + E(ei) = E(pf) + E(ei)

Kinetic energy of the electron and photon in the initial state is

E(p) + E(ei) = E(ef), equation 3

The energy of the electron in the initial state is

E(pi) = hc/λ(i)

The energy of the electron in the final state is

E(pf) = hc/λ(f)

Energy of the photon in the initial state is

E(ei) = y(i)mc2, where y(i) is the frequency of the photon int the initial state

Energy of the electron in the final state is

E(ef) = y(f)mc2

Substitute hc/λ(i) for E(pi), hc/λ(f) for E(pf), y(i)mc² for E(ei) and y(f)mc² for E(ef) in equation 3

Hc/λ(i) + y(i)mc² = hc/λ(f) + y(f)mc², equation 4

Solve the equation for h/λ(f)

h/λ(i) + y(i)mc = h/λ(f) + y(f)mc

h/λ(f) = h/lmda(i) + (y(i) – y(f)c)m

Substitute h/λ(i) + (y(i) – y(f)c)m for h/λ(f) in equation 2 and solve

h/λ(i) -y(i)mu(i) = -h/λ(f) + y(f)mu(f)

h/λ(i) -y(i)mu(i) = -h/λ(i) + (y(f) – y(i))mc + y(f)mu(f)

Rearrange to get all λ(i) terms on one side, we get

2h/λ(i) = m[y(i)u(i) +y(f)u(f) + (y(f) – y(i)c)]

λ(i) = 2h/[m{y(i)u(i) + y(f)u(f) + (y(f) – y(i))c}]

λ(i) = 2h/[m.c{y(i)(u(i)/c) + y(f)(u(f)/c) + (y(f) – y(i))}]

Calculate the Lorentz factor using u(i) = 0.8c for y(i) and u(i) = 0.6c for y(f)

y(i) = 1/[√[1 – (0.8c/c)²] = 5/3

y(f) = 1/√[1 – (0.6c/c)²] = 1.25

Substitute 6.63 x 10⁻³⁴ J.s for h, 0.511eV/c2 = 9.11 x 10⁻³¹ kg for m, 5/3 for y(i), 0.8c for u(i), 1.25 for y(f), 0.6c for u(f), and 3 x 10⁸ m/s for c in the equation derived for λ(i)

λ(i) = 2h/[m.c{y(i)(u(i)/c) + y(f)(u(f)/c) + (y(f) – y(i))}]

λ(i) = 2(6.63 x 10-34)/[(9.11 x 10-31)(3 x 108){(5/3)(0.8) + (1.25)(0.6) + ((1.25) – (5/3))}]

λ(i) = 2.91 x 10⁻¹² m

So, the initial wavelength of the photon was 2.91 x 10-12 m

Energy of the incoming photon is

E(pi) = hc/λ(i)

E(pi) = (6.63 x 10⁻³⁴)(3 x 10⁸)/(2.911 x 10⁻¹²) = 6.833 x 10⁻¹⁴ = 6.8 x 10⁻¹⁴

So the energy of the photon is 6.8 x 10⁻¹⁴ J

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Sasha lifts a couch 8.2 meters from the ground floor of her house to the attic. If the couch has a mass of 120 kg, what is the g

gavmur [86]

As we know that gravitational potential energy is given by

$U = mgH$

here we have

m = mass = 120 kg

$g = 9.81 m/s^2$

h = height = 8.2 m

now from above formula

$U = 120kg (9.81 m/s^2) (8.2 m)$

$U = 9653.04 J$

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A ball rolls across a floor with an acceleration of 0.100 m/s2 in a direction opposite to its velocity. The ball has a velocity

Westkost [7]

Answer:

4.15 m/s

Explanation:

Its given that acceleration is 0.1 m/s² with a direction opposite to the velocity. Since, the direction of acceleration is opposite to the velocity, this gives us a hint that the velocity is decreasing and so acceleration would be negative.

i.e.

acceleration = a = - 0.1 m/s²

Distance covered = S = 6m

Velocity after covering 6 meters = Final velocity = $v_{f}$ = 4 m/s