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Iteru
3 days ago
15

A careful photographic survey of Jupiter’s moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to

heights of 70 km above the surface of this moon. If the value of g on Io is 2.0 m/s2 , estimate the speed with which the liquid sulfur left the volcano.
Physics
1 answer:
Sav [1.1K]3 days ago
7 0

Answer:

529.15 m/s

Explanation:

h = Highest point = 70000 m

g = Gravitational acceleration = 2 m/s²

m = Sulfur's mass

Since both potential and kinetic energies are conserved

mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 2\times 70000}\\\Rightarrow v=529.15\ m/s

The velocity at which the liquid sulfur exited the volcano is 529.15 m/s

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Consider a 4-mg raindrop that falls from a cloud at a height of 2 km. When the raindrop reaches the ground, it won't kill you or
inna [987]

Answer:

The work performed by air resistance totals -0.0782 J

Explanation:

Hello!

According to the principle of conservation of energy, the energy of a raindrop must remain constant.

At the outset, the raindrop possesses only gravitational potential energy:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the raindrop.

g = gravitational acceleration (9.8 m/s²)

h = height.

Let's determine the initial potential energy of the raindrop:

(4 mg should be converted into kg: 4 mg · 1 kg / 1 × 10⁶ mg = 4 × 10⁻⁶ kg)

PE = 4 × 10⁻⁶ kg · 9.8 m/s² · 2000 m

PE = 0.0784 J

As the raindrop descends, some of its potential energy converts into kinetic energy while the rest is lost to the air resistance. Upon reaching the ground, all initial potential energy has been either turned into kinetic energy or spent overcoming air resistance:

initial PE = final KE + Work by air

Where:

KE = kinetic energy.

Work by air = work done by air resistance.

The kinetic energy at ground level is computed as follows:

KE = 1/2 · m · v²

Where:

m = mass

v = velocity

<pThus:

KE = 1/2 · 4 × 10⁻⁶ kg · (10 m/s)²

KE = 2 × 10⁻⁴ J

Now, we can find the work done by air resistance:

initial PE = final KE + Work by air

0.0784 J = 2 × 10⁻⁴ J + Work by air

Work by air = 0.0784 J - 2 × 10⁻⁴ J

Work by air = 0.0782 J

Since work is performed in the opposite direction to movement, this results in a negative value. Therefore, the work done by air resistance is -0.0782 J.

5 0
9 days ago
A 7.5 kg cannon ball leaves a canon with a speed of 185 m/s. Find the average net force applied to the ball if the cannon muzzle
Keith_Richards [1034]

To determine the average net force, we can calculate acceleration using:

x = 0.5*a*t^2

v = a*t

where x=3.6m and v=185 m/s.

Thus,

t=v/a and therefore x = 0.5*a*(v/a)^2 = 0.5 * (v^2)/a

which gives us a= (0.5*v^2)/x

Since we have the known values of v and x, we can compute a by substituting these numbers.

The average net force is then given as:

F = m*a,

with m=7.5kg.


5 0
1 day ago
A simple harmonic wave of wavelength 18.7 cm and amplitude 2.34 cm is propagating along a string in the negative x-direction at
Ostrovityanka [942]

Answer

Given:

Wavelength = λ = 18.7 cm

                  = 0.187 m

Amplitude, A = 2.34 cm

Velocity, v = 0.38 m/s

A)  Calculate the angular frequency.

     f = \dfrac{v}{\lambda}

     f = \dfrac{0.38}{0.187}

     f =2.03\ Hz

Angular frequency,

ω = 2π f

ω = 2π x 2.03

ω = 12.75 rad/s

B) Calculate the wave number:

      K = \dfrac{2\pi}{\lambda}

     K= \dfrac{2\pi}{0.187}

    K =33.59\ m^{-1}

C)

Since the wave is traveling in the -x direction, the sign is positive between x and t

y (x, t) = A sin(k x - ω t)

y (x, t) = 2.34 sin(33.59 x - 12.75 t)

4 0
9 days ago
A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per
Softa [913]

Result: -50.005 kJ

Details:

Provided Data

mass of the system = 10 kg

work done = 0.147 kJ/kg

Elevation change (\Delta h)=-50 m

initial speed (v_1)=15 m/s

Final Speed (v_2)=30 m/s

Specific internal Energy (\Delta U)=-5 kJ/kg

according to the first Law of thermodynamics

Q-W=\Delta H

\Delta H=\Delta KE+ \Delta PE +\Delta U

where KE represents kinetic energy

PE indicates potential energy

U denotes internal Energy

\Delta H=\frac{m(v_2^2-v_1^2)}{2}+mg(\Delta h)+\Delta U

Q=W+\Delta KE+ \Delta PE +\Delta U

Q=0.147\times 10+\frac{10\cdot (30^2-15^2)}{2}+10\cdot 9.81\cdot (-5)-5\times 10

Q = 1.47 + 3.375 - 4.850 - 50

Q = -50.005 kJ

5 0
13 days ago
Assume that the cart is free to roll without friction and that the coefficient of static friction between the block and the cart
Keith_Richards [1034]

Answer:F=\frac{(M+m)g}{\mu _s}

Explanation:

Provided:

The trolley, with mass M, is allowed to roll freely without friction.

The coefficient of friction between the trolley and mass m is \mu _s.

A force F is applied to mass m.

The acceleration of the system is

a=\frac{F}{M+m}

The frictional force will counterbalance the weight of the block.

The frictional force is =\mu _sN

N=ma

\mu _sN=mg

\mu _sma=mg

\mu _s=\frac{g}{a}

F=\frac{(M+m)g}{\mu _s}

6 0
4 days ago
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