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Ivanshal [37]
3 years ago
6

____, one of Saturn's icy moons, is unusual in the solar system in that it has volcanic activity that ejects plumes of icy parti

cles into space.
Physics
1 answer:
blondinia [14]3 years ago
4 0

Answer: Enceladus

Explanation:

Enceladus is a small, icy body with an undergound ocean beneath its crust. Cassini discovered that geyser-like jets spew water vapor and ice particles. It is also the sixth largest moon in Saturn and just about a tenth of the largest moon in Saturn; Titan. It is often regarded as one of the most reflective body in the solar system as a result of its icy surface.

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Use the following half-life graph to answer the following question:
Temka [501]

Answer:

A 1.0 min

Explanation:

The half-life of a radioisotope is defined as the time it takes for the mass of the isotope to halve compared to the initial value.

From the graph in the problem, we see that the initial mass of the isotope at time t=0 is

m_0 = 50.0 g

The half-life of the isotope is the time it takes for half the mass of the sample to decay, so it is the time t at which the mass will be halved:

m'=\frac{50.0 g}{2}=25.0 g

We see that this occurs at t = 1.0 min, so the half-life of the isotope is exactly 1.0 min.

3 0
3 years ago
A girl throws a marshmallow that lands in her friend’s mouth 2 m away. The girl threw the marshmallow at an angle of 30 degrees.
natima [27]

She threw the marshmallow at a speed of around 4.76 m/s.The formula for the horizontal range gives the velocity.

<h3>What is projectile motion?</h3>

The motion of an item hurled or projected into the air, subject only to gravity's acceleration, is known as projectile motion.

The item is known as a projectile, and the course it takes is known as a trajectory. Falling object motion is a simple one-dimensional kind of projectile motion with no horizontal movement.

Given data;

The marshmallow was thrown at a distance of 2 meters

Range,R =  3 m

Initial velocity,u

The angle at which the marshmallow was thrown,θ = 30°

The acceleration due to gravity,g = 9.81 m/s²

The projectile's motion is divided into two parts: horizontal and vertical motion.

\rm R = \frac{u^2sin2 \theta }{g} \\\\ \rm R  = \frac{u^2sin2 \times (30) }{9.81} \\\\ u^2=\frac{2 \times 9.81 }{sin 60^0} \\\\ u^2=22.66 \\\\ u=4.76 \ m/sec

Hence, she throws the marshmallow at a speed of 4.76 m/sec.

To learn more about the projectile motion refer to the link;

brainly.com/question/11049671

#SPJ1

4 0
1 year ago
An oxygen molecule is adsorbed onto a small patch of the surface of a catalyst. It's known that the molecule is adsorbed on of p
FromTheMoon [43]

The given question is incomplete. The complete question is as follows.

An oxygen molecule is adsorbed onto a small patch of the surface of a catalyst. It's known that the molecule is adsorbed on 1 of 36 possible sites for adsorption. Calculate the entropy of this system.

Explanation:

It is known that Boltzmann formula of entropy is as follows.

             s = k ln W

where,   k = Boltzmann constant

              W = number of energetically equivalent possible microstates or configuration of the system

In the given case, W = 36. Now, we will put the given values into the above formula as follows.

                  s = k ln W

                    = 1.38 \times 10^{-23} ln (36)        

                    = 4.945 \times 10^{-23} J/K

Thus, we can conclude that the entropy of this system is 4.945 \times 10^{-23} J/K.

4 0
3 years ago
1) Subatomic particles called muons can be created in the upper atmosphere by collisions of cosmic rays (energetic particles com
Vsevolod [243]

Answer and explanation:

A.

Muon travelled straight down towards the earth. Therefore the tree moves up in the rest frame of muon (option a)

B.

In muon rest frame it travels Zero meters

C.

Distance, d = Velocity, v * Time, s

where, v = 0.9c = 0.9 \times 8 \times 10^8 , s = 2.2 \mu s

d = 0.9 \times 3 \times 10^8 \times 2.2 \times 10^{-6}\\\\d = 594m

D.

Distance from the top of the mountain to the tree is the same as the distance travelled by the tree in the muons rest frame

that is same as in part C which is 594m

E.

Using lorentz contraction

In the rest frame of someone standing on the mountain

the distance is given by

d' = \frac{d}{\gamma} = d\sqrt{1 - \frac{v^2}{c^2}}, where, \frac{1}{\gamma}= \sqrt{1 - \frac{v^2}{c^2}}

d' = 594\sqrt{1 - \frac{(0.9c)^2}{c^2}}

d' = 594\sqrt{1 - 0.81}

d' = 594 \times 0.4359

d' = 258.92m

F.

in the rest frame of someone standing on the mountain,

muon moves straight down

3 0
3 years ago
g An arrow is shot straight up in the air at an initial speed of 15.5 m/s. After how much time is the arrow moving downward at a
Sati [7]

Answer:

2.11 seconds

Explanation:

We use the kinematic equation for the velocity in a constantly accelerated motion under the acceleration of gravity (g):

v_f=v_i-g*t\\-5.2 = 15.5 - 9.8\,t\\9.8 \,t= 15.5 + 5.2\\t = 20.7/9.8\\t = 2.11 \,\,sec

5 0
3 years ago
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