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Schach [20]
2 years ago
8

21

Physics
1 answer:
sergeinik [125]2 years ago
5 0
Distance between the centers of the two objects are the second object
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Which of the following is not a factor in whether a reaction will spontaneously occur? A. Entropy change of the system B. Enthal
Delvig [45]

Answer:

D

Explanation:

pressure change have nothing to do with the spontaneity.

Entropy change , enthalpy change , temperature have roles in deciding spontaneity.

6 0
3 years ago
A cyclotron is used to produce a beam of high-energy deuterons that then collide with a target to produce radioactive isotopes f
serg [7]

Answer:

19080667.0818 m/s

0.637294 m

2.1875\times 10^{15}

Explanation:

m = Mass of deuterons = 3.34\times 10^{-27}\ kg

v = Velocity

K = Kinetic energy = 3.8 MeV

d = Diameter

B = Magnetic field = 1.25 T

q = Charge of electron = 1.6\times 10^{-19}\ C

t = Time = 1 s

i = Current = 350 μA

Kinetic energy is given by

K=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2K}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 3.8\times 10^6\times 1.6\times 10^{-19}}{3.34\times 10^{-27}}}\\\Rightarrow v=19080667.0818\ m/s

The speed of the deuterons when they exit is 19080667.0818 m/s

In this system the centripetal and magnetic force will balance each other

\dfrac{mv^2}{r}=qvB\\\Rightarrow \dfrac{mv^2}{\dfrac{d}{2}}=qvB\\\Rightarrow d=\dfrac{2mv}{qB}\\\Rightarrow d=\dfrac{2\times 3.34\times 10^{-27}\times 19080667.0818}{1.6\times 10^{-19}\times 1.25}\\\Rightarrow d=0.637294\ m

The diameter is 0.637294 m

Current is given by

i=\dfrac{nq}{t}\\\Rightarrow n=\dfrac{it}{q}\\\Rightarrow n=\dfrac{350\times 10^{-6}\times 1}{1.6\times 10^{-19}}\\\Rightarrow n=2.1875\times 10^{15}

The number of deuterons is 2.1875\times 10^{15}

8 0
3 years ago
In the long jump, an athlete launches herself at an angle above the ground and lands at the same height, trying to travel the gr
NikAS [45]

A) 2.64t

B) 2.64h

C) 2.64D

Explanation:

A)

The motion of the athlete is equivalent to the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion (constant acceleration) along the vertical direction

The time of flight of a projectile can be found from the equations of motion, and it is found to be

t=\frac{2u sin \theta}{g}

where

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the time of flight is t.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. Therefore, the time of flight on Mars will be:

t'=\frac{2usin \theta}{g'}=\frac{2u sin \theta}{0.379g}=\frac{1}{0.379}t=2.64t

B)

The maximum height reached by a projectile can be also found using the equations of motion, and it is given by

h=\frac{u^2 sin^2\theta}{2g}

where

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the maximum height is h.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. So, the maximum height reached on Mars will be:

h'=\frac{u^2 sin^2\theta}{2g'}=\frac{u^2 sin^2\theta}{(0.379)2g}=\frac{1}{0.379}h=2.64h

C)

The horizontal distance covered by a projectile is also found from the equations of motion, and it is given by

D=\frac{u^2 sin(2\theta)}{g}

where:

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the horizontal distance covered is D.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. Therefore, the horizontal distance reached on Mars will be:

D'=\frac{u^2 sin(2\theta)}{g'}=\frac{u^2 sin(2\theta)}{(0.379)g}=\frac{1}{0.379}D=2.64D

7 0
3 years ago
How many nanoseconds does it take light to travel 3.50 ft in vacuum??
Tresset [83]

light speed in vacuum = 3.8 * 10^8

Distance (Given) = 3.5 ft

now, time = distance/speed = 3.50 / 3.8 * 10^8 = 9.21 * 10^-9 s = 9.21 nanoseconds

So, your answer is 9.21 nano-seconds...


8 0
4 years ago
List two examples of energy transformation necessary for the sporting event to take place
Morgarella [4.7K]

Electrical to light & electrical to sound (in commentry) & sometimes battery loudspeakers are used so chemical to sound
5 0
4 years ago
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