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lord [1]
3 years ago
5

You throw a 20-N rock vertically into the air from ground level. You observe that when it is a height 14.8m above the ground, it

is traveling at a speed of 25.0 m/s upward.
A) Use the work-energy theorem to find its speed just as it left the ground. What is it?
B) Use the work-energy theorem to find its maximum height. What is it?
Physics
1 answer:
VladimirAG [237]3 years ago
4 0

Answer:

(A) The speed just as it left the ground is 30.25 m/s

(B) The maximum height of the rock is 46.69 m

Explanation:

Given;

weight of rock, w = mg  = 20 N

speed of the rock at 14.8 m, u = 25 m/s

(a) Apply work energy theorem to find its speed just as it left the ground

work = Δ kinetic energy

F x d = ¹/₂mv² - ¹/₂mu²

mg x d = ¹/₂m(v² - u²)

g x d = ¹/₂(v² - u²)

gd = ¹/₂(v² - u²)

2gd = v² - u²

v² = 2gd  + u²

v² = 2(9.8)(14.8) + (25)²

v² = 915.05

v = √915.05

v = 30.25 m/s

B) Use the work-energy theorem to find its maximum height

the initial velocity of the rock = 30.25 m/s

at maximum height, the final velocity = 0

- mg x H = ¹/₂mv² - ¹/₂mu²

- mg x H = ¹/₂m(0) - ¹/₂mu²

- mg x H = - ¹/₂mu²

2g x H = u²

H = u² / 2g

H = (30.25)² / 2(9.8)

H = 46.69 m

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baherus [9]

Answer:

193.4 J/K

Explanation:

For a closed system, the entropy is given as the natural logarithm of microstates.

The number of particles ( grains of sand ) N = 28

The number of boxes ( compartments ) M = 1000

Entropy is the logarithm of number of microstates

S=ln \psi

But the number of microstates is given by

\psi=M^{N} hence S=ln M^{N}=\ln (1000^{28})= \boxed{193.4 J/K}

4 0
3 years ago
Input Voltage = 220V
muminat

Answer:

Correct option is

A

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4 0
2 years ago
A mass spectrometer is being used to separate common oxygen-16 from the much rarer oxygen-18, taken from a sample of old glacial
Nataly_w [17]

Answer:

0.092 m

Explanation:

A charged moving particle immersed in a region with magnetic field follows a circular trajectory at constant speed (uniform circular motion), since the magnetic forces acts perpendicular to the direction of motion of the particle.

Since the magnetic force acts as centripetal force, we can write:

qvB=m\frac{v^2}{r}

where

q is the charge of the particle

v is its velocity

B is the strength of the magnetic field

m is the mass of the particle

r is the radius of the orbit

Solving the equation for r,

r=\frac{mv}{qB}

For the ion of oxygen-16, we have:

m_A=2.66\cdot 10^{-26}kg

q_A = 1.6\cdot 10^{-19}C (it is singly charged)

v_A=2.90\cdot 10^6 m/s

B_A=1.30 T

So the radius of its orbit is

r_A=\frac{m_A v_A}{q_A B_A}=\frac{(2.66\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.371 m

For the ion of oxygen-18, we have:

m_B = \frac{18}{16}m_A = 2.99\cdot 10^{-26}kg

q_B = 1.6\cdot 10^{-19}C (it is singly charged)

v_B=2.90\cdot 10^6 m/s

B_B=1.30 T

So the radius of its orbit is

r_B=\frac{m_B v_B}{q_B B_B}=\frac{(2.99\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.417 m

After each ion has travelled a semicircle, the separation between the two ions will be twice the difference in their radius, so:

d=2(r_B-r_A)=2(0.417-0.371)=0.092 m

3 0
3 years ago
If light has a speed of 122,000 mps in a transparent medium, what is the index of refraction of the medium? A. n = 1.52
castortr0y [4]

Answer:

A.\hspace{3}n=1.52

Explanation:

The refractive index of a medium is a measure to know how much the speed of light within the medium is reduced. It can be calculated with the next equation:

n=\frac{c}{v}   (1)

Where:

c=Speed\hspace{3}of\hspace{3}light\hspace{3}in\hspace{3}vacuum\\n=Refractive\hspace{3}index\\

v=Velocity\hspace{3}of\hspace{3}light\hspace{3}in\hspace{3}the\hspace{3}medium

The speed of light in the vacuum is approximately 300,000 km/s. In order to work with the same units let's do the proper conversion with the velocity of the medium:

122,000\frac{mi}{s} *\frac{1.60934km}{1mi}=196339.48\frac{km}{s}

Finally, replacing the data in (1):

n=\frac{300,000}{196339.48} =1.527965746\approx1.52

3 0
3 years ago
Which of the following best defines insulator?
aliya0001 [1]

Answer:

A substance with low ability or no ability to conduct energy.

Such as Rubber,Silicone,Plastic

6 0
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