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aleksley [76]
2 years ago
12

The sum of the kinetic and potential energies of a system of objects is conserved: Group of answer choices only when no external

force acts on the objects only when the objects move along closed paths only when the work done by the resultant external force is zero none of the above
Physics
1 answer:
uysha [10]2 years ago
7 0

The sum of the kinetic and potential energies of a system of objects is conserved only when no external force acts on the objects.

<h3>Conservation of mechanical energy</h3>

The principle of conservation of mechanical energy states that the total mechanical energy of an isolated system (absence of external force) is always constant.

M.A = P.E + K.E

where;

P.E is potential energy

K.E is kinetic energy

Thus, the sum of the kinetic and potential energies of a system of objects is conserved only when no external force acts on the objects.

Learn more about conservation of mechanical energy here: brainly.com/question/24443465

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Determine the maximum shearing stress in a solid shaft of 1.5-in. diameter as it transmits 75 hp at a speed of 1800 rpm. (Round
erma4kov [3.2K]

Answer:

\tau=3.96\ ksi

Explanation:

Given that

d= 1.5 in                      ( 1 in = 0.0254 m)

d= 0.0381 m

P= 75 hp                      ( 1 hp = 745.7 W)

P= 55927.5 W

N= 1800 rpm

We know that power P is given as

P=\dfrac{2\pi N\ T}{60}

T=Torque

N=Speed

55927.5=\dfrac{2\times \pi \times 1800\ T}{60}

T=296.85 N.m

The maximum shear stress is given as

\tau=\dfrac{16 T}{\pi d^3}

\tau=\dfrac{16\times 296.85}{\pi \times 0.0381^3}

\tau=27.35\ MPa

We know that 1 MPa =0.145 ksi

\tau=3.96\ ksi

3 0
3 years ago
Boyle's Law mainly involves _______.
goblinko [34]
Your answer is B, gases
6 0
3 years ago
Identify the law, write the equation and calculate the answer to the problem below.
lyudmila [28]

Find refractive index first

\\ \rm\Rrightarrow \mu=\dfrac{c}{v}

\\ \rm\Rrightarrow \mu=\dfrac{1.0003}{1.33}

\\ \rm\Rrightarrow \mu =0.75

Now

\\ \rm\Rrightarrow \dfrac{sini}{sinr}=\mu

\\ \rm\Rrightarrow \dfrac{sin45}{sinr}=0.75

\\ \rm\Rrightarrow \dfrac{sin45}{0.75}=sinr

\\ \rm\Rrightarrow sinr=0.94

\\ \rm\Rrightarrow r=sin^{-1}(0.94)

\\ \rm\Rrightarrow r=70^{\circ}

5 0
2 years ago
Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi
ki77a [65]

Answer:

The lowest possible frequency of sound for which this is possible is 1307.69 Hz

Explanation:

From the question, Abby is standing 5.00m in front of one of the speakers, perpendicular to the line joining the speakers.

First, we will determine his distance from the second speaker using the Pythagorean theorem

l₂ = √(2.00²+5.00²)

l₂ = √4+25

l₂ = √29

l₂ = 5.39 m

Hence, the path difference is

ΔL = l₂ - l₁

ΔL = 5.39 m - 5.00 m

ΔL = 0.39 m

From the formula for destructive interference

ΔL = (n+1/2)λ

where n is any integer and λ is the wavelength

n = 1 in this case, the lowest possible frequency corresponds to the largest wavelength, which corresponds to the smallest value of n.

Then,

0.39 = (1+ 1/2)λ

0.39 = (3/2)λ

0.39 = 1.5λ

∴ λ = 0.39/1.5

λ = 0.26 m

From

v = fλ

f = v/λ

f = 340 / 0.26

f = 1307.69 Hz

Hence, the lowest possible frequency of sound for which this is possible is 1307.69 Hz.

5 0
3 years ago
A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c
saw5 [17]

Answer:

The final velocity of the thrower is \bf{3.88~m/s} and the final velocity of the catcher is \bf{0.029~m/s}.

Explanation:

Given:

The mass of the thrower, m_{t} = 70~Kg.

The mass of the catcher, m_{c} = 55~Kg.

The mass of the ball, m_{b} = 0.0480~Kg.

Initial velocity of the thrower, v_{it} = 3.90~m/s

Final velocity of the ball, v_{fb} = 33.5~m/s

Initial velocity of the catcher, v_{ic} = 0~m/s

Consider that the final velocity of the thrower is v_{ft}. From the conservation of momentum,

&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s

Consider that the final velocity of the catcher is v_{fc}. From the conservation of momentum,

&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s

Thus, the final velocity of thrower is 3.88~m/s and that for the catcher is 0.029~m/s.

8 0
2 years ago
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