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V125BC [204]
3 years ago
14

At the instant a 2.0-kg particle has a velocity of 4.0 m/s in the positive x direction, a 3.0-kg particle has a velocity of 5.0

m/s in the positive y direction. What is the speed of the center of mass of the two-particle system
Physics
1 answer:
satela [25.4K]3 years ago
5 0

Answer:

the speed of the center of mass of the two-particle system is 3.4 m/s.

Explanation:

Given that,

Mass of particle 1, m = 2 kg

Velocity of the particle 1, v = 4 m/s in +x direction

Mass of particle 2, m' = 3 kg

Velocity of the particle 2, v' = 5 m/s in +y direction.

The x -coordinate of velocity of the centre of mass is given by :

v_x=\dfrac{mv+m'v'}{m+m'}\\\\v_x=\dfrac{2\times 4+3\times 0}{2+3}\\\\v_x=1.6\ m/s

The y -coordinate of velocity of the centre of mass is given by :

v_y=\dfrac{mv+m'v'}{m+m'}\\\\v_y=\dfrac{2\times 0+3\times 5}{2+3}\\\\v_y=3\ m/s

So, the the speed of the center of mass of the two-particle system given by :

v=\sqrt{v_x^2+v_y^2} \\\\v=\sqrt{1.6^2+3^2} \\\\v=3.4\ m/s

So, the speed of the center of mass of the two-particle system is 3.4 m/s.

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Solve for x. 4(x+1)≤4x+3
alexdok [17]
4x + 4 < 4x + 3 (expand it)
4 < 3 (cancel 4x on both sides)
Since 4 < 3 is not true there is no solution.

Answer: NO SOLUTION.
3 0
3 years ago
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An open organ pipe is 1.6m long. If the speed of sound is 343m/s, what are the pipes: a) fundamental , b) 1st overtone , &amp; c
Yakvenalex [24]

Answer:

a) 107.1875 Hz

b) 214.375 Hz

c) 321.5625 Hz

Explanation:

L = length of the open organ pipe = 1.6 m

v = speed of sound = 343 m/s

f = fundamental frequency

fundamental frequency is given as

f = \frac{v}{2L}

inserting the values

f = \frac{343}{2(1.6)}

f = \frac{343}{2(1.6)}

f = 107.1875 Hz

b)

first overtone is given as

f' = 2f

f' = 2 (107.1875)

f' = 214.375 Hz

c)

first overtone is given as

f'' = 3f

f'' = 3 (107.1875)

f'' = 321.5625 Hz

3 0
3 years ago
HELP ASAP!! WILL MARK BRAINLIEST!! WRITE IN YOUR OWN WORDS!! What is something that you would like to see a physicist develop in
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Answer:

Phones as sunglasses with a mic. I put on my glasses and I say what's the weather today, The sunglasses will tell me the weather and can be charged just like phones

Explanation:

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3 years ago
What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo
Phantasy [73]

Answer:

The magnetic field will be \large{\dfrac{1.4 \times 10^{-4}}{d}} T, '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field (\large{\overrightarrow{B}}) at a distance 'r' due to a current carrying conductor carrying current 'I' is given by

\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}

where '\overrightarrow{dl}' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current 'I', at a distance 'd' is given by

\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}

According to the figure if 'I_{t}' be the current carried by the top wire, 'I_{b}' be the current carried by the bottom wire and '2d' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the \bigotimes symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by \bigodot symbol.

Given \large{I_{t} = 19.5 A} and \large{I_{B} = 12.5 A}

Therefore, the magnetic field (\large{B_{t}}) at 'P' due to the top wire

B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}

and the magnetic field (\large{B_{b}}) at 'P' due to the bottom wire

B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}

Therefore taking the value of \mu_{0} = 4\pi \times 10^{-7} the net magnetic field (\large{B_{M}}) at the midway between the wires will be

\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T

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An inventor claims to have developed a power cycle having a thermal efficiency of 40%, while operating between hot and cold rese
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From Carnot's theorem, for any engine working between these two temperatures:


efficiency <= (1-tc/th) * 100


Given: tc = 300k (from question assuming it is not 5300 as it seems)

For a, th = 900k, efficiency = (1-300/900) = 70%

For b, th = 500k, efficiency = (1-300/500) = 40% 

For c, th = 375k, efficiency = (1-300/375) = 20% 


Hence in case of a and b, efficiency claimed is lesser than efficiency calculated, which is valid case and in case of c, however efficiency claimed is greater which is invalid. 

7 0
3 years ago
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