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castortr0y [4]
3 years ago
14

A particle moving on a plane curve what is the degree of freedom

Physics
1 answer:
kirill115 [55]3 years ago
7 0

A particle confined to move along a curved path has only one degree of freedom. inclined plane are some examples of constrained motion. Every condition of constraint reduces the number of degree of freedom by one.

I hope this helps!

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A sailboat is traveling across the ocean with a speed of 4 m/s when large gusts of wind pick up that starts to accelerate the sa
MatroZZZ [7]
Initial velocity, u = 4 m/s

acceleration due to gusts of wind = 3 m/s^2

time, t = 1 min = 60 s

Let distance travelled = S

From equation of motion,

S = ut +  \frac{1}{2} a {t}^{2}  \\  \\ S = 4 \times 60 +   \frac{1}{2}  \times 3 \times  {60}^{2}  \\  \\ S = 240 + 5400 \\  \\ S = 5640 \: m

Thus, the boat would have traveled 5640m after gusts picked up.
4 0
3 years ago
Read 2 more answers
Hi please help with this question! Need the workings.
andreyandreev [35.5K]
<span>Mass of the copper penny m = 2.6 g Atomic mass of copper = 63.55, Atomic number = 29, So the number of neutrons = Atomic mass - Atomic number = 63 - 29 = 34 a. Neutron mass = 34 x (2.6 / 63.55) = 1.4 grams Copper atoms per mole = 6.040 x 10^23 atoms/mol moles of copper = 2.6 / 63.06 = 0.04123 mol Total atoms in the copper = 6.040 x 10^23 atoms/mol x 0.04123 mol = 0.25 x 10^23 atoms Number of electrons in the copper = 29 per atom Mass of the electron = 9.085 x 10^-28 g b. Electron mass = 0.25 x 10^23 x 29 x 9.085 x 10^-28 = 65.86 x 10^-5 g</span>
6 0
3 years ago
What is the velocity of a quarter drooped from a towards after 10secounds
ArbitrLikvidat [17]
U=0 
<span>t=10 </span>
<span>a=9.8m/s/s </span>
<span>v is velocity (the tower must be very high to be able to fall for 10 seconds!!!) </span>

<span>you work out the result now</span>
6 0
3 years ago
Chen is testing the friction of three surfaces. He pushes the same ball across three different surfaces with the same force and
Law Incorporation [45]

Answer:

im no proffesional

Explanation:

but i tghink you need a proffessional for this one

4 0
3 years ago
Two particles each of mass m and charge q are suspended by strings of length / from a common point. Find the angle e that each s
ozzi

Answer:

\theta =\left (\frac{kq^{2}}{4L^{2}\times mg}  \right )^{\frac{1}{3}}

Explanation:

Let the length of the string is L.

Let T be the tension in the string.

Resolve the components of T.

As the charge q is in equilibrium.

T Sinθ = Fe       ..... (1)

T Cosθ = mg     .......(2)

Divide equation (1) by equation (2), we get

tan θ = Fe / mg

tan\theta =\frac{\frac{kq^{2}}{AB^{2}}}{mg}

tan\theta =\frac{\frac{kq^{2}}{4L^{2}Sin^{\theta }}}}{mg}

tan\theta =\frac{kq^{2}}{4L^{2}Sin^{2}\theta \times mg}

tan\theta\times Sin^{2}\theta =\frac{kq^{2}}{4L^{2}\times mg}

As θ is very small, so tanθ and Sinθ is equal to θ.

\theta ^{3} =\frac{kq^{2}}{4L^{2}\times mg}

\theta =\left (\frac{kq^{2}}{4L^{2}\times mg}  \right )^{\frac{1}{3}}

7 0
3 years ago
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