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Paladinen [302]
3 years ago
11

A particle is moving with SHM of period pie . initially it is 10 cm from The center of the motion and moving in the positive dir

ection with speed of 6cm/s find an equation to describe motion
Physics
1 answer:
Viefleur [7K]3 years ago
7 0

Answer:

y = 10.44cos(2t - 0.291) cm

Explanation:

y = Acos(2πt/T + φ) = Acos(2πt/π + φ) = Acos(2t + φ)

v = y' = -2Αsin(2t + φ)

10 = Acos(2(0) + φ) = Acosφ

6 = -2Αsin(2(0) + φ) = -2Asinφ

6/10 = -2Asinφ/Acosφ = -2tanφ

tanφ = -0.3

φ = -0.291 radians

10 = Acos(-0.291)

A = 10/cos(-0.291) = 10.44

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RideAnS [48]
<span>Answer: skater x km/h cyclist 20 faster x + 20 km/h skater 30 km cyclist 80 km skater time = cyclist time t=d/r 30 / x = 80 /( x + 20 cross multiply 30 ( x + 20 ) = 80 x 30 x + 600 = 80 x 30 x - 80 x = -600 -50 x = -600 / -50 x = 12 km/h 12 km/h skater</span>
3 0
3 years ago
If you act without reason or sound judgement, people will describe you as __________.
EastWind [94]

Answer:

irrational

Explanation:

8 0
2 years ago
What is the mass moment of inertia of a 20kg sphere with a radius of 0.2m about a point on the sphere's perimeter
Kobotan [32]

Answer:

I = M R^2 is the moment of inertia about a point that is a distance R from the center of mass (uniform distributed mass).

The moment  of inertia about the center of a sphere is 2 / 5 M R^2.

By the parallel axis theorem the moment of inertia about a point on the rim of the sphere is  I = 2/5 M R^2 + M R^2 = 7/5 M R^2

I = 7/5 * 20 kg * .2^2 m = 1.12 kg m^2

7 0
2 years ago
A 4.0 m length of gold wire is connected to a 1.5 V battery, and a current of 4.0 mA flows through it. What is the diameter of t
sladkih [1.3K]

Explanation:

Given that,

Length of gold wire, l = 4 m

Voltage of battery, V = 1.5 V

Current, I = 4 mA

The resistivity of gold, \rho=2.44\times 10^{-8}\ \Omega-m

Resistance in terms of resistivity is given by :

R=\dfrac{\rho l}{A}

Also, V = IR

So,

\dfrac{V}{I}=\dfrac{\rho l}{A}

A is area of wire,

\dfrac{V}{I}=\dfrac{\rho l}{\pi r^2}, r is radius, r = d/2 (diameter=d)

\dfrac{V}{I}=\dfrac{\rho l}{\pi (d/2)^2}\\\\\dfrac{V}{I}=\dfrac{4\rho l}{\pi d^2}\\\\d=\sqrt{\dfrac{4\rho l I}{V\pi}} \\\\d=\sqrt{\dfrac{4\times 2.44\times 10^{-8}\times 4\times 4\times 10^{-3}}{1.5\times \pi}} \\\\d=18.2\ \mu m

Out of four option, near option is (C) 17 μm.

6 0
3 years ago
How many air molecules are in a 13.0×12.0×10.0 ft room (28.2 L=1 ft3)? Assume atmospheric pressure of 1.00 atm, a room temperatu
aksik [14]

Answer:

1963.93 Moles

Explanation:

-We know the standard conversion ratio for the volume of a mole is 1 Mole=22.4L

Given volume of rooms as 13.0ft\times12.0ft\times10ft=1560 ft^3

Convert the volume into liters:1560\times28.2l=43992l

#From our conversion ratio above, we get the volume of air molecules in moles as:

V_m=\frac{43992L}{22.4L}\times 1\ mole\\\\=1963.93\ Moles

Hence, the volume of air molecules is 1963.93 Moles

4 0
2 years ago
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