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baherus [9]
3 years ago
15

A student in an undergraduate physics lab is studying Archimede's principle of bouyancy. The student is given a brass cylinder a

nd, using a triple beam balance, finds the mass to be 3.21 kg. The density of this particular alloy of brass is 8.62 g/cm 3 . The student ties a massless string to one end of the cylinder and submerges it into a tank of water where there is an apparent reduction in the weight of the cylinder. With this information, calculate the volume, V , of the cylinder and the tension, T , in the string when it is submerged in the tank of water. The density of water is 1.00 g/cm 3 , and the acceleration due to gravity is gVolume of the cylinder: _____ cm3Tension in the string: ______ N
Physics
1 answer:
Readme [11.4K]3 years ago
3 0

Answer:

V = 0.3724 m³

T = 27.836 N

Explanation:

Given :

m = 3.21 kg  , W= 3.21 * 9.81 m / s² = 31.4901 N

ρ = 8.62 g / cm ³  = 8620 kg / m³

V = m / ρ =  3.21 kg  /  8620 kg / m³

V = 0.3724 m³

when submerged the weight of brass cylinder is equal to the tension in string:

F =  (0.3724m³) * (1000 kg / m³) * (9.81 m/s²²) = 3.653 ≈ 3.65 N

T = 31.4901 N - 3.65 N  

T = 27.836 N

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A horizontal spring is attached to the wall on one end and to a mass on the other end. The mass can slide freely on a frictionle
Radda [10]

Answer:

a) x=0  %T=0,   b) x= A %T=100%,   c) x=-A %T=50%

Explanation:

This is a simple harmonic movement exercise, which is explained by the expression

          x = A cos (wt + Ф)

where angular velocity is related to frequency and period

         w = 2π f = 2π / T

we can write the equation of the oscillation

         x = A cos θ

When seeing the two equations they are equivalent, so what happens with the angle will also happen with time

We are asked for the percentage of the period at three points: at the maximum elongation and at the point of x = 0, in general the distance is measured from the point of the spring without stretching

The period is defined as the time that the system takes to give a complete oscillation, that is, from x = 0 to x = A and return

a) for the unstretched spring point x = 0

In general, both distance and time are measured from this point, so the percentage of time is zero.

         % T = 0

b) for x = A

 let's find the angle

      cos tea = x / A = 1

therefore the angles tea = 2π rad

when the movement reaches the point of 2π radians it begins to repeat so the period is complete

            % T = 100%

c) the point of maximum compression x = -A

let's look for the angles

      cos tea = x / A = -1

therefore the angles tea = π rad

at this point the movement is halfway so it should take half the time

                % T = 50%

6 0
2 years ago
A satellite has a mass of 5832 kg and is in a circular orbit 4.13 × 105 m above the surface of a planet. The period of the orbit
elena55 [62]

Answer:

W = 28226.88 N

Explanation:

Given,

Mass of the satellite, m = 5832 Kg

Height of the orbiting satellite from the surface, h = 4.13 x 10⁵ m

The time period of the orbit, T = 1.9 h

                                            = 6840 s

The radius of the planet, R = 4.38 x 10⁶ m

The time period of the satellite is given by the formula

                             T = 2\pi \sqrt{\frac{(R+h)^{3} }{R^{2} g} }  second

Squaring the terms and solving it for 'g'

                             g = 4 π² \frac{(R+h)^{3} }{R^{2}T^{2}  }   m/s²

Substituting the values in the above equation

                   g = 4 π² \frac{(4.38X10^6+4.13X10^5)^{3} }{(4.38X10^6)^{2}X6840^{2}}  

                                    g = 4.84 m/s²      

Therefore, the weight

                                     w = m x g   newton

                                         = 5832 Kg x 4.84 m/s²

                                         = 28226.88 N

Hence, the weight of the satellite at the surface, W = 28226.88 N                

8 0
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