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erastova [34]
3 years ago
5

A cube of osmium metal 1.500 cm on a side has a mass of 76.31 g at 25 ?c. part a what is its density in g/cm3 at this temperatur

e?
Physics
1 answer:
gayaneshka [121]3 years ago
4 0
The volume of the cube osmium metal is calculated by multiplying the given dimension to itself three times. 
                                    v = (1.5 cm)(1.5 cm)(1.5 cm)
                                    v = 3.375 cm³
Density is calculated by dividing the mass by volume.
                                    d = (76.31 g)/(3.375 cm³)
                                     d = 22.61 g/cm³

Thus, the density of the metal is approximately 22.51 g/cm³.
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True or False: The more mass an object has, the faster it will fall.
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Answer:

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Explanation:

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Answer:

The magnitude of the tangential velocity is v= 0.868 m/s

The magnitude of the resultant acceleration at that point is  a = 4.057 m/s^2

Explanation:

From the question we are told that

      The mass of the uniform disk is m_d = 40.0kg

       The radius of the uniform disk is R_d = 0.200m

       The force applied on the disk is F_d = 30.0N

Generally the angular speed i mathematically represented as

             w = \sqrt{2 \alpha  \theta}

Where \theta is the angular displacement given from the question as

           \theta  = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}

                 =1.257\  rad

   \alpha is the angular acceleration which is mathematically represented as

                    \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

    The moment of inertial is mathematically represented as

                     I = \frac{1}{2} m_dR^2_d

Substituting values

                    I = 0.5 * 40 * 0.200^2

                        = 0.8kg \cdot m^2

Considering the equation for angular acceleration

               \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

Substituting values

               \alph\alpha = \frac{(30.0)(0.200)}{0.8}

                   = 7.5 rad/s^2

Considering the equation for angular velocity

    w = \sqrt{2 \alpha  \theta}

Substituting values

     w =\sqrt{2 * (7.5) * 1.257}

         = 4.34 \ rad/s

The tangential velocity of a given point on the rim is mathematically represented as

                 v = R_d w

Substituting values

                    = (0.200)(4.34)

                     v= 0.868 m/s

The radial acceleration at hat point  is mathematically represented as

            \alpha_r = \frac{v^2}{R}

                  = \frac{0.868^2}{0.200^2}

                 = 3.7699 \ m/s^2

The tangential acceleration at that point is mathematically represented as

               \alpha _t = R \alpha

Substituting values

           \alpha _t = (0.200) (7.5)

                 = 1.5 m/s^2

The magnitude of resultant acceleration at that point is

                 a = \sqrt{\alpha_r ^2+ \alpha_t^2 }

Substituting values

                a = \sqrt{(3.7699)^2 + (1.5)^2}

                   a = 4.057 m/s^2

         

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It is called a pure sound
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Dmitriy789 [7]

The answer is the red sidelight on a powerboat should be visible from the front and from the left (port side).

What are Sidelights?

  • There is various combinations of lights that must be used on a boat when it is dark, and these are:
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  • Masthead Light: The masthead light is a white light that shines forwards and on all sides of the vessel. All powered vessels must use this light.
  • All-Round white light: This light is the major light that is used to join the masthead light and the stern light. This single light is visible to all vessels from all directions.
  • Thus, as a rule for a boat rider, he should show the vision of red light and it should be visible from the front and from the left (port side).

To learn more about Sidelights visit:

brainly.com/question/28205057

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4 0
1 year ago
You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.90 mA when a sin
faust18 [17]

Answer:

Frequency required will be 2421.127 kHz

Explanation:

We have given inductance L=0.450H=0.45\times 10^{-3}H

Current in the inductor i=1.90mA=1.90\times 10^{-3}A

Voltage v = 13 volt

Inductive reactance of the circuit X_l=\frac{v}{i}

X_l=\frac{13}{1.9\times 10^{-3}}=6842.10ohm

We know that

X_l=\omega L=2\pi fL

2\times 3.14\times  f\times 0.45\times 10^{-3}=6842.10

f = 2421.127 kHz

6 0
3 years ago
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