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Dominik [7]
3 years ago
15

a 64 l gas tank is filled completely is 15 degree celsius how much gasoline overflows into can get up to 35 degree celsius while

the car is parked in direct sunlight​
Physics
1 answer:
o-na [289]3 years ago
3 0

Answer:

Volume of gasoline overflow(v)= 40/9 L (I.e. 4.44 L)

Explanation:

Use <u>v1</u><u>/</u><u>T1</u><u>=</u><u>v2</u><u>/</u><u>T2</u>

.....overflow(V)=v2-v1

<u>Note</u><u>;</u> <em>Take</em><em> </em><em>temperature</em><em> </em><em>in</em><em> </em><em>absolute</em><em> </em><em>scale</em><em> </em><em>or</em><em> </em><em>kelvin</em><em> </em><em>scale</em><em> </em>

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a force of 50 newtons pulls a rope attached to a 150 newton sled across a horizontal surface at a constant velocity of 5 meters
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A patient comes to an outpatient laboratory for a physician-ordered fasting test. The patient indicates that he forgot that the
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Which of the following is an example of a properly written testable hypothesis? *
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uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station
Alex787 [66]

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

         

7 0
3 years ago
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