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Nimfa-mama [501]
3 years ago
14

Please help no trolls no links pls

Physics
2 answers:
tigry1 [53]3 years ago
4 0
If it’s mass is heavier it will need more force to travel, making its velocity slower
Novosadov [1.4K]3 years ago
3 0
It would move slower/ faster by changing the mass, which would effect the weight and speed.
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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
If the Fox for an object on an incline is 350N,
Alinara [238K]
Pic? I need a pic thanks
5 0
3 years ago
How do scientists learn about the brain.
velikii [3]
Mainly looking at scans ansd the levels of hormons and chemicals when we use or brain for example when we use the frontal lobe diffren thsades will appear. our stress levels as well. how we react to things problems situations or questions.

7 0
3 years ago
Read 2 more answers
Give an examplea of newtons law of acceleration
coldgirl [10]

The 2nd Law of Motion states acceleration is produced when a n unbalanced force acts on an object (mass) The more mass the object has the more net force has to be used to move it.

Examples:

If you use the same force to push a truck and push a car, the car will have more acceleration than the truck, because the car has less mass.

It's easier to push a empty shopping cart then a full one, because the full one has more mass than the empty one. This means more force is required to push the full shopping cart.

8 0
3 years ago
A uniform meter stick is pivoted at the 50.00 cm mark on the meter stick. A 400.0 gram object is hung at the 20.0 cm mark on the
babunello [35]

Answer:C

Explanation:

Given

mass m_1=400\ gm is at x=20\ cm mark

mass m_2=320\ gm is at x=75\ cm mark

Scale is Pivoted at x=50\ cm mark

For scale to be in equilibrium net torque must be equal to zero

Taking ACW as positive thus

T_{net}=0.4\times g\times (0.5-0.2)-0.32\times g\times(0.75-0.50)

T_{net}=0.12g-0.08g=0.04g

Therefore a net torque of 0.04 g is required in CW sense which a mass 400\ gm can provide at a distance of x_o from pivot

0.04g=0.4\times g\times x_o

x_o=0.1\ m

therefore in meter stick it is at a distance of x=60\ cm

6 0
3 years ago
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