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kap26 [50]
4 years ago
12

Starting from rest, a particle that is confined to move along a straight line is accelerated at a rate of 5.0 m/s2. Which one of

the following statements concerning the slope of the position versus time graph for this particle is true? a) The slope is not constant and decreases with increasing time. b) The slope has a constant value of 5.0 m/s. The slope has a constant value of 5.0 m/s2. c) The slope is not constant and increases with increasing time. d) The slope is both constant and negative.
Physics
1 answer:
Elanso [62]4 years ago
6 0

Answer:

c) The slope is not constant and increases with increasing time.

Explanation:

The equation for the position of this particle (starting from rest is)

s = at^2/2 = 5t^2/2 = 2.5t^2

We can take derivative of this with respect to time t to get the equation of slope:

s' = (2.5t^2)' = 2*2.5t = 5t

As time t increase, the slope would increases with time as well.

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Suppose that two sources of sound produce waves of the same exact amplitude and frequency, but are out of phase by one-half of a
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option D is right

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What is the magnitude of the electric force of attraction between an iron nucleus (q=+26e) and its innermost electron if the dis
iris [78.8K]

Answer:

The magnitude of the force on an electron is 0.069 N.

Explanation:

Given that,

Distance between 1.70 A from the plutonium nucleus, d=1.5\times 10^{-12}\ m

The number of electron in iron nucleus is +26e.

To find,

The magnitude of the force between an iron nucleus.

Solution,

Total charge in the plutonium nucleus is, q=26\times 1.6\times 10^{-19}=4.16\times 10^{-18}\ C. The electric force between charges is given by :

F=\dfrac{kq^2}{d^2}

F=\dfrac{9\times 10^9\times (4.16\times 10^{-18})^2}{(1.5\times 10^{-12})^2}

F = 0.069 N

So, the magnitude of the force on an electron is 0.069 N. Hence, this is the required solution.

4 0
3 years ago
A 6.9-kg wheel with geometric radius m has radius of gyration computed about its mass center given by m. A massless bar at angle
Vilka [71]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is  \alpha =2.538 \  rad/s^2

Explanation:

From the question we are told that

  The mass of the wheel is  m =  6.9  kg

   The radius is  r =  0.69 \  m

    The radius of gyration is  k_G = 0.4\  m

    The angle is  \theta = 47^o

    The force which the massless bar is subjected to F = 22.5 \  N

Generally given that the wheels rolls without slipping on the flat stationary ground surface, it implies that  point A is the  center of rotation.

  Generally the moment of  inertia about A is mathematically represented as

     I_a =  I_G + M* r^2

Here I_G is the moment of inertia about G with respect to the radius of gyration  which is mathematically represented as

    I_G =  M *  k_G

=>I_a = k_G*  M + M* r^2

=>I_a =0.4 * 6.9  + 6.9 * 0.69^2

=>I_a =6.045 \  kg \cdot m^2

Generally the torque experienced by the wheel  is mathematically represented as

       \tau =  F *  cos (47)

=>     \tau =  22.5 *  cos (47)

=>     \tau =  15.34 \ kg \cdot m^2 \cdot s^{-2}

Generally this torque is also mathematically represented as

     \tau = I_a * \alpha

=>   15.34  =  6.045 * \alpha

=>   \alpha =2.538 \  rad/s^2

4 0
3 years ago
A wheel moves in the xy plane in such a way that the location of its center is given by the equations xo = 12t3 and yo = R = 2,
Stella [2.4K]

Answer:

the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s  is   P =  104.04 \hat{i} -314.432 \hat{j}

Explanation:

The free-body  diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;

  • the peripheral velocity that is directed downward (-V_y) along the y-axis
  • the linear velocity (V_x) that is directed along the x-axis

Now;

V_x = \frac{d}{dt}(12t^3+2) = 36 t^2

V_x = 36(1.7)^2\\\\V_x = 104.04\ ft/s

Also,

-V_y = R* \omega

where \omega(angular velocity) = \frac{d\theta}{dt} = \frac{d}{dt}(8t^4)

-V_y = 2*32t^3)\\\\\\-V_y = 2*32(1.7^3)\\\\-V_y = 314.432 \ ft/s

∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s  is   P =  104.04 \hat{i} -314.432 \hat{j}

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