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stellarik [79]
4 years ago
14

1) Johnny completing the spring lab, and makes 12 waves in 10 seconds. What is the

Physics
1 answer:
ankoles [38]4 years ago
5 0

Answer:

1.2 Hz

Explanation:

Frequency: This can be defined as the number of cycles completed in one seconds. The unit of frequency is Hertz (Hz)

From the question above,

f = N/t.................... Equation 1

Where f = frequency, N = number of wave or cycle completed by Johnny, t = time take to complete the waves.

Given: N = 12 waves, t = 10 s

Substitute into equation 1

f = 12/10

f = 1.2 Hz

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A block is pulled along a horizontal surface at a constant speed by a force (14.1i 0 j 5.1k). The direction k is perpendicular t
Pepsi [2]

Answer:

W =84.6\ Nm

Explanation:

given,

F = 14.1 i + 0 j + 5.1 k

displacement = 6 m

Assuming block is moving in x- direction

we know,

 dW = F dx

\int dW = F\int dx

W = F\int_0^6 dx

W = F[x]_0^6

W = 14.1 \times 6

W =84.6\ Nm

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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

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