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nalin [4]
2 years ago
10

Consider taking a time-lapse video of an analog clock that is missing the hour hand. Assume that one frame of video is taken eve

ry X minutes. When you view the compiled video, it appears as though the minute hand is advancing 10 minutes every frame. (5 points/each) What is the frame rate of the time-lapse video if no aliasing occurs
Physics
1 answer:
Vlad [161]2 years ago
4 0

Answer:

the frame rate of the time-lapse video if no aliasing occurs is 10min per frame

Explanation:

Given the data in the question;

No aliasing effect, meaning the clock with missing hour hand will not have any effect

Time required for one frame = 10 minutes

frame rate of time = Time required / number of frame

= 10 min / 1 frame

= 10min per frame

Therefore, the frame rate of the time-lapse video if no aliasing occurs is 10min per frame

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Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2
Bess [88]

Answer:

The acceleration of M_2 is  a =  0.7156 m/s^2

Explanation:

From the question we are told that

    The mass of first block is  M_1 =  2.25 \ kg

    The angle of inclination of first block is  \theta _1 =  43.5^o

    The coefficient of kinetic friction of the first block is  \mu_1  = 0.205

      The mass of the second block is  M_2 = 5.45 \ kg

     The angle of inclination of the second block is  \theta _2 =  32.5^o

      The coefficient of kinetic friction of the second block is \mu _2 = 0.105

The acceleration of M_1 \ and\  M_2 are same

The force acting on the mass M_1 is mathematically represented as

     F_1 = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

=> M_1 a = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

Where T is the tension on the rope

The force acting on the mass M_2 is mathematically represented as    

  F_2 =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

   M_2 a =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

At equilibrium

  F_1 =  F_2

So

 T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

making a the subject of the formula

    a =  \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}

substituting values a =  \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}

    => a =  0.7156 m/s^2

     

3 0
3 years ago
Determine the projection (magnitude and sign), or component, of vector v1 along the direction of vector v2. Your answer could be
professor190 [17]

Answer:

- 1.07 ft

Explanation:

V1 = (-5, 7, 2)

V2 = (3, 1, 2)

Projection of v1 along v2, we use the following formula

=\frac{\overrightarrow{V1}.\overrightarrow{V2}}{V2}

So, the dot product of V1 and V2 is = - 5 (3) + 7 (1) + 2 (2) = -15 + 7 + 4 = -4

The magnitude of vector V2 is given by

= \sqrt{3^{2}+1^{2}+2^{2}}=3.74

So, the projection of V1 along V2 = - 4 / 3.74 = - 1.07 ft

Thus, the projection of V1 along V2 is - 1.07 ft.

so we need to find the direction of v2

7 0
3 years ago
Which condition must exist in order for conduction to occur between two substances
motikmotik
Heat!!!!!!!!!!!!!!!!!!! sorry about the exclamation marks but it wont let me post in less it has more than 20 characteristics
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3 years ago
What may be a serious side effect of anti-depressants specific in adolescents?
Klio2033 [76]

I think its suicidal ideation......

I think

3 0
3 years ago
Plz help!!!
shtirl [24]

Answer:

I think the answer to your question is true

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3 years ago
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