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Svetradugi [14.3K]
3 years ago
8

What is the approximate efficiency of the engine?

Physics
2 answers:
Elodia [21]3 years ago
8 0

Answer:91% I think

Explanation:

MArishka [77]3 years ago
4 0

Answer:

B) 91%

Explanation:

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The psychologist known for latent learning and cognitive maps is _________. A. Robert Rescorla B. Edward Tolman C. William James
Lubov Fominskaja [6]

Answer:

B

Explanation:

B. Edward Tolman

4 0
3 years ago
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When a wave passes from one medium to another, its _________ remains constant.
Mashcka [7]
The frequency will not change
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3 years ago
SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the
Firlakuza [10]

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

5 0
3 years ago
A block (mass = 61.2 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, wh
kolezko [41]

Answer:

The angular velocity is  w = 53.35 \ rounds /minute

Explanation:

From the question we are told that

    The mass of the block is  m = 61.2kg

     The of the pulley is  M = 14.2 kg

      The radius of the pulley is  R = 1.5m

       The radius  of the cord around the pulley is  r = 1.5 m

       The distance of the block to the floor is  d = 8.0 m

         

From the question we are told that the moment of inertia of the pulley is

          I  = \frac{1}{2} MR^2 kg \cdot m^2

Substituting value  

         I = \frac{1}{2}  * 14.2 * (1.5)^2

         I = 15.975 kg \cdot m^2

Using the Newtons law we can express the force acting on the vertical axis as

              ma = mg -T

         =>  T = mg -ma

Now when the pulley is rotated that  torque generated on the massless cord as a r result of the tension T and the radius of the cord around the pulley is mathematically represented as

                  \tau = I \alpha

     Here \alpha is the angular acceleration

           Here \tau is the torque which can be equivalent to

              \tau = T r

  Substituting this above

            Tr = I \alpha      

Substituting for T

         (mg - ma ) r =  I\  r \alpha

Here a is the  linear acceleration which is mathematically represented as

           a = r\alpha

    (mg - m(r\alpha ) ) r =  I\  r \alpha

     mgr = I\alpha  + m(r\alpha ) r

    mgr = \alpha  [ I + mr^2]

   making \alpha the subject

          \alpha  = \frac{mgr}{I -mr ^2}          

   Substituting values

            \alpha  = \frac{61.2 * 1.5 * 9.8}{15.975 + (61.2 ) * (1.5)^2}

             \alpha =5.854 rad /s^2

Now substituting into the equation above to obtain the acceleration

             a = 5.854 * 1.5

                a=8.78 m/s^2

This acceleration is a = \frac{v}{t}

and v is the linear velocity with is mathematically represented as

         v = \frac{d}{t}

Substituting this into the formula acceleration

        a = \frac{d}{t^2}

making t the subject

         t = \sqrt{\frac{d}{a} }

substituting value

      t = \sqrt{\frac{8}{8.78}}

     t = 0.9545 \ s

Now the linear velocity is

       v = \frac{8}{0.9545}

       v = 8.38 m/s

The angular velocity is  

       w = \frac{v}{r}

So

       w = \frac{8.38}{1.5}

        w = 5.59 rad/s

Generally 1 radian is equal to  0.159155 rounds or turns

        So  5.59 radian is  equal to x

Now x is mathematically obtained as

         x = \frac{5.59 * 0.159155}{1}

            = 0.8892 \ rounds

 Also

      60  second =  1 minute

So   1 second  = z      

Now z is mathematically obtained as

         z = \frac{ 1}{60}

            z = 0.01667 \ minute

Therefore

              w = \frac{0.8892}{0.01667}

              w = 53.35 \ rounds /minute

           

8 0
3 years ago
I need the correct answer for this question please​
kirill115 [55]
Try C. a downward gravitational force exerted by Earth
7 0
3 years ago
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