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Setler [38]
3 years ago
9

Are sloths secondary consumer

Physics
2 answers:
ankoles [38]3 years ago
6 0
<span>Primary consumers, because sloths do not eat any meat, they eat the plants.</span>
ehidna [41]3 years ago
5 0
No, they are primary consumers, because they don't eat meat, they eat plants
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What two things determine whether a molecule is polar?
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The Biot-Savart force law does not apply if the velocity is parallel to the field direction.
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3 years ago
Haley is driving down a straight highway at 73 mph. A construction sign warns that the speed limit will drop to 55 mph in 0.50 m
skad [1K]

Answer: a= -1029.97 m/s^{2}

Explanation:

We know Haley is driving with an initial velocity of V_{o}=73mi/h and then she has to change to a velocity of V_{f}=55mi/h in a distance d=0.50 mi, and we also know we are dealing with constant acceleration  a.

Therefore, the following equation will be useful:

{V_{f}}^{2}={V_{o}}^{2}+2ad (1)

Clearing a:

a=\frac{{V_{f}}^{2}-{V_{o}}^{2}}{2d} (2)

a=\frac{{(55mi/h)}^{2}-{(73mi/h)}^{2}}{2(0.50 mi)} (3)

a=-2304 mi/h^{2} (4)

Knowing 1h=3600 s and 1mi=1609.34 m:

a= -1029.97 m/s^{2} this is the constant acceleration that will bring Haley to lower speed

7 0
4 years ago
A small Keplerian telescope has an objective with a 1.33 m focal length. Its eyepiece is a 2.82 cm focal length lens. It is used
forsale [732]

Answer:

The angle is  \phi =  0.45 0 ^o

Explanation:

From the question we are told that  

     The objective  focal length   f  =  1.33 \ m

     The  eyepiece focal length is  f_o =  2.82 \ cm  = 0.0282 \ m

      The diameter of the sunlight is  d  = 25000km =  2.5 *10^{7} \ m

       The distance of the sun from from the earth is  D =  1.5 *10^8 km  =  1.5 *10^{11} \ m

  Generally the magnification of the object is mathematically evaluated as

        m  =  -\frac{f_o }{f_e }

The negative  sign is because the lens of the telescope is  diverging light

 substituting values  

      m  =  -\frac{1.33 }{0.0282 }

      m  =  - 47.16 3

Now we can obtain the angle made by the object (sunlight ) with respect to the telescope  as follows  

        tan  \theta  = \frac{d}{D}

substituting values

      tan  \theta  = \frac{2.5 *10^{7}}{1.5*10^{11}}

      tan  \theta  =  0.0001667

      \theta=  tan^{-1}[0.0001667]

     \theta= 0.00955^o

The  magnification can  also be mathematically represented as

      m  =  \frac{\phi }{\theta }

Where \phi is the angle the image made with telescope

Since the negative sign indicate direction of light movement we will remove it from the calculation below

      =>   47.163  =  \frac{\phi}{0.00955}

     =>    \phi =  0.45 0 ^o

8 0
3 years ago
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