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photoshop1234 [79]
3 years ago
14

You work in a materials testing lab and your boss tells you to increase the temperature of a sample by 37.1 ∘c . the only thermo

meter you can find at your workbench reads in degrees fahrenheit.
Physics
1 answer:
Ivahew [28]3 years ago
7 0
Each Celsius degree is the size of 1.8 Fahrenheit degrees. So you need dip your Fahrenheit thermometer into the sample, see where you're starting, and then warm it up to a temperature that reads (37.1 x 1.8) = 66.8 Fahreheit degrees higher.
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18 m/s.

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Where does energy originate ina food web
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The three forces shown act on a particle. what is the direction of the resultant of these three forces?
melisa1 [442]
Missing figure: http://d2vlcm61l7u1fs.cloudfront.net/media/f5d/f5d9d0bc-e05f-4cd8-9277-da7cdda3aebf/phpJK1JgJ.png

Solution:
We need to find the magnitude of the resultant on both x- and y-axis.

x-axis) The resultant on the x-axis is
F_x = 65 N\cdot cos 30^{\circ} - 30 N - 20 N\cdot sin 20^{\circ} = 19.45 N
in the positive direction.

y-axis) The resultant on the y-axis is
F_y = 65 N \cdot sin 30^{\circ} - 20 N \cdot cos 20^{\circ} = 13.70 N
in the positive direction.

Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using
\tan \alpha =  \frac{F_y}{F_x} = \frac{13.70 N}{19.45 N}=0.7
from which we find 
\alpha=35^{\circ}
7 0
3 years ago
A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hit
lapo4ka [179]

Answer:

The  angle is  \theta  =  15.48^o

Explanation:

From the question we are told that  

     The distance of the dartboard from the dart is  d  =  3.66  \ m

     The time taken is  t =  0.455 \ s

   

The  horizontal component of the speed of the dart is mathematically represented as

      u_x =  ucos \theta

where u is the the velocity at dart is lunched

  so

      distance =  velocity \ in \ the\  x-direction  *  time

substituting values

      3.66 =   ucos  \theta *  (0.455)

 =>   ucos \theta =  8.04  \ m/s

From projectile kinematics the time taken by the dart can be mathematically represented as

         t  =  \frac{2usin \theta }{g}

=>    usin \theta =  \frac{g  * t}{2 }

       usin \theta =  \frac{9.8  * 0.455}{2 }

      usin \theta = 2.23

=>   tan \theta =  \frac{usin\theta }{ucos \theta }  =  \frac{2.23}{8.04}

       \theta  =  tan^{-1} [0.277]

      \theta  =  15.48^o

     

4 0
3 years ago
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