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IgorC [24]
3 years ago
14

An arrow is shot from a bow at an angle of 35 degrees above horizontal with an initial speed of 50 m/s what is the arrows horizo

ntal x and y components ?

Physics
2 answers:
ZanzabumX [31]3 years ago
4 0

Answer: Horizontal component of arrow :40.955 m/s

Vertical component of arrow :28.675 m/s

Explanation:

The initial speed of the arrow = u = 50 m/s

The horizontal component of the arrow =u_x=u\cos\theta

The horizontal component of the arrow =u_y=u\sin\theta

Angle between the velocity vector and x component = 35°

Horizontal component of arrow :

\cos\theta=\frac{u_x}{u}

\cos35^o=0.8191=\frac{u_x}{50}

u_x=0.8191\times 50=40.955 m/s

Vertical component of arrow :

\sin\theta=\frac{u_y}{u}

\sin35^o=0.5735=\frac{u_y}{50}

u_y=0.5735\times 50=28.675 m/s

Horizontal component of arrow :40.955 m/s

Vertical component of arrow :28.675 m/s

Alexandra [31]3 years ago
3 0
These kinds of problems can be broken down to a simple right triangle where we want the side lengths.  Knowing the hypotenuse (50) and the angle, we can get the  other sides with trigonometry.  These sides are then the components of the original vector.  I drew it up for you here.

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A traveling electromagnetic wave in a vacuum has an electric field amplitude of 81.1 V/m. Calculate the intensity S of this wave
ICE Princess25 [194]

Answer:

U = 4.39 J

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here we know that

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6 0
4 years ago
The weights of soy patties sold by a diner are normally distributed. A random sample of 15 patties yields a mean weight of 3.8 o
shepuryov [24]

Complete Question

The weights of soy patties sold by a diner are normally distributed. A random sample of 15 patties yields a mean weight of 3.8 ounces with a sample standard deviation of 0.5 ounces. At the 0.05 level of significance,perform a hypothesis test to see if the true mean weight is  less than 4 ounce.

Answer:

Yes the true mean weight is less than 4 ounce

Explanation:

  From the question we are told that

     The random sample is n = 15

       The mean weight is \= x = 3.8\  ounce

        The standard deviation is  \sigma = 0.5 \ ounce

         The  level of significance is  \alpha  = 0.05

       

So

   The  null hypothesis is  H_o : \mu \ge 4

     The alternative hypothesis is H_a : \mu < 4

Generally the critical value which a bench mark to ascertain whether the null hypothesis is  true or false is mathematically represented as

            t_{0.05} = 1.79

This value is  obtained from  the critical value table

Generally the test statistics is mathematically represented as

                Test \ Statistics (ST)  = \frac{\= x - \mu }{\frac{\sigma}{\sqrt{n} }  }

           =>  ST = \frac{3.8 -4 }{\frac{0.5}{\sqrt{20} } }

                ST = - 1.79

So since ST is less than t_{0.05}  then the null hypothesis would be rejected and the alternative hypothesis would be accepted so

  Thus the true mean weight is less than 4

4 0
3 years ago
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