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Basile [38]
3 years ago
5

ONLINE CALCULATOR .A force of 187 pounds makes an angle of 73 degrees 36 ' with a second force. The resultant of the two forces

makes an angle of 29 degrees 1 ' to the first force. Find the magnitudes of the second force and of the resultant.
Physics
1 answer:
saul85 [17]3 years ago
3 0

Answer:

The magnitudes of the second force is   Z = 129.9 N

The magnitudes of the  resultant force is   R = 256.047 N

Explanation:

From the question we are told that  

    The force is  F = 187 \ lb

     The angle made with second force \theta_o = 73 ^o 36' =  73 + \frac{36}{60}  =  73.6^o

     The angle between the resultant force and the first force \theta _1  = 29 ^o 1 ' = 29 + \frac{1}{60}  = 29.0167^o

For us to solve problem we are going to assume that

     The magnitude of the second force is  Z N

     The magnitude of the resultant force is R N

According to Sine rule

                \frac{F}{sin (\theta _o - \theta_1 }  = \frac{Z}{\theta _1}

Substituting values

             \frac{187}{sin(73.3 - 29.01667)} =\frac{Z}{sin (29.01667)}  

             267.82 =\frac{Z}{0.4851}  

              Z = 129.9 N

According to cosine rule

       R = \sqrt{F ^2 + Z^2 + 2(F) (Z) cos (\theta _o) }

Substituting values

     R = \sqrt{187^2 + 129.9 ^2  + 2 (187 ) (129.9) cos (73.6)}

     R = 256.047 N

 

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Answer:

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Explanation:

Given the following data;

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We know that pie, π = 22/7

To find the mass, we would use the following formula;

F = 1/2π√(k/m)

Where;

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Substituting into the formula, we have;

10 = 1/2 * 22/7 * √250/m

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Cross-multiplying, we have;

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Dividing both sides by 22, we have;

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A particle with a charge of − 5.10 nC is moving in a uniform magnetic field of B⃗ =−( 1.20 T )k^. The magnetic force on the part
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Answer:

Explanation:

Given that,

Charge q=-5.10nC

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And the magnetic force

F =−( 3.30×10−7N )i+( 7.60×10−7N )j

Let the velocity be V(xi + yj + zk)

Then, the force is given as

Note i×i=j×j×k×k=0

i×j=k. j×i=-k

j×k=i. k×j=-i

k×i=j. i×k=-j

F= q(v×B)

−( 3.30×10−7N )i+( 7.60×10−7N )j =

q(xi + yj + zk) × -1.2k

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( -1.2x i×k - 1.2y j×k - 1.2z k×k)

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( 1.2xj - 1.2y i )

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( -1.2y i + 1.2x j)

So comparing comparing coefficients

let compare x axis component

-( 3.30×10−7N )i=-1.2qy i

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Let compare y-axisaxis

7.6×10−7N j = 1.2qx j

7.6×10−7N = 1.2qx

x= 7.6×10^-7/-1.2q

x= 7.6×10^-7/1.2×-5.10×10^-9)

x=-124.18m/s

a. Then, the velocity of the x component is x= -124.18m/s

b. Also, the velocity component of the y axis is =-53.92m/s

c. We will compute

V•F

V=-124.18i -53.92j

F=−( 3.30×10−7 N )i+( 7.60×10−7 N )j

Note

i.j=j.i=0. Also i.i = j.j =1

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(-124.18i-53.92j)•−(3.30×10−7N)i+(7.60×10−7 N )j =

4.1×10^-5 - 4.1×10^-5=0

V•F=0

d. Angle between V and F

V•F=|V||F|Cosx

0=|V||F|Cos

Cosx=0

x= arccos(0)

x=90°

Since the dot product is zero, from vectors , if the dot product of two vectors is zero, then the vectors are perpendicular to each other

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