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mdurga7799p50yjg
@mdurga7799p50yjg
June 2021
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plz answer the above question .......APPROPRIATE ANSWER ONLY..........I will mark u as brainliest.....OKAY!!!!!
Answers & Comments
flower161
Let the force of smaller magnitude be A and the force of larger magnitude be B.
here,it is clearly given that the resultant force is R perpendicular to small force. hence, we can say that b is the hyopetenuse .
therefore, b² = R² + A²
=> R² = b² - A²
= 8²
= 64 .......................... ( 1 ).
here, it is also given that
A + B = 16
therefore , B = 16 - A ..........................( 2 ).
Now, by substituting ( 2 ) in ( 1 ).we get,
( 16 - A ) ²- A² = 64
=> 256 - 32A + A² - A² = 64
32A = 256 - 64 = 192
A = 192 / 32
= 6
hence, we get the result
B = 16 - A = 16 - 6 = 10
therefore, the magnitude of the two vectors are 6 and 10.
1 votes
Thanks 1
flower161
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Answers & Comments
here,it is clearly given that the resultant force is R perpendicular to small force. hence, we can say that b is the hyopetenuse .
therefore, b² = R² + A²
=> R² = b² - A²
= 8²
= 64 .......................... ( 1 ).
here, it is also given that
A + B = 16
therefore , B = 16 - A ..........................( 2 ).
Now, by substituting ( 2 ) in ( 1 ).we get,
( 16 - A ) ²- A² = 64
=> 256 - 32A + A² - A² = 64
32A = 256 - 64 = 192
A = 192 / 32
= 6
hence, we get the result
B = 16 - A = 16 - 6 = 10
therefore, the magnitude of the two vectors are 6 and 10.