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MAXImum [283]
3 years ago
11

A lawn mower is pushed a distance of 100 ft. Along a horizontal path by a constant force of 60 lbs. Yhe handle of the lawn mower

is held at a downward angle of 45° toward the horizontal. Explain what each value in this situation represents in relation to the formula w=F•PQ=||F||•||PQ||costheta. Then find the work done pushing the lawn mower
Mathematics
1 answer:
olga2289 [7]3 years ago
6 0
For this case we have the following equation:
 w = || F || • || PQ || costheta
 Where,
 || F ||: force vector module
 || PQ ||: distance module
 costheta: cosine of the angle between the force vector and the distance vector.
 Substituting values:
 w = (60) * (100) * (cos (45))
 w = 4242.640687 lb.ft
 Answer:
 
The work done pushing the lawn mower is: 
 w = 4242.640687 lb.ft
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PLZZZ HELP I DONT UNDERSTAND THIS AT ALLLLLL. :( :(
NISA [10]

Answer:

W÷9=z

Step-by-step explanation:

18÷9=2

45÷9=5

81÷9=9

Z is the quotient of W when divided by 9.

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3 years ago
Many games depend on how a ball bounces. For example, if different basketballs rebounded differently, one basketball would bounc
Crank

Answer:

Basketball = 0.743

Step-by-step explanation:

Given

Tennis:

Starting Height = 200 cm

Rebound Height = 111 cm

Soccer Balls;

Starting Height = 200 cm

Rebound Height = 120 cm

Basketball:

Starting Height = 72 inches

Rebound Height = 53.5 inches

Squash:

Starting Height = 100 inches

Rebound Height = 29.5 inches

For measuring the bounciness of a ball, one needs that starting Height of and the rebound Height of that ball which have been listed out above.

Calculating the rebound ratio of each balls.

Rebound Ratio = Rebound Height/Starting Height

Tennis: 111/200= 0.556

Soccer Balls: 120/200 = 1.667

Basketball: 53.5/72 = 0.743

Squash: 29.5/100 = 0.295

From the rebounding ratio calculated above, it can be seen that basketball has the highest rebound ratio of 0.743 and is the bounciest of all whole Squash has the least rebound of 0.295 ratio, hence it is the least bounce of all.

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3 years ago
Delia's relative long jump length was recorded as 0. What does this mean?
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She either didn't have legs or never did long jump. 
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4 years ago
Determine whether the given vectors are orthogonal, parallel or neither. (a) u=[-3,9,6], v=[4,-12,-8,], (b) u=[1,-1,2] v=[2,-1,1
nevsk [136]

Answer:

a) u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168

Since the dot product is not equal to zero then the two vectors are not orthogonal.

|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}

|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}

cos \theta = \frac{uv}{|u| |v|}

\theta = cos^{-1} (\frac{uv}{|u| |v|})

If we replace we got:

\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi

Since the angle between the two vectors is 180 degrees we can conclude that are parallel

b) u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5

|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}

|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}

cos \theta = \frac{uv}{|u| |v|}

\theta = cos^{-1} (\frac{uv}{|u| |v|})

\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557

Since the angle between the two vectors is not 0 or 180 degrees we can conclude that are either.

c) u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0

Since the dot product is equal to zero then the two vectors are orthogonal.

Step-by-step explanation:

For each case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.

Part a

u=[-3,9,6], v=[4,-12,-8,]

The dot product on this case is:

u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}

|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}

And finally we can calculate the angle between the vectors like this:

cos \theta = \frac{uv}{|u| |v|}

And the angle is given by:

\theta = cos^{-1} (\frac{uv}{|u| |v|})

If we replace we got:

\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi

Since the angle between the two vectors is 180 degrees we can conclude that are parallel

Part b

u=[1,-1,2] v=[2,-1,1]

The dot product on this case is:

u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}

|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}

And finally we can calculate the angle between the vectors like this:

cos \theta = \frac{uv}{|u| |v|}

And the angle is given by:

\theta = cos^{-1} (\frac{uv}{|u| |v|})

If we replace we got:

\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557

Since the angle between the two vectors is not 0 or 180 degrees we can conclude that are either.

Part c

u=[a,b,c] v=[-b,a,0]

The dot product on this case is:

u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0

Since the dot product is equal to zero then the two vectors are orthogonal.

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