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Ivan
3 years ago
13

you exert a force of 350N on his vehicle and manage to push it 15m and completely out of the snow. How much work was done on the

vehicle?
Mathematics
1 answer:
Nikitich [7]3 years ago
4 0

Answer:

Step-by-step explanation:

work done=force*displacement

=350 N *15m

=5250 Joule

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Oksanka [162]
I use the sin rule to find the area

A=(1/2)a*b*sin(∡ab)

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sin(∡B)=[2*A]/[(AB)*(BC)]

we know that
A=5√3
BC=4
AB=5
then

sin(∡B)=[2*5√3]/[(5)*(4)]=10√3/20=√3/2
(∡B)=arc sin (√3/2)= 60°

 now i use the the Law of Cosines 

c2 = a2 + b2 − 2ab cos(C)

AC²=AB²+BC²-2AB*BC*cos (∡B)

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2) we know that

a/sinA=b/sin B=c/sinC

and

∡K=α

∡M=β

ME=b

then

b/sin(α)=KE/sin(β)=KM/sin(180-(α+β))

KE=b*sin(β)/sin(α)

A=(1/2)*(ME)*(KE)*sin(180-(α+β))

sin(180-(α+β))=sin(α+β)

A=(1/2)*(b)*(b*sin(β)/sin(α))*sin(α+β)=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)

A=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)

KE/sin(β)=KM/sin(180-(α+β))

KM=(KE/sin(β))*sin(180-(α+β))--------- > KM=(KE/sin(β))*sin(α+β)

the answers part 2) are

side KE=b*sin(β)/sin(α)
side KM=(KE/sin(β))*sin(α+β)
Area A=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)

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3 years ago
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Answer:

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To find the variable we need to take out the magazines from the price. 112-16=96. Now we're left with 12x=96. Now all we have to do is divide 96 by 12, which equals 8.

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8 0
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<em>See attached</em>

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Answer:

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3 years ago
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