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kati45 [8]
4 years ago
6

A 55 newton force applied in a object moves the object 10 meters as the same direction as the force what is the value of work do

ne on this object
Mathematics
2 answers:
katovenus [111]4 years ago
5 0

Answer: 550 Joules

Step-by-step explanation:

We know that when force(Newtons) moves an object some distance (meters) in the same direction as the force, then the formula to calculate work (Joules) done is given by :-

W=F\times d, where F is the force and d is the distance.

Given: Force : F= 55 Newtons

Distance: d = 10 meters

\Rightarrow W=F\times d=55\times10=550\text{ Joules}.

Hence, the value of work done on this object = 550 Joules.

vladimir1956 [14]4 years ago
4 0
I think it would be 550 lbs not sure if this helps though.
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The ratio of girls to boys in the coed soccer league is 3:7. Choose True or
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Answer:

True

Step-by-step explanation:

3:7 is girls to boys, and to find the total, do 7+3. After that, you'd get 10, and the ratio would be 3:10, so TRUE

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X2 is the GCF of this polynomial.
nadezda [96]

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4 years ago
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I suck at math can someone help please??
Zielflug [23.3K]
6/7 × 4 would be the correct answer. This is because 6/7 × 4= 3 3/7 which is equal to 4/7 × 6= 3 3/7. No other expression would equal 4/7 × 6
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3 years ago
3.<br> If r=9.b=5, and g=-6, what does (r +b-8)(b + g) equal?
VLD [36.1K]

Answer:

-6

Step-by-step explanation:

r = 9, b = 5, g = -6

Plug these values into (r + b - 8)(b + g)

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Combine like terms in the brackets

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8 0
3 years ago
Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E
Ipatiy [6.2K]

Answer:

(A) y=ke^{2t} with k\in\mathbb{R}.

(B) y=ke^{2t}/2-1/2 with k\in\mathbb{R}

(C) y=k_1e^{2t}+k_2e^{-2t} with k_1,k_2\in\mathbb{R}

(D) y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5 with k_1,k_2\in\mathbb{R},

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then 0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow  \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}. With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form e^{2t} with t real.

(B) Proceeding and the previous item, we obtain 1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow  \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2. Which is not a vector space with the usual operations (this is because -1/2), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set y=e^{mt} \Rightarrow y''=m^2e^{mt} and we obtain the characteristic equation 0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2 and then the general solution is y=k_1e^{2t}+k_2e^{-2t} with k_1,k_2\in\mathbb{R}, and as in the first items the set of solutions form a vector space.

(D) Using C, let be y=me^{3t} we obtain that it must satisfies 3^2m-4m=1\Rightarrow m=1/5 and then the general solution is y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5 with k_1,k_2\in\mathbb{R}, and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).  

4 0
4 years ago
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