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2 years ago

8

A deliveryman moves 10 cartons from the sidewalk, along a 10-meter ramp to a loading dock, which is 1.5 meters above the sidewal

k. If each carton has a mass of 25 kg, what is the total work done by the deliveryman on the cartons to move them to the loading dock?

1 answer:

docker41 [41]2 years ago

7 0

Answer:

3750 J

Step-by-step explanation:

We are given that

Total number of cartons=10

Height of loading dock from the side walk=1.5 m

Length of ramp=10 m

Mass of each carton=25 kg

We have to find the work done by the delivery man on the cartons to move them to the loading dock.

Work done= $Mgh$

Where M=Total mass of object

h=Height of object above the ground

g=Acceleration due to gravity= $10m/s^2$

Total mass of cartons=M= $10\times 25=250 kg$

h=1.5 m

Substitute the values then, we get

Work done= $250\times 1.5\times 10=3750$ J

Hence, the total work done by the deliverman on the cartons to move them to the loading dock=3750J

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Consider the product (x + y + z)2. If the expression is multiplied out and like terms collected, the result is: x2 + y2 + z2 + 2

VLD [36.1K]

Answer:

Part a: The coefficient of $v^9w^2x^5y^7z^2$ is $1.766 \times 10^{13}$

Part b: The number of terms are 23751.

Step-by-step explanation:

part a

From the given equation the

$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx$

So the coefficient of any term $x^ny^nz^n$ is given as

$\dfrac{2!}{n!n!n!}$

Similarly for the generic equation of coefficient of the term $x_1^{r_1}x_2^{r_2}x_3^{r_3}......x_k^{r_k}$ in the equation of form $(x_1+x_2+x_3+x_4......x_k)^n$ is given as

$\dfrac{n!}{r_1!r_2!r_3!....r_k!}$

So now the coefficient of $v^9w^2x^5y^7z^2$ is given as

$=\dfrac{25!}{9!2!5!7!2!}\\=1.766 \times 10^{13}$

The coefficient of $v^9w^2x^5y^7z^2$ is $1.766 \times 10^{13}$

Part b:

The number of terms is given as $\left (\ {{m+n-1} \atop {n}} \right )$

where m is the number of variables which are 5 here

n is the power which is 25 so the number of variables is given as

$\left (\ {{5+25-1} \atop {25}} \right )\\\left (\ {{29} \atop {25}} \right )\\\dfrac{29!}{25!4!}=23751$

So the number of terms are 23751.

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Answer:

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Step-by-step explanation:

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Answer:

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