Which statement correctly explains mechanical advantage?

This is a 2D problem (parabolic) so we have to think that way. We have to split up the problem into its 2 dimensions to solve it. Think "y-stuff" and "x-stuff".

In the y-stuff category:

v₀ = 0 (initial upwards velocity is 0 since we are told the penny is thrown horizontally)

Δx = -10.0 m (this displacement is negative because the penny lands 10.0 m below the point from which it was thrown)

a = -9.8 m/s/s

t = ? (we need to find the time in this dimension so we can use it in the x dimension to find the displacement, our unknown)

In the "x-stuff" category:

v₀ = 7.25 m/s (this is given)

Δx = ???

a = 0 (acceleration in this dimension is ALWAYS 0)

t = (we will solve for this in the y-dimension and plug it in here).

In the y dimension:

Δx = v₀t + $\frac{1}{2}at^2$ and plugging in from the y-dimension info:

$-10.0=0t+\frac{1}{2}(-9.8)t^2$ which simplifies to

$-10.0=-4.9t^2$ so

$t=\sqrt{\frac{-10.0}{-4.9} }$ which, to 2 significant digits is

t = 1.4 seconds

Now we will do the same in the x-dimension, using t = 1.4:

Δx = v₀t + $\frac{1}{2}at^2$ and filling in the x-stuff:

Δx = $7.25(1.4)+\frac{1}{2}(0)(1.4)^2$ Notice that the stuff after the + sign goes to 0 cuz of the multiplication of 0, so what we are left with is another form of the d = rt equation:

Δx = 7.25(1.4) + 0 so

Δx = 1.0 × 10¹ m (That's rounded correctly to 2 sig dig's: 10 m from the base of the building).

6 0

3 years ago

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8. A Ferris wheel with diameter 18 m and rotates with constant speed of 3.8 ms.

icang [17]

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6 0

3 years ago

The graph at the right shows the force needed to pull a bow back as the string is pulled further and further.

Sindrei [870]

A. 9 J

In a force-distance graph, the work done is equal to the area under the curve in the graph.

In this case, we need to extrapolate the value of the force when the distance is x=30 cm. We can easily do that by noticing that there is a direct proportionality between the force and the distance:

$F=kx$

where k is the slope of the line. We can find k, for instance chosing the point at x=5 cm and F=10 N:

$k=\frac{F}{x}=\frac{10 N}{5 cm}=2 N/cm$

And now we can calculate the work by calculating the area under the curve until x=30 cm, F=60 N:

$W=\frac{1}{2} (height) (base)= \frac{1}{2}(60 N)(0.30 m)=9 J$

B. 24.5 m/s

The mass of the arrow is m=30 g=0.03 kg. The kinetic energy of the arrow when it is released is equal to the work done by pulling back the bow for 30 cm:

$W=K=\frac{1}{2}mv^2$

where m is the mass of the arrow and v is its speed. By re-arranging the formula and using W=9 J, we find the speed:

$v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2\cdot 9J}{0.03 kg}}=24.5 m/s$

8 0

3 years ago