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Elena-2011 [213]

3 years ago

8

i know that when wind blows aginst a sail and propels the boat but if you had a whole bunch of leaf blowers blowing at the sail

would it move?

2 answers:

yuradex [85]3 years ago

8 0

Answer:

Yes

Explanation:

Since the sail boat uses wind to move, and since you have a whole bunch of leaf blowers which are producing wind, then yes it will move.

oksian1 [2.3K]3 years ago

6 0

Answer: Yes it would move

Explanation:

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Block 1 of mass m1 slides along a frictionless floor and into a one-dimensional elastic collision with stationary block 2 of mas

Elan Coil [88]

<span>Answers: (a) 2.0 m/s (b) 4 m/s

Method:

(a) By conservation of momentum, the velocity of the center of mass is unchanged, i.e., 2.0 m/s.

(b) The velocity of the center of mass = (m1v1+m2v2) / (m1+m2)

Since the second mass is initially at rest, vcom = m1v1 / (m1+m2)

Therefore, the initial v1 = vcom (m1+m2) / m1 = 2.0 m/s x 6 = 12 m/s

Since the second mass is initially at rest, v2f = v1i (2m1 /m1+m2 ) = 12 m/s (2/6) = 4 m/s </span>

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A particle initially located at the origin has an acceleration of = 2.00ĵ m/s2 and an initial velocity of i = 8.00î m/s. Find (a

sergejj [24]

a. The particle has position vector

$\vec r(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)t\right)\,\vec\imath+\left(\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\right)\,\vec\jmath$

$\vec r(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath$

b. Its velocity vector is equal to the derivative of its position vector:

$\vec v(t)=\vec r'(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath$

c. At $t=7.00\,\mathrm s$ , the particle has position

$\vec r(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(7.00\,\mathrm s)\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)^2\,\vec\jmath$

$\vec r(7.00\,\mathrm s)=\left(56.0\,\vec\imath+49.0\,\vec\jmath\right)\,\mathrm m$

That is, it's 56.0 m to the right and 49.0 m up relative to the origin, a total distance of $\|\vec r(7.00\,\mathrm s)\|=\sqrt{(56.0\,\mathrm m)^2+(49.0\,\mathrm m)^2}=74.4\,\mathrm m$ away from the origin in a direction of $\theta=\tan^{-1}\dfrac{49.0\,\mathrm m}{56.0\,\mathrm m}=41.2^\circ$ relative to the positive $x$ axis.

d. The speed of the particle at $t=7.00\,\mathrm s$ is the magnitude of the velocity at this time:

$\vec v(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)\,\vec\jmath$

$\vec v(7.00\,\mathrm s)=\left(8.00\,\vec\imath+14.0\,\vec\jmath\right)\dfrac{\rm m}{\rm s}$

Then its speed at this time is

$\|\vec v(7.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+(14.0\dfrac{\rm m}{\rm s}\right)^2}=16.1\dfrac{\rm m}{\rm s}$

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