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2 years ago

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1. A tourist accidentally drops a camera from a 40.0 m high bridge and it falls for 2.85 seconds. If g = 9.81 m/s2 and air resis

tance are disregarded, what is the speed of the camera as it hits the water?

1 answer:

attashe74 [19]2 years ago

6 0

Answer:

28 m/s

Explanation:

v(final)^2=v(initial)^2+2a(delta y)

=sqrt(2*-9.8 m/s^2*-40)

28 m/s

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2 years ago

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3 years ago

(d) Suppose you use a spring to launch a payload horizontally from the asteroid so that the payload ends up far from the asteroi

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ks= 133.2 N/m

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Assuming that we can neglect the gravitational potential energy of the mass, and that no other forces acting on the payload, total mechanical energy must be conserved.
This energy, at any time, is part elastic potential energy (stored in the spring) and part kinetic energy.
When the spring is initially compressed, the payload is at rest, so all energy is elastic potential.
Once the spring has returned to its natural state, all this elastic potential energy must have been turned into kinetic energy.
If the payload is launched horizontally, and no gravity is present,this means that its final speed will be horizontal only also, according to Newton's First Law.
So, we can write the following equation:

$\Delta U + \Delta K = 0 (1)$

where ΔU = -1/2*k*(Δx)² (2)
and ΔK = 1/2*m*v² (3)
Replacing in (2) and (3) by the givens, and simplifying, we can find the stiffness ks as follows:

$k_{s} =\frac{m*v^{2}}{\Delta x^{2}} = \frac{29 kg*(3m/s)^{2}}{(1.4m)^{2}} = 133.2 N/M (4)$

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2 years ago