Physics, 11.07.2019 07:30 kelli151

Acrow is flying horizontally with a constant speed of 2.70 m/s when it releases a clam from its beak. the clam lands on the rocky beach 2.10 s later. just before the clam lands, what is (a) its horizontal component of velocity, and (b) its vertical component of velocity? (c) how would your answers to parts (a) and (b) change if the speed of the crow were increased? explain.

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Answer from: kitttimothy55

a) 2.70 m/sec

b) 20.6 m/sec

c) As the speed of crow increases the horizontal component of speed of the clam will also increases but there is no change in the vertical component of speed.

Given :

Speed = 2.70 m/sec

Solution :

a) There is no gravity or no other acceleration in horizontal direction. Therefore velocity in horizontal direction is,

$\rm v_h = 2.70\; m/sec$

b) To find velocity in vertical direction we can use kinematics

$\rm v_v = u_v + a_vt$ ---- (1)

Where,

$\rm u_v = 0$ ,

$\rm a_v = 9.81 \; m/sec$

t = 2.10 sec

Put the values of $\rm u_v,\;a_v\;and \;t$ in equation (1),

$\rm v_v = 0+9.81\times2.1$

$\rm v_v = 20.6 \; m/sec$

c) As the speed of crow increases the horizontal component of speed of the clam will also increases but there is no change in the vertical component of speed.

For more information, refer the link given below

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Answer from: sonyav732

Part a)

in horizontal direction there is no gravity or no other acceleration

so in horizontal direction the speed of clam will remain same

v_x = 2.70 m/s

Part b)

In vertical direction we can use kinematics

v_f = v_i + at

v_f = 0 + 2.1 * 9.8

v_f = 20.6 m/s

part c)

if the speed of crow will be increased then the horizontal speed of the clam will also increase but there is no change in the vertical speed

Answer from: Quest

the simple machine has three main features. they are:

loadfulcrumeffort

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

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Acrow is flying horizontally with a constant speed of 2.70 m/s when it releases a clam from its beak...