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Natasha_Volkova [10]

3 years ago

8

Bodies weighing 1 kilogram and 5 kilograms lie on a smooth horizontal surface. If a traction force of 0.6 N acts on another 5 kg

body, determine the acceleration of these bodies.

1 answer:

natima [27]3 years ago

6 0

0.6/5,1,5

so calculate it

not so sure though

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mafiozo [28]

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5 0

3 years ago

Read 2 more answers

The y-component of the force F which a person exerts on the handle of the box wrench is known to be 70 lb. Determine the x-compo

Yuri [45]

Answer:

<em>Fx = 121.24lb</em>

<em>F = 140lb</em>

Explanation:

Since we are not given the angles subtended by the force, we can assume it to be 30 degrees.

The y component of the force expressed by the formula:

Fy = Fsintheta

Given the y-component of the force F to bee 70lb

70lb = Fsintheta

Get magnitude of the force

F = 70/sin theta

F = 70/sin 30

F = 70/0.5

F = 140lb

Get the x-component of the force

Fx = Fcos theta

Fx = 140cos 30

Fx = 140(0.8660)

Fx = 1,212.4lb

<em>Hence the x-component of the force sis 121.24lb</em>

<em></em>

<em>Note that the angle used was assumed. Other values can as well be used</em>

5 0

3 years ago

Two ice skaters, Daniel (mass 70.0 kg) and Rebecca (mass 45.0 kg), are practicing. Daniel stops to tie his shoelace and, while a

wel

Answer:

a) $v=7.32m/s$

b) $\alpha =-35$ º

c) $ΔK=-1094.62J$

Explanation:

From the exercise we know that the collision between Daniel and Rebecca is elastic which means they do not stick together

So, If we analyze the collision we got

$p_{1x}=p_{2x}$

To simplify the problem, lets name D for Daniel and R for Rebecca

a) $p_{D1x}+p_{R1x}=p_{D2x}+p_{R2x}$

Since Daniel's initial velocity is 0

$m_{R}v_{Rx}=m_{D}v_{D2x}+m_{R}v_{R2x}$

$v_{D2x}=\frac{m_{R}*v_{R1x}-m_{R}*v_{R2x}}{m_{D}}$

$v_{D2x}=\frac{(45kg)(14m/s)-(45kg)(8cos(55.1)m/s)}{(70kg)}=6m/s$

Now, lets analyze the movement in the vertical direction

$p_{1y}=p_{2y}$

Since $p_{1y}=0$

$0=m_{D}v_{D2y}+m_{R}v_{R2y}$

$v_{D2y}=-\frac{m_{R}v_{2Ry}}{m_{D}}=-\frac{(45kg)(8sin(55.1)m/s)}{(70kg)}=-4.21m/s$

Now, we can find the magnitude of Daniel's velocity after de collision

$v_{D}=\sqrt{(6m/s)^2+(-4.21m/s)^2}=7.32m/s$

b) To know whats the direction of Daniel's velocity we need to solve the arctan of the angle

$\alpha =tan^{-1}(\frac{v_{y}}{v_{x}})=tan^{-1}(\frac{-4.21}{6})=-35$ º

c) The change in the total kinetic energy is:

ΔK= $K_{2}-K_{1}$

ΔK= $\frac{1}{2}[(45kg)(8m/s)^2+(70kg)(7.32m/s)^2-(45kg)(14m/s)^2]=-1094.62J$

That means that the kinetic energy decreases

5 0

3 years ago

While accelerating at 5.22 m/s/s an object changes its velocity from 6.73 m/s to 29.88 m/s. Over what

yan [13]

Answer:

Explanation:

v² = u² + 2as

s = (v² - u²) / 2a

s = (29.88² - 6.73²) / (2(5.22))

s = 81.1802203065...

s = 81.18 m

4 0

3 years ago