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3 years ago

12

una caja de 25 kg se mueve durante 10 s. sobre una superficie horizontal , sin rozamiento a partir del reposo hasta alcanzar una

velocidad de 50 cm/s ¿cuál es la magnitud de la fuerza neta que actúa sobre dicha caja?

2 answers:

Lena [83]3 years ago

7 0

Please translate in English. id be happy to help but i can't :/

almond37 [142]3 years ago

6 0

I dont know the answer i need points

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Question in pic.. thanks!

11Alexandr11 [23.1K]

That's false. Displacement would be (r2 - r1) .

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3 years ago

Compare and contrast the particle motion in each state of matter

Vladimir79 [104]

Answer:

Gases, liquids and solids are all made up of microscopic particles, but the behaviors of these particles differ in the three phases. ... gas are well separated with no regular arrangement. liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.

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3 years ago

4. What is the acceleration of the car in each section?<br> b<br> с<br> d<br> a

makvit [3.9K]

Answer:

0-4 acceleration comes at 12 m/s where (B) stagnates at 12 m/s and remains for 4 seconds (C) is breaks being activated slowing the car to 6 m/s in 2 seconds and (D) over the course of 4 seconds brings the car to 10 m/s.

Explanation:

3 0

3 years ago

Read 2 more answers

A string that is under 54.0 N of tension has linear density 5.20 g/m . A sinusoidal wave with amplitude 2.50 cm and wavelength 1

kicyunya [14]

Answer:

8.89288275 m/s

Explanation:

F = Tension = 54 N

$\mu$ = Linear density of string = 5.2 g/m

A = Amplitude = 2.5 cm

Wave velocity is given by

$v=\sqrt{\frac{F}{\mu}}\\\Rightarrow v=\sqrt{\frac{54}{5.2\times 10^{-3}}}\\\Rightarrow v=101.90493\ m/s$

Frequency is given by

$f=\frac{v}{\lambda}\\\Rightarrow f=\frac{101.90493}{1.8}\\\Rightarrow f=56.61385\ Hz$

Angular frequency is given by

$\omega=2\pi f\\\Rightarrow \omega=2\pi 56.61385\\\Rightarrow \omega=355.71531\ rad/s$

Maximum velocity of a particle is given by

$v_m=A\omega\\\Rightarrow v_m=0.025\times 355.71531\\\Rightarrow v_m=8.89288275\ m/s$

The maximum velocity of a particle on the string is 8.89288275 m/s

5 0

3 years ago

A rock of mass 200 g is attached to a 0.75 m long string and swung in a vertical plane.

Ainat [17]

Hello!

a) Assuming this is asking for the minimum speed for the rock to make the full circle, we must find the minimum speed necessary for the rock to continue moving in a circular path when it's at the top of the circle.

At the top of the circle, we have:

- Force of gravity (downward)

*Although the rock is still connected to the string, if the rock is swinging at the minimum speed required, there will be no tension in the string.

Therefore, only the force of gravity produces the net centripetal force:

$\Sigma F = F_g\\\\F_c = F_g\\\\\frac{mv^2}{r} = mg$

We can simplify and rearrange the equation to solve for 'v'.

$\frac{v^2}{r} = g\\\\v^2 = gr\\\\v = \sqrt{gr}$

Plugging in values:

$v = \sqrt{9.8 * 0.75} = \boxed{2.711 m/s^}$

b)
Let's do a summation of forces at the bottom of the swing. We have:
- Force due to gravity (downward, -)

- Tension force (upward, +)

The sum of these forces produces a centripetal force, upward (+).

$\Sigma F = T - F_g\\\\F_c = T - F_g\\\\\frac{mv^2}{r} = T - mg$

Rearranging for 'T":
$T = \frac{mv^2}{r} + mg\\\\$

Plugging in the appropriate values:
$T = \frac{(0.2)(2.711^2)}{(0.75)} + 0.2(9.8) = \boxed{3.92 N}$

5 0

2 years ago

Read 2 more answers