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2 years ago

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A factory robot drops a 10 kg computer onto a conveyor belt running at 3.1 m/s. The materials are such that μs = 0.50 and μk = 0

.30 between the belt and the computer. How far is the computer dragged before it is riding smoothly on the belt?

1 answer:

frutty [35]2 years ago

8 0

Answer:

x = 1.63 m

Explanation:

mass (m) = 10 kg

μk = 0.3

velocity (v) = 3.1 m/s

Assuming that most of the computers weight is applied on the belt instantaneously, we can apply the constant acceleration equation below

x = $v^{2}/2a$

where a = μk.g , therefore

x = $v^{2}$ /2μk.g

x = (3.1 x 3.1)/(2 x 0.3 x 9.8)

x = 1.63 m

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calculate the force between two objects that have masses of 20 kg and 100 kg separated by a distance of 2.6 m

olga55 [171]

The gravitational force between the objects is $1.97\cdot 10^{-8} N$

Explanation:

The magnitude of the gravitational force between two objects is given by the equation:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

$m_1, m_2$ are the masses of the two objects

r is the separation between the objects

For the two objects in this problem:

$m_1 = 20 kg\\m_2 = 100 kg$

And their distance is

r = 2.6 m

So, the gravitational force between them is

$F=\frac{(6.67\cdot 10^{-11})(20)(100)}{2.6^2}=1.97\cdot 10^{-8} N$

Learn more about gravitational force:

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brainly.com/question/12785992

#LearnwithBrainly

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3 years ago

se lanza un cuerpo desde el origen con velocidad horizontal de 40 m/s, y con un ángulo de 60º. calcular la máxima altura y el al

EastWind [94]

Answer:

1. $h = 244.8 m$

2. $x = 564.8 m$

Explanation:

1. La altura máxima se puede calcular usando la siguiente ecuación:

$v_{f}^{2} = v_{0}^{2} - 2gh$ (1)

Where:

$v_{f_{y}}$ : es la velocidad final = 0 (en la altura máxima)

$v_{0_{y}}$ : es la velocidad inicial horizontal en "y"

g: es la gravedad = 9.81 m/s²

h: es la altura máxima =?

La velocidad incial en "y" se puede calcular de la siguiente manera:

$tan(\theta) = \frac{v_{0_{y}}}{v_{0_{x}}}$

$v_{0_{y}} = tan(60)*40 m/s = 69.3 m/s$

Resolviendo la ecuación (1) para "h" tenemos:

$h = \frac{v_{0_{y}}^{2}}{2g} = \frac{(69.3 m/s)^{2}}{2*9.81 m/s^{2}} = 244.8 m$

2. Para calcular el alcance horizontal podemos usar la ecuación:

$x = v_{x}*t$

Primero debemos encontrar el tiempo cuando la altura es máxima ( $v_{f_{y}}$ = 0).

$v_{f_{y}} = v_{0_{y}} - gt$

$t = \frac{v_{0_{y}}}{g} = \frac{69.3 m/s}{9.81 m/s^{2}} = 7.06 s$

Ahora, como el tiempo de subida es el mismo que el tiempo de bajada, el tiempo máximo es:

$t_{m} = 2*7.06 s = 14.12 s$

Finalmente, el alcance horizontal es:

$x = 40 m/s*14.12 s = 564.8 m$

Espero que te sea de utilidad!

7 0

2 years ago

When fat comes in contact with sodium hydroxide, it produces soap and glycerin. Determine whether this is a physical change or a

DochEvi [55]

A quick, easy way to decide whether there was a chemical change

is to look and see whether there are NEW substances after the

event, that weren't there when it started.

This particular scenario started out with fat and sodium hydroxide (lye).

And then, suddenly, POOF ! Soap and glycerin showed up. Where did

THOSE come from ? They came from the molecules in fat and lye,

getting broken up and recombined to make different substances.

THAT's exactly a chemical change.

6 0

3 years ago

Read 2 more answers

Plz help me!!!! A spring is connected to a wall as shown below. A mass on a horizontal surface is connected to the springs and p

kirill [66]

Answer:

320 N/m

Explanation:

F = k·Δx

where

F is the restoring force of the spring

k is the proportionality constant called the ‘spring constant’

Δx is the change in the spring’s position due to the deformation.

You need the k so

25 cm= 0.25m

k=F/∆x = 80/0.25 = 320 N/m

6 0

3 years ago

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.

8_murik_8 [283]

The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

$V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}$

1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,

$V_a = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

The distance from the center of the square to one of the corners is $\sqrt2 L/2 = 0.035m$

$V_a = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.035} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.035} = 0$

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.

2) $V_b = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

$r_1 = 0.05\sqrt2m\\r_2 = 0.05m$

$V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between energies. This is the work-energy theorem. So,[tex]W = U_b - U_a$

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_3}{r} = Vq_3$

$W = q_3(V_b - V_a) = q_3(V_b - 0)\\W = (-2\times10^{-6})(-\frac{2.9\times10^{-6}}{\pi\epsilon_0})\\W = \frac{5.8\times10^{-12}}{\pi\epsilon_0}$

4 0

3 years ago