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2 years ago

Una rueda gira con una frecuencia de 530 rpm. Determina la velocidad angular, el periodo y la frecuencia.

Physics

1 answer:

Vlad1618 [11]2 years ago

6 0

Answer:

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Explanation:

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A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9 degrees below th

zhannawk [14.2K]

Answer: $V_{f}=2.96m/s$

Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.

Where the weight $w$ of the block has an x-component and y-component:

$w_{x}=wsin(\theta)$ (1)

$w_{y}=wcos(\theta)$ (2)

As well as the Normal Force $N$ :

$N_{x}=Nsin(\theta)$ (3)

$N_{y}=Ncos(\theta)$ (4)

In addition, we know $N=w$ , then $\sum F_{y}=0$

In the X-component:

$\sum F_{x}=m.a$

$m.a=w_{x}$ (5)

Substituting (1) in (5):

$wsin(\theta)=m.a$ (6)

In addition, we know $w=m.g$ , where $m$ is the mass of the block and $g$ the gravity acceleration, which is equal to $9.8m/{s}^{2}$

So:

$m.g.sin(\theta)=m.a$ (7)

$a=g.sin(\theta)$ (8)

$a=5.88m/{s}^{2}$ (9) >>>>This is the acceleration of the block

On the other hand, we have the following equation that expresses a <u>relation between</u> the distance $d$ with the acceleration $a$ and time $t$ :

$d=\frac{1}{2}a{t}^{2}$ (10)

We already know the value of $d$ and calculated $a$ , we have to find $t$ :

$t=\sqrt{\frac{2d}{a}}$ (11)

$t=\sqrt{\frac{2(0.75m)}{5.88m/{s}^{2}}}$ (12)

$t=0.50s$ (13) >>>This is the time it takes to the block to go from the initial velocity $V_{o}$ to its final velocity $V_{f}$

If the acceleration is the variation of the velocity in time, we can use the following equation to find $V_{f}$ :

$V_{f}-V_{o}=a.t$ (13)

If $V_{o}=0$

$V_{f}=a.t$ (14)

$V_{f}=(5.88m/{s}^{2})(0.50s)$ (15)

Finally we get the value of the Final Velocity of the block:

$V_{f}=2.96m/s$

6 0

3 years ago

The device is a burner from an electric stove. It is used to transfer heat to a pot by A. Conduction B. Convection C. Radiation

Irina-Kira [14]

I’m pretty positive that it’s A. Conduction. Just refer to the definition of it.

3 0

3 years ago

Read 2 more answers

Solve the following system of equations by using either substitution or elimination.

strojnjashka [21]

I believe that the answer to the question provided above are the following;

x = 29.8410

y = 16.6794

z = -1.2642
Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.

5 0

3 years ago

A hot-water stream at 80 ℃ enters a mixing chamber with a mass flow rate of 0.5 kg/s where it is mixed with a stream of cold wat

rewona [7]

Answer:

$\dot{m_{2}}=0.865 kg/s$

Explanation:

$\dot{m_1}= 0.5kg/s$

from steam tables , at 250 kPa, and at

T₁ = 80⁰C ⇒ h₁ = 335.02 kJ/kg

T₂ = 20⁰C⇒ h₂ = 83.915 kJ/kg

T₃ = 42⁰C ⇒ h₃ = 175.90 kJ/kg

we know

$\dot{m_{in}}=\dot{m_{out}}$

$\dot{m_{1}}+\dot{m_{2}}=\dot{m_{3}}$

according to energy balance equation

$\dot{m_{in}}h_{in}=\dot{m_{out}}h_{out}$

$\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}$

$\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=(\dot{m_{1}}+\dot{m_{2}})h_{3}\\(0.5\times 335.02)+(\dot{m_{2}}\times 83.915)=(0.5+\dot{m_{2}})175.90\\\dot{m_{2}}=0.865 kg/s$

4 0

3 years ago

Can anyone solve these for my by using unit vectors? Can you also please show your work

Oxana [17]

4. The Coyote has an initial position vector of $\vec r_0=(15.5\,\mathrm m)\,\vec\jmath$ .

4a. The Coyote has an initial velocity vector of $\vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath$ . His position at time $t$ is given by the vector

$\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2$

where $\vec a$ is the Coyote's acceleration vector at time $t$ . He experiences acceleration only in the downward direction because of gravity, and in particular $\vec a=-g\,\vec\jmath$ where $g=9.80\,\frac{\mathrm m}{\mathrm s^2}$ . Splitting up the position vector into components, we have $\vec r=r_x\,\vec\imath+r_y\,\vec\jmath$ with

$r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t$

$r_y=15.5\,\mathrm m-\dfrac g2t^2$

The Coyote hits the ground when $r_y=0$ :

$15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s$

4b. Here we evaluate $r_x$ at the time found in (4a).

$r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m$

5. The shell has initial position vector $\vec r_0=(1.52\,\mathrm m)\,\vec\jmath$ , and we're told that after some time the bullet (now separated from the shell) has a position of $\vec r=(3500\,\mathrm m)\,\vec\imath$ .