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3 years ago

How is the axe an example of a wedge (think about simple machines)?

Physics

1 answer:

vfiekz [6]3 years ago

5 0

Answer:

Heavier on one side lighter on the other side.

Explanation:

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One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has

Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = $\sqrt{6gL}$

Hence, v0 = 6.86 m/s

4 0

3 years ago

Question 4 (1 point)

Valentin [98]

cardiovascular endurance :))

4 0

3 years ago

Two traveling sinusoidal waves are described by the wave functions y1 = 4.85 sin [(4.35x − 1270t)] y2 = 4.85 sin [(4.35x − 1270t

Tamiku [17]

Answer:

Approximately $9.62$ .

Explanation:

$y_1 = 4.85\, \sin[(4.35\, x - 1270\, t) + 0]$ .

$y_2 = 4.85\, \sin[(4.35\, x - 1270\, t) + (-0.250)]$ .

Notice that sine waves $y_1$ and $y_2$ share the same frequency and wavelength. The only distinction between these two waves is the $(-0.250)$ in $y_2\!$ .

Therefore, the sum $(y_1 + y_2)$ would still be a sine wave. The amplitude of $(y_1 + y_2)\!$ could be found without using calculus.

Consider the sum-of-angle identity for sine:

$\sin(a + b) = \sin(a) \cdot \cos(b) + \cos(a) \cdot \sin(b)$ .

Compare the expression $\sin(a + b)$ to $y_2$ . Let $a = (4.35\, x - 1270)$ and $b = (-0.250)$ . Apply the sum-of-angle identity of sine to rewrite $y_2\!$ .

$\begin{aligned}y_2 &= 4.85\, \sin[(\underbrace{4.35\, x - 1270\, t}_{a}) + (\underbrace{-0.250}_{b})]\\ &= 4.85 \, [\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}$ .

Therefore, the sum $(y_1 + y_2)$ would become:

$\begin{aligned}& y_1 + y_2\\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t) \\ &\quad \quad \quad\;+\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}$ .

Consider: would it be possible to find $m$ and $c$ that satisfy the following hypothetical equation?

$\begin{aligned}& (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)\\&= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}$ .

Simplify this hypothetical equation:

$\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\&=\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)\end{aligned}$ .

Apply the sum-of-angle identity of sine to rewrite the left-hand side:

$\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\[0.5em]&=m\, \sin(4.35\, x - 1270\, t)\cdot \cos(c) \\ &\quad\quad + m\, \cos(4.35\, x - 1270\, t)\cdot \sin(c) \\[0.5em] &=\sin(4.35\, x - 1270\, t)\cdot (m\, \cos(c)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot (m\, \sin(c)) \end{aligned}$ .

Compare this expression with the right-hand side. For this hypothetical equation to hold for all real $x$ and $t$ , the following should be satisfied:

$\displaystyle 1 + \cos(-0.250) = m\, \cos(c)$ , and

$\displaystyle \sin(-0.250) = m\, \sin(c)$ .

Consider the Pythagorean identity. For any real number $a$ :

${\left(\sin(a)\right)}^{2} + {\left(\cos(a)\right)}^{2} = 1^2$ .

Make use of the Pythagorean identity to solve this system of equations for $m$ . Square both sides of both equations:

$\displaystyle 1 + 2\, \cos(-0.250) + {\left(\cos(-0.250)\right)}^2= m^2\, {\left(\cos(c)\right)}^2$ .

$\displaystyle {\left(\sin(-0.250)\right)}^{2} = m^2\, {\left(\sin(c)\right)}^2$ .

Take the sum of these two equations.

Left-hand side:

$\begin{aligned}& 1 + 2\, \cos(-0.250) + \underbrace{{\left(\cos(-0.250)\right)}^2 + {\left(\sin(-0.250)\right)}^2}_{1}\\ &= 1 + 2\, \cos(-0.250) + 1 \\ &= 2 + 2\, \cos(-0.250) \end{aligned}$ .

Right-hand side:

$\begin{aligned} &m^2\, {\left(\cos(c)\right)}^2 + m^2\, {\left(\sin(c)\right)}^2 \\ &= m^2\, \left( {\left(\sin(c)\right)}^2 + {\left(\cos(c)\right)}^2\right)\\ &= m^2\end{aligned}$ .

Therefore:

$m^2 = 2 + 2\, \cos(-0.250)$ .

$m = \sqrt{2 + 2\, \cos(-0.250)} \approx 1.98$ .

Substitute $m = \sqrt{2 + 2\, \cos(-0.250)}$ back to the system to find $c$ . However, notice that the exact value of $c\!$ isn't required for finding the amplitude of $(y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)$ .

(Side note: one possible value of $c$ is $\displaystyle \arccos\left(\frac{1 + \cos(0.250)}{\sqrt{2 \times (1 + \cos(0.250))}}\right) \approx 0.125$ radians.)

As long as $\! c$ is a real number, the amplitude of $(y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)$ would be equal to the absolute value of $(4.85\, m)$ .

Therefore, the amplitude of $(y_1 + y_2)$ would be:

$\begin{aligned}|4.85\, m| &= 4.85 \times \sqrt{2 + 2\, \cos(-0.250)} \\&\approx 9.62 \end{aligned}$ .

8 0

3 years ago

in the figure shown, if the mass of the block is 4kg and the coefficient of static friction is 0.5 and the coefficient of kineti

Elan Coil [88]

Answer:

Ff = 19.6 N

Explanation:

So since its saying whats the minimum F to move the block, we will use static friction (0.5).

We will use the equation for force of friction, which is Ff = uFn

Ff = (0.5)(4)(9.8)

Ff = 19.6 N

this is the minumum force needed to move the block, as that is the frictional force. You would need to apply a minimum force of 19.6 N to move the block

3 0

3 years ago

(HELP!! 10pts) a train travels 75km in 1hr, and then 68km in 2hrs. what is its average speed?

blondinia [14]

Average speed = total distance travelled ÷ total time taken

AS = (75km + 68km) ÷ (1hr + 2hr)

As = 143km ÷ 3hr

AS = 47.66667 km/hr

AS = 47.7 km/hr (3sf)

8 0

3 years ago