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3 years ago

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A block with mass 0.470 kg sits at rest on a light but not long vertical spring that has spring constant 85.0 N/m and one end on

the floor. A second identical block is dropped onto the first from a height of 4.40 m above the first block and sticks to it. What is the maximum elastic potential energy stored in the spring during the motion of the blocks after the collision?

1 answer:

a_sh-v [17]3 years ago

4 0

Answer: elastic potential energy = 20.27 J

Explanation:

Given that the

Mass M = 0.470 kg

Height h = 4.40 m

Spring constant K = 85 N/m

The maximum elastic potential will be equal to the maximum kinetic energy experienced by the block.

But according to conservative of energy, the maximum kinetic energy is equal to the maximum potential energy experienced by the block of mass M.

That is

K .E = P.E = mgh

Where g = 9.8m/s^2

Substitutes all the parameters into the formula

K.E = 0.470 × 9.8 × 4.4

K.E = 20.27 J

Where K.E = maximum elastic potential energy stored in the spring during the motion of the blocks after the collision which is 20.27J.

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What is the relationship between the horizontal and vertical components of velocity for a projectile launched

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Answer: The horizontal velocity of a projectile is constant (a never changing in value. The vertical velocity of a projectile changes by 9.8 m/s each second.

Explanation: I hope that helped!

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3 years ago

What is the SI unit for kinetic energy

trapecia [35]

The easiest way to build a unit for energy is to remember that
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That unit is so complicated that it's been given a special,
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3 years ago

Read 2 more answers

Fill in the blanks for the following:

storchak [24]

Answer:

<em>a. 4.21 moles</em>

<em>b. 478.6 m/s</em>

<em>c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>

Explanation:

Volume of container = 100.0 L

Temperature = 293 K

pressure = 1 atm = 1.01325 bar

number of moles n = ?

using the gas equation PV = nRT

n = PV/RT

R = 0.08206 L-atm- $mol^{-1}$ $K^{-1}$

Therefore,

n = (1.01325 x 100)/(0.08206 x 293)

n = 101.325/24.04 = <em>4.21 moles</em>

The equation for root mean square velocity is

Vrms = $\sqrt{\frac{3RT}{M} }$

R = 8.314 J/mol-K

where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol

Vrms = $\sqrt{\frac{3*8.314*293}{0.0319} }$ = <em>478.6 m/s</em>

<em>For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship</em>

$\frac{Voxy}{Vnit}$ = $\sqrt{\frac{Mnit}{Moxy} }$

where

Voxy = root mean square velocity of oxygen = 478.6 m/s

Vnit = root mean square velocity of nitrogen = ?

Moxy = Molar mass of oxygen = 31.9 g/mol

Mnit = Molar mass of nitrogen = 14.00 g/mol

$\frac{478.6}{Vnit}$ = $\sqrt{\frac{14.0}{31.9} }$

$\frac{478.6}{Vnit}$ = 0.66

Vnit = 0.66 x 478.6 = <em>315.876 m/s</em>

<em>the root mean square velocity of the oxygen gas is </em>

<em>478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>

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2 years ago

Acceleration = change of velocity divided by time interval = Δv/Δt.

MariettaO [177]

Answer:

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3 years ago

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matrenka [14]

Answer:

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The question assumes that the skier is in a free-fall motion. In other words, the skier travels with a constant horizontal velocity and accelerates downwards at $g$ ( $g \approx \rm -9.81\; m\cdot s^{-2}$ near the surface of the earth.) At $t$ seconds after the skier goes beyond the edge of the ramp, the position of the skier will be:

$x$ -coordinate: $30.556t$ meters (constant velocity;)
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To eliminate $t$ from this expression, solve the equation between $t$ and $x$ for $t$ . That is: express $t$ as a function of $x$ .

$x = 30.556\;t\implies \displaystyle t = \frac{x}{30.556}$ .

Replace the $t$ in the equation of $y$ with this expression:

$\begin{aligned} y = &-\frac{9.81}{2}\cdot t^{2}\\ &= -\frac{9.81}{2} \cdot \left(\frac{x}{30.556}\right)^{2}\\&= -0.0052535\;x^{2}\end{aligned}$ .

Plot the two functions:

$y = -x$ ,
$\displaystyle y= -0.0052535\;x^{2}$ ,

and look for their intersection. Refer to the diagram attached.

Alternatively, equate the two expressions of $y$ (right-hand side of the equation, the part where $y$ is expressed as a function of $x$ .)

$-0.0052535\;x^{2} = -x$ ,

$\implies x = 190.35$ .

The value of $y$ can be found by evaluating either equation at this particular $x$ -value: $x = 190.35$ .

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4 0

3 years ago