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Akimi4 [234]
3 years ago
14

how could you measure the amount of elastic potential energy in a streched rubber band? think about the definition of energy whe

n planning your investigation
Physics
1 answer:
irina1246 [14]3 years ago
6 0
<h2>The elastic potential energy gained by the rubber band is the work done by you in elongating it. Positive work is done on the rubber band, so it gains energy. While work is done by you (i.e. negative work is done on you), so you lose some energy. The total energy is conserved. </h2><h2> </h2><h2>In simple words, the energy gained by the rubber band is the energy lost by you while elongating it.</h2>
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An overhead door is guided by wheels at a and b that roll in horizontal and vertical tracks. when θ = 40°, the velocity of wheel
Karo-lina-s [1.5K]

I think the situation is modeled by the scenario in the attached image. Some specific values seem to be missing (like the height of door d)...

The door forms a right triangles that satisfies

\tan\theta=\dfrac ab\implies\sec^2\theta\dfrac{\mathrm d\theta}{\mathrm dt}=\dfrac{b\frac{\mathrm da}{\mathrm dt}-a\frac{\mathrm db}{\mathrm dt}}{b^2}

We also have

\tan\theta=\dfrac ab\implies\cos\theta=\dfrac bd

so if you happen to know the height of the door, you can solve for b and a.

d is fixed, so

a^2+b^2=d^2\implies2a\dfrac{\mathrm da}{\mathrm dt}+2b\dfrac{\mathrm db}{\mathrm dt}=0\implies\dfrac{\mathrm da}{\mathrm dt}=-\dfrac ba\dfrac{\mathrm db}{\mathrm dt}

We can solve for the angular velocity \dfrac{\mathrm d\theta}{\mathrm dt}:

\dfrac{\mathrm d\theta}{\mathrm dt}=\cos^2\theta\dfrac{b\left(-\frac ba\frac{\mathrm db}{\mathrm dt}\right)-a\frac{\mathrm db}{\mathrm dt}}{b^2}=-\dfrac1a\dfrac{\mathrm db}{\mathrm dt}

At the point when \theta=40^\circ and \dfrac{\mathrm db}{\mathrm dt}=1.8 ft/s, we get

\dfrac{\mathrm d\theta}{\mathrm dt}=-\dfrac{1.8}a\dfrac{\rm deg}{\rm s}=-\dfrac{1.8}{d\sin40^\circ}\dfrac{\rm deg}{\rm s}

6 0
2 years ago
A chemical formula uses the chemical symbol and dots representing electrons.
bogdanovich [222]

Answer: FALSE

Explanation:

7 0
2 years ago
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A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (
Xelga [282]

Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants (v_{oy} = 0) zero, so let's use the equation

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     y= - ½ g t²

     t = √2y/g

b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

     x = vox t

     x = v₀ₓ √2y/g

c) Speeds before touching the ground

     vₓ = vox = constant

     v_{y} = v_{oy} - gt

     v_{y} = 0 - g √2y/g

    v_{y}  = - √2gy

    tan θ = Vy / vx

    θ = tan⁻¹ (vy / vx)

    θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

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3 years ago
If a person believes that dreams have hidden meaning he or she would agree with Freud's ideas about _ content
uysha [10]
A.lucid because it makes more sense yo the answer how about you do it yourself
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What will happen to the current if the voltage is reduced to one half? ​
stira [4]
We use v=IR and assuming the resistance doesn’t change we can also say that the voltage and current (I) are directly proportional which means the voltage also decreases by 1/2
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