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3 years ago

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a solid sphere is rolling without slipping along a horizontal surface with a speed of 5.5 meters per second when it starts up a

ramp that makes an angle of 25 degrees with the horizontal what is the speed of the sphere after it has Road 3M up the ramp

1 answer:

iVinArrow [24]3 years ago

8 0

Answer:

3.53 m/s

Explanation:

KE₁ + RE₁ = KE₂ + RE₂ + PE

½ mv₁² + ½ Iω₁² = ½ mv₂² + ½ Iω₂² + mgh

For a solid sphere, I = ⅖ mr²

½ mv₁² + ½ (⅖ mr²) ω₁² = ½ mv₂² + ½ (⅖ mr²) ω₂² + mgh

½ mv₁² + ⅕ mr² ω₁² = ½ mv₂² + ⅕ mr² ω₂² + mgh

Rolling without slipping means v = ωr.

½ mv₁² + ⅕ mv₁² = ½ mv₂² + ⅕ mv₂² + mgh

⁷/₁₀ mv₁² = ⁷/₁₀ mv₂² + mgh

⁷/₁₀ v₁² = ⁷/₁₀ v₂² + gh

Substitute:

⁷/₁₀ (5.50 m/s)² = ⁷/₁₀ v₂² + (9.8 m/s²) (3.00 m sin 25.0°)

v₂ = 3.53 m/s

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At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir

Agata [3.3K]

Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

$\omega = \omega_{o} + \alpha \cdot t$

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

Where:

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

$\omega_{o}$ , $\omega$ - Initial and final angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

$\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}$

$\theta-\theta_{o} = 134.88\,rad$

The final angular angular speed can be found by the equation:

$\omega = \omega_{o} + \alpha \cdot t$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)$

$\omega = 87.4\,\frac{rad}{s}$

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

$\Delta \theta = 134.88\,rad + 435\,rad$

$\Delta \theta = 569.88\,rad$

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

The angular acceleration is now cleared:

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$

Given that $\omega_{o} = 87.4\,\frac{rad}{s}$ , $\omega = 0\,\frac{rad}{s}$ and $\theta-\theta_{o} = 435\,rad$ , the angular deceleration is:

$\alpha = \frac{ \left(0\,\frac{rad}{s}\right)^{2}-\left(87.4\,\frac{rad}{s} \right)^{2}}{2\cdot \left(435\,rad\right)}$

$\alpha = -8.780\,\frac{rad}{s^{2}}$

Now, the time interval of the Deceleration Phase is obtained from this formula:

$\omega = \omega_{o} + \alpha \cdot t$

$t = \frac{\omega - \omega_{o}}{\alpha}$

If $\omega_{o} = 87.4\,\frac{rad}{s}$ , $\omega = 0\,\frac{rad}{s}$ and $\alpha = -8.780\,\frac{rad}{s^{2}}$ , the time interval is:

$t = \frac{0\,\frac{rad}{s} - 87.4\,\frac{rad}{s} }{-8.780\,\frac{rad}{s^{2}} }$

$t = 9.954\,s$

The total time needed for the grinding wheel before stopping is:

$t_{T} = 2.40\,s + 9.954\,s$

$t_{T} = 12.354\,s$

The grinding wheel stop at t = 12.354 seconds.

c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

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4 years ago

A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At

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Answer:

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Explanation:

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E = σ/(2ε₀)

Where;

E is the electric field

σ is the surface charge density

ε₀ is the electric constant.

Formula to calculate σ is;

σ = Q/A

Where;

Q is the total charge of the sheet

A is the sheet's area.

We are told the elastic sheet is a square with a side length as d, thus ;

A = d²

So;

σ = Q/d²

Putting Q/d² for σ in the electric field equation to obtain;

E = Q/(2ε₀d²)

Now, we can see that E is inversely proportional to the square of d i.e.

E ∝ 1/d²

The electric field at P has some magnitude E. We now double the side length of the sheet to 2L while keeping the same amount of charge Q distributed over the sheet.

From the relationship of E with d, the magnitude of electric field at P will now have a quarter of its original magnitude which is;

E_new = E/4

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3 years ago

What is the derivative of x^2 - x + 3 at the point x = 5?

Hunter-Best [27]

Answer:

9

Explanation:

d/dx (x² - x + 3)

= 2x - 1

when x = 5,

2x-1

= 2(5) - 1

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3 years ago

A transformer consists of 290 primary windings and 824 secondary windings. Part A: If the potential difference across the primar

jasenka [17]

Answer:

Part 1) Voltage in secondary windings is 61.08 Volts

Part 2) Current in secondary windings is 0.53 Amperes

Explanation:

The potential developed in the primary and secondary winding of a transformer are related as

$\frac{N_{p}}{N_{s}}=\frac{V_{p}}{V_{s}}$

where

Np no of turns in primary coil

Ns no of turns in secondary coil

Vp Voltage of turns in primary coil

Vs Voltage of turns in secondary coil

Applying values in the formula we get

$\frac{290}{824}=\frac{21.5}{V_{s}}\\\\\therefore V_{s}=21.5\times \frac{824}{290}=61.08V$

Part 2)

Using Ohm's law the current is given by

$I=\frac{V_{s}}{R}\\\\I=\frac{61.089}{115}=0.53A$

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3 years ago