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tensa zangetsu [6.8K]

3 years ago

15

Formula for working maximum velocity?

2 answers:

Varvara68 [4.7K]3 years ago

8 0

Answer:

v=Aω.

Explanation:

hope this helps you!!

MA_775_DIABLO [31]3 years ago

3 0

Answer:

Now, we know that velocity is maximum when y=0, i.e., displacement is zero and acceleration is zero, which means the system is in equilibrium. Therefore, at a point in simple harmonic motion, the maximum velocity can be calculated using the formula v=Aω.

Explanation:

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During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive

Katyanochek1 [597]

Answer:

The force is $F = 1041.7N$

Explanation:

The moment of Inertia I is mathematically evaluated as

$I = MR_A^2$

Substituting $1.9kg$ for M(Mass of the wheel) and $\frac{66cm}{2} * \frac{1m}{100cm} = 0.33m$ for $R_A$ (Radius of wheel)

$I = 1.9 * 0.33^2$

$= 0.207kgm^2$

The torque on the wheel due to net force is mathematically represented as

$\tau = FR_B - F_rR_A$

Substituting 135 N for $F_r$ (Force acting on sprocket), $\frac{8.7cm}{2} * \frac{1m}{100cm} = 0.0435m$ for $R_B$ (radius of the chain) and F is the force acting on the sprocket due to the chain which is unknown for now

$\tau = F (0.0435) - 135 (0.33)$

This same torque due to the net force is the also the torque that is required to rotate the wheel to have an angular acceleration of $\alpha = 3.70 rad/s^2$ and this torque can also be represented mathematically as

$\tau = \alpha I$

Now equating the two equation for torque

$F (0.0435) - 135 (0.33) = \alpha I$

Making F the subject

$F = \frac{\alpha I + (135*0.33) }{0.0435}$

Substituting values

$F = \frac{(3.70 * 0.207) + (135*0.33)}{0.0435}$

$= 1041.7N$

4 0

3 years ago

You have a rod with a length of 146.4 cm. You prop up one end on a brick which is 3.8 cm thick. Your uncertainty in measuring th

AfilCa [17]

Answer:

$\partial \theta = 0.003$

Explanation:

we know that

$sin\theta = \frac{3.8}{146.4}$

$\theta = sin^{-1} \frac{3.8}{146.4}$

$\theta = 1.484°$

$\theta = 1.484° *\frac{\pi}{180} = 0.0259 radians$

as we see that $sin\theta = \theta$

relative error $\frac{\partial \theta}{\theta} = \frac{\partial X}{X_1} +\frac{\partial X}{X_2}$

Where X_1 IS HEIGHT OF ROCK

$X_2$ IS THE HEIGHT OF ROAD

$\partial X$ = uncertainity in measuring distance

$\partial X = 0.05$

Putting all value to get uncertainity in angle

$\frac{\partial \theta}{0.0259} = \frac{0.05}{3.8} +\frac{0.05}{146.4}$

solving for $\partial \theta$ we get

$\partial \theta = 0.003$

3 0

3 years ago

Optical tweezers use light from a laser to move single atoms and molecules around. Suppose the intensity of light from the tweez

WITCHER [35]

Answer:

Explanation:

a )

If it is totally absorbed pressure is calculated as follows .

Pressure = I / c where I is intensity of light falling .

= 1000 / 3 x 10⁸

= 3.33 x 10⁻⁶ N / m²

b ) weight of tritium atom

= 3 x 1.67 x 10⁻²⁷ kg

acceleration = force / mass

= 3.33x 10⁻⁶ / 3 x 1.67 x 10⁻²⁷

= .6646 x 10²¹ m /s²

= 66.46 x 10¹⁹ m / s²

7 0

3 years ago

A car travels 30 km north in 25 min. and 40 km east in 35 min. What is the total time in hours? Be careful to carry over the pro

inna [77]

25+35=60mins so 1.0 hours

3 0

3 years ago

Read 2 more answers

A 4.0-m-diameter playground merry-go-round, with a moment of inertia of

HACTEHA [7]

Answer:

$7.1$ ms⁻¹

Explanation:

$d$ = diameter of merry-go-round = 4 m

$r$ = radius of merry-go-round = $\frac{d}{2}$ = $\frac{4}{2}$ = 2 m

$I$ = moment of inertia = 500 kgm²

$w_{i}$ = angular velocity of merry-go-round before ryan jumps = 2.0 rad/s

$w_{f}$ = angular velocity of merry-go-round after ryan jumps = 0 rad/s

$v$ = velocity of ryan before jumping onto the merry-go-round

$m$ = mass of ryan = 70 kg

Using conservation of angular momentum

$Iw_{i} - m v r = (I + mr^{2})w_{f}$

$(500)(2.0) - (70) v (2) = (I + mr^{2})(0)$

$1000 = 140 v$

$v = 7.1$ ms⁻¹

5 0

3 years ago