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Juli2301 [7.4K]
2 years ago
11

A flywheel of diameter 1.2 m has a constant angular acceleration of 5.0 rad/s2. The tangential acceleration of a point on its ri

m is:_____.
a. 12 m/s2.
b. 6.0 m/s2.
c. 5.0 m/s2.
d. 5.0 rad/s2.
e. 3.0 m/s2.
Physics
1 answer:
Anna71 [15]2 years ago
3 0

Answer:

3 m/s²

Explanation:.

To solve this, we use the formula of tangential acceleration. The tangential acceleration is given by the following equation:

a(t) = r * a(a)

where

a(t) is the tangential acceleration, m/s²

r is the radius, m

a(a) is the angular acceleration, rad/s²

Again, we know that the radius can be gotten by saying r = d/2, thus, r = 1.2/2 = 0.6 m

We then multiply this by the angular acceleration, to get our tangential acceleration

a(t) = 0.6 * 5 = 3 m/s²

Therefore, the Tangential acceleration of a point on the flywheel rim has been found to be 3 m/s²

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sukhopar [10]

Answer:

Mass and height

Explanation:

Gravitational potential energy is energy an object possesses because of its position in a gravitational field. The most common use of gravitational potential energy is for an object near the surface of the Earth where the gravitational acceleration can be assumed to be constant at about 9.8 m/s2.

Which is represented as;

PE_g=mgh

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Here we see that mass of object and height is directly proportional to the gravitational potential energy.

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2 years ago
A 1.80-m string of weight 0.0126 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v
Veseljchak [2.6K]

Answer:

W = 0.135 N

Explanation:

Given:

- y (x, t) = 8.50*cos(172*x -2730*t)

- Weight of string m*g = 0.0126 N

- Attached weight = W

Find:

The attached weight W given that Tension and W are equal.

Solution:

The general form of standing mechanical waves is given by:

                            y (x, t) = A*cos(k*x -w*t)  

Where k = stiffness and w = angular frequency

Hence,

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- Tension in the string T:

                            T = Y*V^2

where Y: is the mass per unit length of the string.

- The tension T and weight attached W are equal:

                           T = W = Y*V^2 = (w/L*g)*V^2

                            W = (0.0126 / 1.8*9.81)*(13.78)^2

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</span>
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Answer:

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\longrightarrow \:  \:  \sf\Delta x .\Delta p =  \dfrac{h}{4\pi}

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34}} {4 \times  \frac{22}{7} }

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\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34} } { 8  \times 24 \times  {10}^{ - 15} }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34} } { 192 \times  {10}^{ - 15} }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34}  \times  {10}^{15} } { 192}

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\longrightarrow \:  \:  \sf\Delta p =  \dfrac{4382 \times  {10}^{ - 21}   } { 192}

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\longrightarrow \:  \:  \sf\Delta p = 2.2822 \times  {10}^{1} \times  {10}^{ - 21}

\longrightarrow \:  \: \underline{ \boxed{ \red{  \bf\Delta p = 2.2822 \times  {10}^{ - 20}  \:  kg/ms}}}

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