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2 years ago

The reciprocal of frequency is

1 answer:

Colt1911 [192]2 years ago

8 0

Frequency=1/T, where T is the time period.
Reciprocal of frequency = T or it can be written as T = 1/f, where f is frequency.

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If the mass of an object increases, how is its acceleration affected, assuming the net force acting on the object remains the sa

vovikov84 [41]

Based on Newton's second law of motion, the net force applied to an object is equal to the product of the mass of the object and the acceleration it experiences. That is,

F = ma

If we are to assume that the net force is constant and that the mass is increased, the acceleration should therefore decrease in order to make constant the value at the right-hand side of the equation.

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2 years ago

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kompoz [17]

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3 years ago

4. Using the bone density of 2.0 kg/m3, calculate the mass of an adult femur bone that has a volume of 0.00027 m3.

kolbaska11 [484]

Answer:

$\boxed{\sf Mass \ of \ an \ adult \ femur \ bone = 0.00054 \ kg}$

Given:

Bone density = 2.0 kg/m³

Volume of bone (V) = 0.00027 m³

To Find:

Mass of an adult femur bone (m).

Explanation:

$\sf \implies Density = \frac{Mass (m)}{Volume (V)} \\ \\ \sf \implies \frac{Mass}{Volume} = Density \\ \\ \sf \implies Mass = Density \times Volume \\ \\ \sf \implies Mass = 2.0 \ kg/ \cancel{m^{3}} \times 0.00027 \ \cancel{m^{3}} \\ \\ \sf \implies Mass = 0.00054 \ kg$

6 0

3 years ago

A girl stands on the edge of a merry-go-round of radius 1.71 m. If the merry go round uniformly accerlerates from rest to 20 rpm

Mashutka [201]

Answer:

$a = 0.53 m/s^2$

Explanation:

initially the merry go round is at rest

after 6.73 s the merry go round will accelerates to 20 rpm

so final angular speed is given as

$\omega = 2\pi f$

$\omega = 2\pi ( \frac{20}{60})$

$\omega = 2.10 rad/s$

so final tangential speed is given as

$v = r\omega$

$v = 1.71 (2.10) = 3.58 m/s$

now average acceleration of the girl is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{3.58 - 0}{6.73}$

$a = 0.53 m/s^2$

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3 years ago

A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.1 rad/s in 2.99 s.(a) fi

elena55 [62]

The angular acceleration of a rotating object is given by
$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$
where
$\omega_f$ is the final angular speed of the object
$\omega_i$ is its initial angular speed
$\Delta t$ is the time taken to accelerate

For the wheel in our problem, $\omega_f=11.1 rad/s$ , $\omega_i = 0$ and $\Delta t=2.99 s$ , so its angular acceleration is
$\alpha= \frac{11.1 rad/s-0}{2.99 s}=3.71 rad/s^2$

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2 years ago