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3 years ago

A child on a playground swing makes a complete to-and-fro swing each two seconds. The frequency of the swing is

1 answer:

4 0

Frequency=f
T=period of motion
f=1/T therefore f=1/2=0.500Hz or 0.5Hz

Hope this helps. Any questions please just ask. Thank you.

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Derive the kinetic equations for Vmax and KM using the transit time and net rate constant method as discussed in class and follo

Pepsi [2]

Answer:

Rate = vmax k3/k2+k3

Explanation:

The rate of reaction when the enzyme is saturated with substrate is the maximum rate of reaction, is referred to as Vmax.

This is usually expressed as the Km ie. Michaelis constant of the enzyme, an inverse measure of affinity. For practical purposes, Km is the concentration of substrate which permits the enzyme to achieve half Vmax.

Please kindly check attachment for the step by step solution of the given problem.

6 0

3 years ago

I dolt understand what to do next please help

lidiya [134]

Divide 24 by 12.
24/t = 12
24/12 = t
24/12 = 2
t = 2

6 0

3 years ago

An automatic dryer spins wet clothes at an angular speed of 5.2 rad/s. Starting from rest, the dryer reaches its operating speed

Dvinal [7]

<h2>Time taken by dryer to come up to speed is 1.625 seconds.</h2>

Explanation:

We have equation of motion v = u + at

Initial velocity, u = 0 rad/s

Final velocity, v = 5.2 rad/s

Time, t = ?

Acceleration, a = 3.2 rad/s²

Substituting

v = u + at

5.2 = 0 + 3.2 x t

t = 1.625 s

Time taken by dryer to come up to speed is 1.625 seconds.

6 0

3 years ago

What is tan θ for the given triangle?

yulyashka [42]

Tan = opposite/adjacent
= 20/15
=4/3

7 0

3 years ago

Read 2 more answers

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0

3 years ago