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KIM [24]
2 years ago
7

Why is thesize saved prior to entering the for loop? 2. what is the running time of removefirsthalf if lst is an arraylist? 3. w

hat is the running time of removefirsthalf if lst is a linkedlist?
Physics
1 answer:
Mrac [35]2 years ago
8 0
<span>After entering the loop, it should use the correct list size and the loop will be affected if the remove call changes the size of the list. If lst is an Arraylist the running time of removefirsthalf is O (n^2). So when the beginning is removed the next element will move forward. If lst is a LinkedList which is a dynamic structure the running is O (n) for removefirsthalf</span>
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At its lowest setting a centrifuge rotates with an angular speed of ω1 = 250 rad/s. When it is switched to the next higher setti
dalvyx [7]

Answer:

Part(a): The angular acceleration is 5.63~rad~s^{-2}.

Part(b): The angular displacement is 2629~rad.

Explanation:

Part(a):

If \omega_{1},~\omega_{2}~and~\alpha be the initial angular speed, final angular speed and angular acceleration  of the centrifuge respectively, then from rotational kinematic equation, we can write

\alpha = \dfrac{\omega_{2} - \omega_{1}}{t}......................................................(I)

where 't' is the time taken by the centrifuge to increase its angular speed.

Given, \omega_{i} = 250~rad~s^{-1}, \omega_{f} = 750~rad~s^{-1} and t = 9.5~s. From equation (I), the angular acceleration is given by

\alpha = \dfrac{750 - 250}{9.5}~rad~s^{-2} = 5.63~rad~s^{-2}

Part(b):

Also the angular displacement (\Delta \theta) can be written as

&&\Delta \theta = \omega_{1}~t + \dfrac{1}{2}\alpha~t^{2}\\&or,& \Delta \theta = (250 \times 9.5 + \dfrac{1}{2} \times 5.63 \times 9.5^{2})~rad = 2629~rad

8 0
2 years ago
We know that every object exerts an attraction on every other object and the heavier the object ___ the attraction.
hichkok12 [17]
I don't know what the exact word is, but I do know that the bigger an objects mass is the more it will attract other objects toward it, mainly smaller objects with less mass. it might be gravity or something around those lines....is it a multiple choice question? 
6 0
2 years ago
A rectangular glass block of dimensions 8.0 cm by 15.0 cm by 3.0 cm weighs 10.5 N. Calculate the minimum and the maximum pressur
BaLLatris [955]
The pressure can be solved using the following formula:

P = mg / A

Where: P = pressure
mg = mass * gravity = W = 10.5 N
A = l*w

Solving for A (maximum):

A = 8*3
A = 24 sq. cm

Solving for A (minimum):

A = 15*8
A = 120 sq. cm

Substituting into the first equation:

Pmax = 10.5 N / 24 sq.cm
Pmax = 0.4375 Pa

Pmin = 10.5 N / 120 sq.cm
<span>Pmin = 0.0875 Pa</span>
3 0
2 years ago
A 5.5Kg block is hanging from a rope that is wrapped around the outside of a 13Kg flywheel disk witha radius of 33cm that is hag
Sergeeva-Olga [200]

Answer:

3.9m/s^{2}

Explanation:

Using second law of motion

a =\frac {m1 * g - \frac {T}{r}}{m1 + 0.5 * m2} where m1 is mass of block, m2 is mass of flywheel, g is acceleration due to gravity whose value is taken as 9.81 m/s^{2}, T is torque and r is radius

Substituting 5.5 Kg for m1, 13 Kg for m2, 0.33 m for r, 2.5 Nm for T we obtain

a = \frac {5.5 \times 9.81 - \frac {2.5}{0.33}}{(5.5 + 0.5 \times13)}=3.9m/s^{2}

8 0
2 years ago
A fire engine is moving south at 35 m/s while blowing its siren at a frequency of 400 Hz.
vodomira [7]

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Here,

f_d=frequency received by detector

f_s=frequency of wave emitted by source

v_d=velocity of detector

v_s=velocity of source

v=velocity of sound wave

Replacing we have that,

f_d = 400(\frac{(343+18)}{(343-35)})

f_d=422 Hz

Therefore the frequencty that will hear the passengers is 422Hz

8 0
2 years ago
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