Two cars collide in a perfectly inelastic collision. What is the final velocity of the cars to the right?

A disk has 128 tracks of 32 sectors each, on each surface of eight platters. The disk spins at 3600 RPM and takes 15 ms to move

Serga [27]

Answer:

the longest time needed to read an arbitrary sector located anywhere on the disk is 2971.24 ms

Explanation:

Given the data in the question;

first we determine the rotational latency

Rotational latency = 60/(3600×2) = 0.008333 s = 8.33 ms

To get the longest time, lets assume the sector will be found at the last track.

hence we will access all the track, meaning that 127 transitions will be done;

so the track changing time = 127 × 15 = 1905 ms

also, we will look for the sectors, for every track rotations that will be done;

128 × 8.33 = 1066.24 ms

∴The Total Time = 1066.24 ms + 1905 ms

Total Time = 2971.24 ms

Therefore, the longest time needed to read an arbitrary sector located anywhere on the disk is 2971.24 ms

7 0

3 years ago

Is the force of the mosquito on the car larger than, smaller than, or equal to the force of the car on the mosquito?

Goshia [24]

Equal to, mosquitos are in no way compared to a the force of a vehicle

7 0

3 years ago

Read 2 more answers

Electric field lines moves away from positive to wards negative?

Archy [21]

The vector force on the unit positive charge placed at any location in the field defines the strength of the electric field at that point. The charge used to determine field intensity (field strength) is known as the test charge. Now, a field line is defined as a line to which the previously mentioned field strength vectors are tangents at the relevant places. When we study positive charge field lines, the field strength vectors point away from the positive charge. If there is a negative charge anywhere in the vicinity, the field lines that began from the positive charge will all terminate at the negative charge if the value of the negative charge is the same as the value of the positive charge. Remember that the number of field lines originating from positive charge is proportional to the charge's value, and similarly, the number of field lines terminating at negative charge is proportionate to the charge's value. As a result, if all charges are equivalent, all lines originating from the positive charge terminate at the negative charge. If the value of the positive charge is greater than the value of the negative charge, the number of lines ending at the negative charge will be proportionally less than the number of lines beginning at the positive charge. The remaining lines that do not end at the negative charge will go to infinity. If the positive charge is less, all lines from it terminate at a negative charge, and any other reasonable number of ines terminate at a negative charge from infinity. We should also keep in mind that the number of lines that run perpendicular to the field direction across a surface of unit area is proportional to the field strength at that location. As a result, lines are dense in the strong field zone and sparse in the low intensity region.

6 0

2 years ago

Two wheels initially at rest roll the same distance without slipping down identical planes. Wheel B has twice the radius, but th

OlgaM077 [116]

Answer:

Their translational kinetic energies are the same

Explanation:

The translational kinetic energy of an object is given by the formula:

$KE = 0.5 mv^2$

Where m = the mass of the object and

v = the linear speed of the object

From the question, it is stated that wheel A has the same mass as wheel B, that is $m_A = m_B$

Linear speed is also a function of the distance covered. Since both wheels cover the same distance within the same interval, we can conclude that $v_A = v_B$

Both wheels A and B have equal speed and mass, this means that their translational kinetic energy is the same.

Note that translational kinetic energy is not a function of the radius

8 0

3 years ago

A force of 48 newtons is required to start a 5.0 kg box moving across a horizontal concrete floor. What is the coefficient of st

Soloha48 [4]

1) The coefficient of static friction is 0.980

2) The coefficient of kinetic friction is 0.908

Explanation:

1)

In the first situation, the box is still at rest. There are two forces acting on the box:

- The force of push, F, forward

- The force of static friction, $F_f$

Since the box is in equilibrium we have

$F_f = F$ (1)

The value of frictional force changes from zero to a maximum value which is given by:

$F_f = \mu_s mg$ (2)

where

$\mu_s$ is the coefficient of static friction

m is the mass of the box

g is the acceleration of gravity

So, if the force F needed to put the box in motion is 48 N, it means that this is also the maximum value of the force of friction. So, we can combine eq.(1) and (2) to find the coefficient of static friction:

$\mu_s = \frac{F}{mg}$