What will be the acceleration of a 40-kilogram object that is pushed with a net force of 80 newtons?

A particle of mass m=5.00 kilograms is at rest at t=0.00 seconds. a varying force f(t)=6.00t2−4.00t+3.00 is acting on the partic

olga_2 [115]

Answer:

The speed v of the particle at t=5.00 seconds = 43 m/s

Explanation:

Given :

mass m = 5.00 kg

force f(t) = 6.00t2−4.00t+3.00 N

time t between t=0.00 seconds and t=5.00 seconds

From mathematical expression of Newton's second law;

Force = mass (m) x acceleration (a)

F = ma

$a = \frac{F}{m}$ ...... (1)

acceleration (a) $= \frac{dv}{dt}$ ......(2)

substituting (2) into (1)

Hence, F $= \frac{mdv}{dt}$

$\frac{dv}{dt} = \frac{F}{m}$

$dv = \frac{F}{m} dt$

$dv = \frac{1}{m}Fdt$

Integrating both sides

$\int\limits {} \, dv = \frac{1}{m} \int\limits {F(t)} \, dt$

The force is acting on the particle between t=0.00 seconds and t=5.00 seconds;

$v = \frac{1}{m} \int\limits^5_0 {F(t)} \, dt$ ......(3)

Substituting the mass (m) =5.00 kg of the particle, equation of the varying force f(t)=6.00t2−4.00t+3.00 and calculating speed at t = 5.00seconds into (3):

$v = \frac{1}{5} \int\limits^5_0 {(6t^{2} - 4t + 3)} \, dt$

$v = \frac{1}{5} |\frac{6t^{3} }{3} - \frac{4t^{2} }{2} + 3t |^{5}_{0}$

$v = \frac{1}{5} |(\frac{6(5)^{3} }{3} - \frac{4(5)^{2} }{2} + 3(5)) - 0|$

$v = \frac{1}{5} |\frac{6(125)}{3} - \frac{4(25)}{2} + 15 |$

$v = \frac{1}{5} |\frac{750}{3} - \frac{100}{2} + 15 |$

$v = \frac{1}{5} | 250 - 50 + 15 |$

$v = \frac{215}{5}$

v = 43 meters per second

The speed v of the particle at t=5.00 seconds = 43 m/s

6 0

3 years ago

Define electric current and drift velocity.

Blizzard [7]

Answer:

Current- the flow of free charges, such as electrons and ions

Drift velocity- the average speed at which these charges move

3 0

3 years ago

Read 2 more answers

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

A 0.5 kg ball is tied to a swung in a circle with a radius of 0.400m.If the ball is accelerating at 58.8 m/s^2,what is the tange

jeka57 [31]

If the acceleration is centripetal then the formula is a=v^2/r
Plug in 58.8=v^2/.400 solve for v
V=12.12 m/s

5 0

3 years ago

The wattage marked on a lightbulb is not an inherent property of the bulb but depends on the voltage to which it is connected, u

Mashutka [201]

Answer:

The current that flows through the lamp is 0.5 A.

Explanation:

A lamp functions like a resistor and the real power absorbed by a resistor is given by the product of the voltage drop across it's terminals and the current that flows through it. If we wish to find the current that this lamp draws we should divide the wattage given (60 W) by the voltage drop provided (120 V). We then have:

i = P/V = 60/120 = 0.5 A.

5 0

3 years ago