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3 years ago

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An object of mass m moves in one dimension with velocity given by v a/x for a a constant. Find the force on the object as a func

tion of position and the position time. To find the force, you may use either Newton's 2nd law or energy conservation. Assume that the object is initially at the origin.

1 answer:

balandron [24]3 years ago

8 0

Answer:

$F = -m\frac{a^2}{x^3}$

Explanation:

As we know that velocity as a function of position is given as

$v = \frac{a}{x}$

now in order to find the acceleration we can use that rate of change in velocity is known as acceleration

So we will have

$a = \frac{dv}{dt}$

$a = \frac{d}{dt}(\frac{a}{x})$

$a = -\frac{a}{x^2}\frac{dx}{dt}$

$a = - \frac{a^2}{x^3}$

Now by Newton's 2nd law we have

$F = ma$

$F = -m\frac{a^2}{x^3}$

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Add these measurements, using significant digit rules:<br><br> 1.0090 cm + 0.02 cm = cm

marin [14]

Answer:

1.029

Explanation:

1.0090 can also be looked at as "1.009"

0.02 can also be looked at as "0.020"

I think of it as 20+9 which is 29. There for your answer should be 1.029. There are no measurement rules applying to this equation since they are both in centimeters. So you don't have to convert anything.

7 0

3 years ago

The 10-lb block has a speed of 4 ft/s when the force of f=(8t2)f=(8t2) lb is applied. determine the velocity of the block when t

KatRina [158]

The velocity of the block when t == 2 s is 60.7 ft./sec.

Equations of Motion.

Here the friction is $F_f = \mu_k N$ = 0.2 N

$+ \uparrow \sum F_y = ma_y; \quad N – 10 = \frac { 10 } { 32.2 }(0) \quad N = 10 lb \\ \begin{aligned} \underrightarrow{ + } \sum F_x = ma_x; \quad 8t^2 – 0.2(10 &) = \frac { 10 } { 32.2 }a \\ & a = 3.22(8t^2 – 2) ft/s^2 \end{aligned}$

Kinematics.

The velocity of the block as a function of t can be determined by

integrating dv = adt using the initial condition v = 4 ft./s at t = 0.

$\int_{ 4 ft/s }^{ v } dv = \int_0^t 3.22(8t^2 – 2)dt \\ \begin{aligned} v – &4 = 3.22 (\frac 8 3 t^3 – 2t) \\ & v = \{8.5867t^3 – 6.44t + 4 \} ft/s \end{aligned}$

The displacement as a function of t can be determined by integrating

ds = vdt using

the initial condition s = 0 at t = 0

$\int_0^s ds = \int_0^t (8.5867t^3 – 6.44t + 4)dt \\ s = \{2.1467t^4 – 3.22t^2 + 4t \} ft$

at t = 2 sec

s = 30 ft.

Thus, at s = 30 ft.,

$\begin{aligned} v &= 8.5867(2.0089^3) – 6.44(2.0089) + 4 \\ &= 60.67 ft/s \\ &= 60.7 ft/s \end{aligned}$

Kinematics is a subfield of physics, developed in classical mechanics, that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without considering the forces that cause them to move.

Kinematics, as a field of study, is often referred to as the "geometry of motion" and is occasionally seen as a branch of mathematics. A kinematics problem begins by describing the geometry of the system and declaring the initial conditions of any known values of position, velocity and/or acceleration of points within the system.

Then, using arguments from geometry, the position, velocity and acceleration of any unknown parts of the system can be determined. The study of how forces act on bodies falls within kinetics, not kinematics. For further details, see analytical dynamics.

Learn more about kinematics here : brainly.com/question/24486060

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5 0

1 year ago

Can anyone tell me what's the base quantities for Force, Pressure and Charge?

mr Goodwill [35]

Force, pressure, and charge are all what are called <em>derived units</em>. They come from algebraic combinations of <em>base units</em>, measures of things like length, time, temperature, mass, and current. <em>Speed, </em>for instance, is a derived unit, since it's a combination of length and time in the form [speed] = [length] / [time] (miles per hour, meters per second, etc.)

Force is defined with Newton's equation F = ma, where m is an object's mass and a is its acceleration. It's unit is kg·m/s², which scientists have called a <em>Newton</em>. (Example: They used <em>9 Newtons</em> of force)

Pressure is force applied over an area, defined by the equation P = F/A. We can derive its from Newtons to get a unit of N/m², a unit scientists call the <em>Pascal</em>. (Example: Applying <em>100 Pascals </em>of pressure)

Finally, charge is given by the equation Q = It, where I is the current flowing through an object and t is how long that current flows through. It has a unit of A·s (ampere-seconds), but scientist call this unit a Coulomb. (Example: 20 <em>Coulombs</em> of charge)

4 0

3 years ago

A raindrop of mass 0.5 * 10^-4 kg is falling verctically under the influence of gravity. The air drag on the raindrop is fdrag =

Elina [12.6K]

Answer:

The displacement of the air drop after 3 second is 18.27 m.

Explanation:

Mass of the rain drop = m = $0.5\times 10^{-4} kg$

Weight of the rain drop = W

Duration of time = t = 3 seconds

$W=m\times g$

Drag force on rain drop = $D=0.2\times 10^{-5} v^2$

$W=0.5\times 10^{-4} kg\time 10 m/s^2=0.5\times 10^{-3} N$

Motion of the rain drop:

$F=m\times a$

Net force on the rain drop , F= W - D

$W-D=m\times a$

$0.5\times 10^{-3} N-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times a$

$0.5\times 10^{-3} kg m/s^2-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times \frac{v}{t}$

$0.006v^2+0.05v-1.5=0$

v = 12.18 m/s

Initial velocity of the rain drop = u = 0 (since, it is starting from rest)

v=u+at (First equation of motion)

$12.18 m/s=0m/s+a\times 3 s$

$a=4.06 m/s^2$

$s=ut+\frac{1}{2}at^2$ (second equation of motion)

$s=0\times 3s+\frac{1}{2}\times 4.06m/s^2\times (3 s)^2$

s = 18.27 m

The displacement of the air drop after 3 second is 18.27 m.

6 0

2 years ago

Describe how radio telescopes function.

FrozenT [24]

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2. A receiver and amplifier to boost the very weak radio signal to a measurable level. These days the amplifiers are extremely sensitive and are normally cooled to very low temperatures to minimise interference due to the noise generated by the movement of the atoms in the metal (called thermal noise).

3. A recorder to keep a record of the signal. Most radio telescopes nowadays record directly to some form of computer memory disk as astronomers use sophisticated software to process and analyse the data.

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3 years ago