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3 years ago

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Suppose that a particle's position is described by r(t)=(t+3)i+(t2+2)j. Give an equation (in the form of a formula involving x a

nd y set equal to 0 ) whose solutions consist of the path of the particle. Find the velocity vector for the particle: v(t)

1 answer:

DiKsa [7]3 years ago

6 0

Answer:

The velocity for the particle is (1i+2tj).

Explanation:

Given that,

The particle position is

$r(t)=(t+3)i+(t^2+2)j$ ...(I)

In form of x and y

$x(t)=t+3$ ....(II)

$y(t)=t^2+2$ ...(III)

From equation (II)

$t=x-3$

Put the value of t in equation (III)

$y=(x-3)^2+2$

$y=x^2+9-6x+2$

$x^2-6x-y+11=0$

We need to calculate the velocity

Velocity is the rate of change of the position of the particle

$v(t)=\dfrac{dr}{dt}$

$\vec{v(t)}=1i+2t j$

Hence, The velocity for the particle is (1i+2tj).

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A 200g of iron at 120 degrees and a 150 g piece of copper at -50 degrees are dropped into an insulated beaker containing 300 g o

kodGreya [7K]

Answer:

T = 15.03°C

Explanation:

given data:

copper specific heat = Sc = 0.385 J/g °C

iron specific iron = Si = 0.450 J/g °C

specific heat of ethanol = Se = 2.46 J/g °C

net heat loss is equal to zero

$(m*S*\Delta T)_{copper} +(m*S*\Delta T)_ {iron} +(m*S*\Delta T)_ {ethanol} = 0$

$150*0.385 *( T - (-50)) + 200*0.450*(T - 120) + 300*2.46 * (T -20) = 0$

$57.75( T - (-50)) + 0.90(T - 120) +738(T -20) = 0$

$57.75T + 2887.5 + 0.90T - 108 + 738T - 14760 = 0$

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$796.65T= 11980.5$

T = 15.03°C

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3 years ago

For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener

Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

A)

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

$a=9.8 m/s^2$

Therefore, the velocity after

t = 2 s

is:

$v=0+(9.8)(2)=19.6 m/s$ (downward)

B)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

C)

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

D)

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

$KE'=\frac{1}{2}(6)(19.6)^2=1152 J$

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

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A heat engine uses thermal energy that flows from a heat source with a temperature of 325 K to a cold sink, which has a temperat

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The answer is "58.5"

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3 years ago

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A rocket is launched and the observer measuring the rocket altitude is standing at a distance of 200 feet from the launcher. Whe

Gre4nikov [31]

Answer:

100 foot

Explanation:

Using the formula for calculating range to calculate the speed first as shown;

<h3>Range R = U²sin 2theta/g</h3>

U is the speed

theta is the observe angle

g is the acceleration due to gravity.

200 = U²sin 2(45)/9.8

Usin90 = 200 * 9.8

U² = 1960

U = √1960

U = 44.27 m/s

Get the required altitude

Altitude H = u²/2g

H = 44.27²/2(9.8)

H = 1,959.8329/19.6

<em>H = 99.99 feet</em>

<em>Hence the altitude of the rocket to the nearest foot is 100 foot</em>

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3 years ago

1.What is the kinetic energy of an object with a velocity of 7m / s and a mass of 14kg

Vlada [557]

Answer:

Explanation:

The formula for KE is

$KE=\frac{1}{2}mv^2$ and we fill in to solve for KE:

$KE=\frac{1}{2}(14)(7.0)^2$ I used 2 significant digits in the velocity because 1 isn't accurate enough. Therefore, my answer will be rounded to 2 sig fig's:

KE = 340 J (rounded from 343). And we do the exact same thing in #2:

$KE=\frac{1}{2}(5.0)(5.0)^2$ (Used 2sig fig's for both the mass and the velocity) to get

KE = 63 J (rounded from 62.5)

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3 years ago