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3 years ago

The temperature of a system drops by 30°F during a cooling process. Express this drop in temperature in K, R, and °C.

Physics

1 answer:

babunello [35]3 years ago

8 0

Answer:

272.05 K; 489.69 °R; -1.11 °C

Explanation:

Fahrenheit can be converted to degrees Celsius using the following formula:

°C = 5/9 x (°F - 32) = -1.11 °C

The temperature in Kelvin is calculated from the temperature in degrees Celsius as follows:

K = °C + 273.15 = 272.05 K

The temperature on the Rankine scale is calculated from Kelvin as follows:

°R = 1.8K = 1.8(272.05) = 489.69 °R

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4 years ago

(a) Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t)

dolphi86 [110]

Answer:

$r(t)=(\frac{10t^3}{3}+t)i+(-sint+t+1)j+(\frac{-cos2t}{4}+\frac{1}{4})k$

Explanation:

a(t)=20t i+sin(t) j +cos(2t) k

v(t)= $\int\limits^a_b {a(t)} \, dt$

= $(10t^2+c_1)i+(-cost+c_2)j+\frac{sin2t}{2}k$ ---------------eqn 1

given v(0)=i

i= $c_1i+(-1+c_2)j+(0+c_3)k$

$c_1=1$ $c_2=1$ $c_3=0$

from equation 1

V(t)= $(10t^2+c_1)i+(-cost+c_2)j+\frac{sin2t}{2}k$ ----------eqn 2

now r(t)= $\int\limits^a_b {v (t)} \, dt$

$(\frac{10t^3}{3}+t+c_1)i+(-sint+t+c_2)j+(\frac{-cos2t}{4}+c_3)k$

given r(0)=j

0i+1j+ok = $c_1i+c_2j+(\frac{-1}{4}+c_3)k$

$c_1=0$ $c_2=0$ $c_3 = \frac{1}{4}$

$r(t)=(\frac{10t^3}{3}+t)i+(-sint+t+1)j+(\frac{-cos2t}{4}+\frac{1}{4})k$

8 0

3 years ago

Two identical cars are driving in opposite directions at the same speed. their kinetic energies have ____.

Minchanka [31]

The equation for determining the kinetic energy of the system is,

KE = 0.5mv²

where m is the mass, v is velocity and KE is the kinetic energy. When both objects have the same mass, and velocity but different direction, they will have the same kinetic energy.

This is because the exponent in the velocity is squared cancelling out the effect of the sign appearing before the number.

6 0

4 years ago

A rubber ball and a lump of clay have equal mass. They are thrown with equal speed against a wall. The ball bounces back with ne

gregori [183]

Answer:

The ball experiences the greater momentum change

Explanation:

The momentum change of each object is given by:

$\Delta p = m \Delta v= m (v-u)$

where

m is the mass of the object

v is the final velocity

u is the initial velocity

Both objects have same mass m and same initial velocity u. So we have:

- For the ball, the final velocity is

$v=-u$

Since it bounces back (so, opposite direction --> negative sign) with same speed (so, the magnitude of the final velocity is still u). So the change in momentum is

$\Delta p=m(v-u)=m((-u)-u)=-2mu$

- For the clay, the final velocity is

$v=0$

since it sticks to the wall. So, the change in momentum is

$\Delta p = m(v-u)=m(0-u)=-mu$

So we see that the greater momentum change (in magnitude) is experienced by the ball.

3 0

3 years ago