A plane travels at a constant speed of 250 m/s as it flies once around a horizontal circle whose radius is 2542m.

Newton’s empirical law of cooling/warming of an object is given by ( ), T Tm k dt dT = − where k is a constant of proportionalit

pychu [463]

Answer:

The time for the cake to cool off to room temperature is

approximately 30 minutes.

Let $T_{0}$ = $70^{0}$ F be the temperature and T that of the body

Explanation:

Our Tm = 70, the initial-value problem is

$\frac{DT}{dt}$ = k(T − 70), T(0) = 300

Solving the equation, we get

$\frac{DT}{t-70}$ = kdt

In [T-70]= kt + $C_{1}$

T = 70 + $C_{2}$ $e^{kt}$

Finding he value for $C_{2}$ using the initial value of T (0)= 300, therefore we get:

300=70+ $C_{2}$

$C_{2}$ = 230 therefore

T= 70+ 230 $e^{kt}$

Finding the value for k using T (3) = 200, therefore we get

T (3) = 200

$e^{3k}$ = $\frac{13}{23}$

K = $\frac{1}{3}$ in $\frac{13}{23}$

= -0.19018

Therefore

T(t) = 70+230 $e^{-0.19018}$

4 0

2 years ago

Olivia is on a swing at the playground at which point is her kinetic energy increasing and her potential energy decreasing

Nimfa-mama [501]

Whenever an object is falling, its potential energy
is decreasing and its kinetic energy is increasing.

Olivia's potential energy is decreasing and her kinetic energy
is increasing as she moves toward the right side of the picture,
all the way from W, through X, to the bottom of the arc.

7 0

3 years ago

Read 2 more answers

Learning Goal: To understand the nature of a sound wave, including its properties: frequency, wavelength, loudness, pitch, and t

Lady_Fox [76]

Answer:

Sound wave is a longitudinal wave that propagates in a medium

Explanation:

Part A: (C) Sound wave is propagation of pressure fluctuations in a medium.

Part B: (C) Pressure fluctuations travel along the direction of propagation of the sound wave.

Part C: (A) Yes air play a role in the propagation of the human voice from one end of the lecture hall to the other.

5 0

3 years ago

How many earths could fit inside jupiter

Snezhnost [94]

1,300 or more.

hope this helped :)

8 0

3 years ago

Find the energy in Joules required to lift a 55.0 Megagram object a distance of 500 cm.

fredd [130]

Energy to lift something =

(mass of the object) x (gravity) x (height of the lift).

BUT ...

This simple formula only works if you use the right units.

Mass . . . kilograms
Gravity . . . meters/second²
Height . . . meters

For this question . . .

Mass = 55 megagram = 5.5 x 10⁷ grams = 5.5 x 10⁴ kilograms

Gravity (on Earth) = 9.8 m/second²

Height = 500 cm = 5.0 meters

So we have ...

Energy = (5.5 x 10⁴ kilogram) x (9.8 m/s²) x (5 m)

= 2,696,925 joules .

That's quite a large amount of energy ... equivalent to
straining at the rate of 1 horsepower for almost exactly an
hour, or burning a 100 watt light bulb for about 7-1/2 hours.

The reason is the large mass that's being lifted.
On Earth, that much mass weighs about 61 tons.

7 0

2 years ago