Let's say the Moon's complete cycle of phases is about 28 days (4 weeks) ... just so the arithmetic is easier. (Close enough. It's actually 29.53 days.)

If it's halfway through the waxing crescent, then it'll reach first quarter in 3 days or so.

From there, it'll take the next 7 days to become Full, and another 7 days to reach last quarter.

Total from half-waxing-crescent to last quarter . . . about 17 days.

4 0

4 years ago

A projectile starts from rest and moves 4.8 m down a frictionless ramp inclined at 14◦ with the horizontal. The acceleration of

Sunny_sXe [5.5K]

Answer:166.32 m/s 2

Explanation:

6 0

4 years ago

An object is taken from an oven at 350o F and left to cool in a room at 70o F. If the temperature fell to 250o F in one hour, wh

Oksana_A [137]

Answer:

117.83° F

Explanation:

Using Newton's Law of Cooling which can be expressed as:

$\dfrac{dT}{dt}= k(T-T_1)$

The differential equation can be computed as:

$\dfrac{dT}{dt}= k(T-70)$

$\dfrac{dT}{(T-70)}= kdt$

$\int \dfrac{dT}{(T-70)}= \int kdt$

$In|T-70| = kt +C$

$T- 70 = e^{kt+C} \\ \\ T = 70+e^{kt+C} \\ \\ T = 70 + C_1e^{kt} --- (1)$

where;

$C_1 = e^C$

At the initial condition, T(0)= 350

$350 = 70 C_1^{k*0}$

$350 -70 = C_1$

$280 = C_1$

replacing $C_1$ = 280 into (1)

Hence, the differential equation becomes:

$T(t) = 70 + 280 e^{kt}$

when;

time (t) = 1 hour

T(1) = 250

Since;

$250 = 70 + 280 e^{k*1}$

$180 = 280e^k \\ \\ \dfrac{180}{280}= e^k$

$k = In (\dfrac{180}{280})$

k = -0.4418

Therefore;

T(t) = 70 + 280e^{(-0.4418)}t

After 4 hours, the temperature is:

T(t) = 70 + 280e^{(-0.4418)}4

T(4) = 117.83° F

7 0

3 years ago

A mercury thermometer is constructed as

Tamiku [17]

Answer:

The change in height of the mercury is approximately 2.981 cm

Explanation:

Recall that the formula for thermal expansion in volume is:

$\frac{\Delta V}{V_0} =\alpha_V\, \Delta\, T\\\Delta V = V_0\,\, \alpha_V\,\,\Delta C$

from which we solved for the change in volume $\Delta V$ due to a given change in temperature $\Delta T$

We can estimate the initial volume of the mercury in the spherical bulb of diameter 0.24 cm ( radius R = 0.12 cm) using the formula for the volume of a sphere:

$V_0=\frac{4}{3} \pi \, R^3\\V_0=\frac{4}{3} \pi \, (0.12\,cm)^3\\V_0=0.007238\,cm^3$

Therefore, the change in volume with a change in temperature of 36°C becomes:

$\Delta V = V_0\,\, \alpha_V\,\,\Delta C\\\Delta V = 0.007238229\, cm^3\,(0.000182)\,(36)\\\Delta V=0.0000474248\, cm^3$

Now, we can use this difference in volume, to estimate the height of the cylinder of mercury with diameter 0.0045 cm (radius r= 0.00225 cm):

$V_{cyl}=\pi r^2\,h\\h =\frac{V_{cyl}}{\pi r^2} \\h=\frac{0.0000474248\, cm^3}{\pi \, (0.00225\,cm)^2} \\h=2.98188 \,cm$

8 0

3 years ago