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3 years ago

Suppose you saw a waxing gibbous moon. What phase would you expect one week later?

1 answer:

3 0

The gibbous moon spends roughly a week waxing, from the First Quarter until the Full Moon. (This is the second week following a New Moon.)

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You are measuring the mass of different chemicals to get ready to conduct an experiment.

vova2212 [387]

Answer:

B) grams

The SI unit for mass is grams.

3 0

2 years ago

Read 2 more answers

What do the three variables (f,m, a) in the equation mean?

Anna [14]

Answer:

f=force m=mass and a=acceleration

7 0

2 years ago

Help what is the answer pls

Vladimir [108]

I think the answer should be: “100.4957 N”

7 0

2 years ago

Light is incident along the normal to face AB of a glass prism of refractive index 1.54. Find αmax, the largest value the angle

marusya05 [52]

To solve this problem it is necessary to use the concepts related to Snell's law.

Snell's law establishes that reflection is subject to

$n_1sin\theta_1 = n_2sin\theta_2$

Where,

$\theta =$ Angle between the normal surface at the point of contact

n = Indices of refraction for corresponding media

The total internal reflection would then be given by

$n_1 sin\theta_1 = n_2sin\theta_2$

$(1.54) sin\theta_1 = (1.33)sin(90)$

$sin\theta_1 = \frac{1.33}{1.54}$

$\theta = sin^{-1}(\frac{1.33}{1.54})$

$\theta = 59.72\°$

Therefore the $\alpha_{max}$ would be equal to

$\alpha = 90\°-\theta$

$\alpha = 90-59.72$

$\alpha = 30.27\°$

Therefore the largest value of the angle α is 30.27°

3 0

3 years ago

Wagon wheel. While working on your latest novel about settlers crossing the Great Plains in a wagon train, you get into an argum

OverLord2011 [107]

Answer:

I = 16.7kgm²

Explanation:

Since, Torque is given by,

\tau = F*r = I*\alpha

here, I = Moment of inertia = ??

\alpha = angular acceleration of wheel = a/r

F = tangential tension acting on the wheel = T

a = acceleration of bag of sand = 2.95 m/s^2

r = radius of wheel = d/2 = 120/2 = 60 cm = 0.60 m

from force balance on sand bag,

mg - T = m*a

T = m*(g-a)

m = mass of sand bag = 20 kg

So, I = T*r/\alpha = m*(g-a)*r/(a/r)

Using known values:

I = 20*(9.81 - 2.95)*0.60/(2.95/0.60) = 16.74

I = 16.7 kgm² = Moment of inertia of wheel experimentally

also, Moment of inertia of wheel theoretically(I') = M*r²

given, M = mass of wheel = 70 kg

I' = 70*0.60²= 25.2 kgm² = Moment of inertia of wheel theoretically

7 0

2 years ago