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gulaghasi [49]
3 years ago
6

As denizens of the surface of a spinning planet, we are always in uniform circular motion. Imagine you are in Nairobi (on the Ea

rth's equator) at noon on a Monday. Answer the following questions only considering the rotation of the earth and NOT the Earth's circular motion around the sun. The radius of the earth is 6371 km. A day is 24 hours.
Assuming that you are at at theta = 0 rad at noon on Monday and moving in the positive direction, what is your position theta in rad on Tuesday at 11:00 am?
Physics
1 answer:
Klio2033 [76]3 years ago
7 0

Answer:

6.02 radian

Explanation:

At 12 noon position is zero radian on Monday

at 12 midnight the position is π/2

at 11 am position on Tuesday is 2π/24 x 23 ( after 23 hours ) = 6..02 radian.

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A puck of mass 0.5100.510kg is attached to the end of a cord 0.8270.827m long. The puck moves in a horizontal circle without fri
yanalaym [24]

Answer: 2.75 1/sec

Explanation:

The only external force (neglecting gravity) acting on the puck, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:

T = mv² / r (1)

Our unknown, is the frequency at which the puck can go around the circle, which is the inverse of the period Tp.

By definition, a period is the time needed by the puck to complete one entire circle.

By definition also , angular velocity is the rate of change of the angle advanced, so we can express this way:

ω = ∆θ / ∆t  

The angle advanced during one period, is exactly (by angle definition) 2 π radians.

So, we can always write the angular velocity, ω, as follows:

ω = 2π / Tp = 2πf

Now, there is a relationship between linear and angular velocity, that can be found applying simply the definition of velocity and of an angle too, as follows:

v = ∆s / ∆t = r ∆θ/∆t = ω r

Replacing in (1), we have:

T = mω2 r2 / r = m ω2r (2)

We have just found that ω= 2πf, so, replacing in (2) :

T = m (2π)2 f2 r  

Solving for f:

f = 1/2π√(T/mr) = 1/2π 17.28 1/sec = 2.75 1/sec

6 0
3 years ago
The mass of a spacecraft is about 435 kg . An engine designed to increase the speed of the spacecraft while in outer space provi
svlad2 [7]

Answer:

Δ v =  125 m/s

Explanation:

given,

mass of space craft = 435 Kg

thrust = 0.09 N

time = 1 week

       = 7 x 24  x 60 x 60 s

change in speed of craft = ?

Assuming no external force is exerted on the space craft

now,

T= m_s a

a=\dfrac{T}{m_s}

a =\dfrac{0.09}{435}

a = 2.068 x 10⁻⁴ m/s²

using equation of motion

Δ v = a t

Δ v = 2.068 x 10⁻⁴ x 7 x 24 x 60 x 60

Δ v =  125 m/s

3 0
3 years ago
A family drives north for 30km then turns east for 20km. The family then decided to turn west for 5km before finally stopping to
salantis [7]

Answer:

A

Explanation:

They drove 30km north. The displacement adds up to 25km therefore making the distance greater

Hope this helps!

6 0
3 years ago
A balloon is filled to a volume of 7.00*10^2 mL at a temperature of 20.0°C. The balloon is then cooled at constant pressure to a
wolverine [178]

The final volume of the gas is 238.9 mL

Explanation:

We can solve this problem by using Charle's law, which states that for a gas kept at constant pressure, the volume of the gas (V) is proportional to its absolute temperature (T):

\frac{V}{T}=const.

Which can be also re-written as

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where

V_1, V_2 are the initial and final volumes of the gas

T_1, T_2 are the initial and final temperature of the gas

For the gas in the balloon in this problem, we have:

V_1 = 7.00\cdot 10^2 mL = 700 mL is the initial volume

T_1=20.0^{\circ}C+273=293 K is the initial absolute temperature

V_2 is the final volume

T_2 = 1.00\cdot 10^2 K = 100 K is the final temperature

Solving for V_2,

V_2 = \frac{V_1 T_2}{T_1}=\frac{(700)(100)}{293}=238.9 mL

Learn more about ideal gases:

brainly.com/question/9321544

brainly.com/question/7316997

brainly.com/question/3658563

#LearnwithBrainly

6 0
3 years ago
A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length 53.2 m. At the insta
rusak2 [61]
From an energy balance, we can use this formula to solve for the angular speed of the chimney

ω^2 = 3g / h sin θ

Substituting the given values:
ω^2 = 3 (9.81) / 53.2 sin 34.1
ω^2 = 0.987 /s

The formula for radial acceleration is:
a = rω^2

So,
a = 53.2 (0.987) = 52.494 /s^2

The linear velocity is:
v^2 = ar
v^2 = 52.949 (53.2) = 2816.887

The tangential acceleration is:
a = r v^2
a = 53.2 (2816.887)
a = 149858.378 m/s^2

If the tangential acceleration is equal to g:
g = r^2 3g / sin θ
Solving for θ
θ = 67° 
7 0
3 years ago
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