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4 years ago

Kepler's second law implies what about planetary motion?

2 answers:

5 0

His 2nd law says that the line from the sun to the planet sweeps over equal areas in equal periods of time. This implies that the planet has to move faster in its orbit when it's closer to the sun (like the Earth is in January), and slower when it's farther from the sun (like the Earth is in July).

romanna [79]4 years ago

4 0

It implies the following thing. If you draw a vector connecting the Sun and a planet this vector will be covering equal areas in equal time spans. Say, each second, this vector covers N square meters, each 2 seconds - 2N square meters and so on

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Determine the maximum shearing stress in a solid shaft of 1.5-in. diameter as it transmits 75 hp at a speed of 1800 rpm. (Round

erma4kov [3.2K]

Answer:

$\tau=3.96\ ksi$

Explanation:

Given that

d= 1.5 in ( 1 in = 0.0254 m)

d= 0.0381 m

P= 75 hp ( 1 hp = 745.7 W)

P= 55927.5 W

N= 1800 rpm

We know that power P is given as

$P=\dfrac{2\pi N\ T}{60}$

T=Torque

N=Speed

$55927.5=\dfrac{2\times \pi \times 1800\ T}{60}$

T=296.85 N.m

The maximum shear stress is given as

$\tau=\dfrac{16 T}{\pi d^3}$

$\tau=\dfrac{16\times 296.85}{\pi \times 0.0381^3}$

$\tau=27.35\ MPa$

We know that 1 MPa =0.145 ksi

$\tau=3.96\ ksi$

3 0

4 years ago

If HST has a tangential speed of 7,750 m/s, how long is HST’s orbital period? The radius of Earth is 6.38 × 106 m. s

Sladkaya [172]

the orbital period is 5170 s

6 0

3 years ago

It is estimated that uranium is relatively common in the earth's crust, occurring in amounts of 4 g / metric ton. A metric ton i

timama [110]

Answer:

The mass of Uranium present in a 1.2mg sample is $4.8 \cdot 10^{-6}\,mg$

Explanation:

The ration between Uranium mass and total sample mass is: $\frac{4g}{1000kg} =\frac{4g}{1000000g}=\frac{1}{250000}$

For a sample of mass 1.2 mg, the amount of uranium is:

$1.2\, mg \cdot \frac{1}{250000}=4.8 \cdot 10^{-6}\,mg$

7 0

3 years ago

Simon is riding a bike at 12 km/h away from his friend Keesha.He throws a ball at 5 km/h back to Keesha, who is standing still o

Tems11 [23]

I notice that even though we're working with frames of reference
here, you never said which frame the '5 km/hr' is measured in.

In fact ! You didn't even say which frame the '12 km/hr' of his
bike is measured in.

So there are several different ways this could go. I'll do it the way
I THINK you meant it, but that doesn't guarantee anything.

-- Simon is riding his bike at 12 km/hr relative to the sidewalk,
away from Keesha.

-- He throws a ball at Keesha, at 5 km/hr relative to his own face.

-- Keesha sees the ball approaching her at (12 - 5) = 7 km/hr
relative to the ground and to her.

5 0

3 years ago

A 0.540-kg bucket rests on a scale. Into this bucket you pour sand at the constant rate of 56.0 g/s. If the sand lands in the bu

romanna [79]

Answer:

a) 12.8212 N

b) 12.642 N

Explanation:

Mass of bucket = m = 0.54 kg

Rate of filling with sand = 56.0 g/ sec = 0.056 kg/s

Speed of sand = 3.2 m/s

g= 9.8 m/sec2

<u>Condition (a);</u>

Mass of sand = Ms = 0.75 kg

So total mass becomes = bucket mass + sand mass = 0.54 +0.75=1.29 kg

== > total weight = 1.29 × 9.8 = 12.642 N

Now impact of sand = rate of filling × velocity = 0.056 × 3.2 = 0.1792 kg. m /sec2=0.1792 N

Scale reading is sum of impact of sand and weight force ;

i-e

scale reading = 12.642 N+0.1792 N = 12.8212 N

<u>Codition (b);</u>

bucket mass + sand mass = 0.54 +0.75=1.29 kg

==>weight = mg = 1.29 × 9.8 = 12.642 N (readily calculated above as well)

6 0

3 years ago