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3 years ago

7

If a train conductor is describing the speed of the train as well as the direction the train is going, then the train conductor

is said to be describing the train's

1 answer:

natita [175]3 years ago

5 0

Then, train conductor is said to be describing the train's VELOCITY........

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You're driving your new sports car at 80 mph over the top of a hill that has a radius of curvature of 540 m. What fraction of yo

luda_lava [24]

Answer:

75.84%

Explanation:

We were given Speed of the sports car, v as 80 mph , we can convert to m/s for unit consistency.

v=80mph= 35.76 m/s

The radius of curvature is given as , r = 540 m

✓ the normal weight can be denoted as Wn

✓ the apparent weight of the person can be denoted as Wa

Wn= normal weight= mg

Wa=apparent weight = (mg - mv^2/r)

g= acceleration due to gravity= 9.8m/s^2

The apparent weightand normal weight has a ratio of

Mn/Ma= [mg - mv^2/r]/mg ........eqn(1)

If we simplify eqn(1) we have

Mn/Ma=[g - vr^2/g].............eqn(2)

Then substitute the given values

Mn/Ma=9.8 - [(35.76^2)/540]/ 9.8

=0.758×100%

Mn/Ma=75.84%

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4 0

3 years ago

Europa, a satellite of Jupiter, is believed to have a liquid ocean of water (with a possibility of life) beneath its icy surface

Goryan [66]

Answer:

so maximum velocity for walk on the surface of europa is 0.950999 m/s

Explanation:

Given data

legs of length r = 0.68 m

diameter = 3100 km

mass = 4.8×10^22 kg

to find out

maximum velocity for walk on the surface of europa

solution

first we calculate radius that is

radius = d/2 = 3100 /2 = 1550 km

radius = 1550 × 10³ m

so we calculate no maximum velocity that is

max velocity = √(gr) ...............1

here r is length of leg

we know g = GM/r² from universal gravitational law

so G we know 6.67 × $10^{-11}$ N-m²/kg²

g = 6.67 × $10^{-11}$ ( 4.8×10^22 ) / ( 1550 × 10³ )

g = 1.33 m/s²

now

we put all value in equation 1

max velocity = √(1.33 × 0.68)

max velocity = 0.950999 m/s

so maximum velocity for walk on the surface of europa is 0.950999 m/s

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3 years ago