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NARA [144]
3 years ago
6

A situation known as ( blank)

Physics
2 answers:
shutvik [7]3 years ago
7 0
Apparent weightlessness
Lisa [10]3 years ago
6 0

Answer:

Apparent weightlessness

Explanation:

You might be interested in
Police radar equipment is used to detect the speed of objects. In one trial, the radar equipment records a stationary tree as tr
Brut [27]

Answer:

reliability

accuracy

Explanation:

If a reading of a measurement is consistently the same then the measurement is reliable.

If a reading of measurement is close the actual value of the measurement then the reading is accurate.

Here, a stationary tree shows reading 6 mph once and 0 mph another instant. So, neither the reading of a measurement is consistent not the reading of measurement is close the actual value.

Hence, the radar has problems in its reliability and accuracy

8 0
3 years ago
How do you do this question?
umka2103 [35]

Explanation:

The moment of inertia of each disk is:

Idisk = 1/2 MR²

Using parallel axis theorem, the moment of inertia of each rod is:

Irod = 1/2 mr² + m (R − r)²

The total moment of inertia is:

I = 2Idisk + 5Irod

I = 2 (1/2 MR²) + 5 [1/2 mr² + m (R − r)²]

I = MR² + 5/2 mr² + 5m (R − r)²

Plugging in values:

I = (125 g) (5 cm)² + 5/2 (250 g) (1 cm)² + 5 (250 g) (5 cm − 1 cm)²

I = 23,750 g cm²

7 0
3 years ago
A distant large asteroid is detected that might pose a threat to Earth. If it were to continue moving in a straight line at cons
Vlada [557]

Answer:

The minimum speed required is 5.7395km/s.

Explanation:

To escape earth, the kinetic energy of the asteroid must be greater or equal to its gravitational potential energy:

K.E\geq P.E

or

\dfrac{1}{2}mv^2 \geq  G\dfrac{Mm}{R}

where m is the mass of the asteroid, R= 24,000,000\:m is its distance form earth's center, M = 5.9*10^{24}kg is the mass of the earth, and G = 6.7*10^{-11}m^3/kg\: s^2 is the gravitational constant.

Solving for v we get:

v \geq \sqrt{\dfrac{2GM}{R} }

putting in numerical values gives

v \geq \sqrt{\dfrac{2(6.7*10^{-11})(5.9*10^{24})}{(24,000,000)} }

\boxed{v\geq 5739.5m/s}

in kilometers this is

v\geq5.7395m/s.

Hence, the minimum speed required is 5.7395km/s.

5 0
3 years ago
Hurry up i need help.
spayn [35]

Answer:

The answer would be B) The Same

Explanation:

Not gonna lie I checked my class notes but I figured this would help :)

Good luck!!!

8 0
3 years ago
A pin fin of uniform, cross-sectional area is fabricated of an aluminum alloy (k = 160 W/m-K). The fin diameter is D = 4 mm, and
frozen [14]

Answer:

Given that

D= 4 mm

K = 160 W/m-K

h=h = 220 W/m²-K

ηf = 0.65

We know that

m=\sqrt{\dfrac{hP}{KA}}

For circular fin

m=\sqrt{\dfrac{4h}{KD}}

m=\sqrt{\dfrac{4\times 220}{160\times 0.004}}

m = 37.08

\eta_f=\dfrac{tanhmL}{mL}

0.65=\dfrac{tanh37.08L}{37.08L}

By solving above equation we get

L= 36.18 mm

The effectiveness for circular fin given as

\varepsilon =\dfrac{2\ tanhmL}{\sqrt{\dfrac{hD}{K}}}

\varepsilon =\dfrac{2\ tanh(37.08\times 0.03618)}{\sqrt{\dfrac{220\times 0.004}{160}}}

ε = 23.52

5 0
3 years ago
Read 2 more answers
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