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3 years ago

Which of the following is not an advantage that attractive people enjoy in interpersonal relationships?

Physics

2 answers:

Gnoma [55]3 years ago

7 0

Answer: Option (C) is the correct answer.

Explanation:

When two or more persons know each other and help each other by fulfilling physical or emotional needs then this type of relationship is known as interpersonal relationship.

Also, it is known that people who are attractive are actually more popular in their school, college, work place etc.

But it is not necessary that an attractive person will be honest in nature.

Hence, we can conclude that out of the given options, being perceived as more honest is not an advantage that attractive people enjoy in interpersonal relationships.

Licemer1 [7]3 years ago

5 0

Psychology on Egenuity Oct 5th 2018 says answer is C

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2 years ago

Prove that the weight of an object on moon is 1/6th of that on earth

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Answer:

The mass of moon is 1/100 times and its radius 1/4 times that of earth. As a result, the gravitational attraction on the moon is about one sixth when compared to earth. Hence, the weight of an object on the moon is 1/6th its weight on the earth.

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2 years ago

How many planets are in our Solar system?<br> I'll give brainliest

Dima020 [189]

Answer:

The following is Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune

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Explanation:

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2 years ago

A block starting from rest slides down the length of an 18 plank with an acceleration of 4.0 meters per second. How long does th

Snowcat [4.5K]

Answer:

$\boxed{\text{\sf \Large 3.0 s}}$

Explanation:

Use distance formula

$\displaystyle d=ut+\frac{1}{2} at^2$

$u= \text{\sf initial velocity}\\d= \text{\sf distance}\\a= \text{\sf acceleration}\\t= \text{\sf time taken}$

$\displaystyle 18=0 \times t+\frac{1}{2} \times 4 \times t^2$

$t=3$

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3 years ago

A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel

Dafna1 [17]

Answer:

$1.69\cdot 10^{10}J$

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

$E=-G\frac{mM}{2r}$

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

$M=5.98\cdot 10^{24}kg$ is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

$r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m$

So the initial total energy is

$E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J$

When the satellite hits the ground, it is now on Earth's surface, so

$r=R=6370 km=6.37\cdot 10^6 m$

so its gravitational potential energy is

$U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J$

And since it hits the ground with speed

$v=1.90 km/s = 1900 m/s$

it also has kinetic energy:

$K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J$

So the total energy when the satellite hits the ground is

$E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J$

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

$\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J$

8 0

3 years ago