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2 years ago

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astronauts brought back 500 lb of rock samples from the moon. how many kilograms did they bring back? 1 kg

1 answer:

Nadya [2.5K]2 years ago

8 0

Astronauts brought back 500 lbs rock samples from the moon. They brought back 227 kilograms of sample.

An Astronaut is a person who has been educated, prepared, and put into space as part of a human spaceflight mission to serve as a commander or crew member. The word is frequently used to refer to scientists as well as professional space explorers, while being usually reserved for all of those who travel into space on a daily basis.

According to the given question, Astronaut bought back 550 lbs sample,

We know that,

2.20lbs = 1 kg.

1 lbs = 1/2.20 kg

1 lbs = 0.45 kg

Hence, 500 lbs equals 227kgs according to the given relation.

So, Astronauts brought back 227 kilograms of sample.

Learn more about Astronaut here, brainly.com/question/11244838

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A toddler pushes his 7.0 kg toy box at a relatively constant velocity across the tiled floor of the family room applying a horiz

xxMikexx [17]

Answer:The coefficient of friction between the box and the floor, = 1.456 × 10⁻²

Explanation:

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3 years ago

What are 3 wether instuerments i know anemometer and barometer but the third one and what there job!!

BaLLatris [955]

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3 years ago

A bicyclist of mass 60kg supplies 340W of power while riding into a 5 m/s head wind. The frontal area of the cyclist and bicycle

NikAS [45]

Answer:

87.1 mph

Explanation:

We are given that

Mass,m=60 kg

Power,P=340 W

Speed,v=5 m/s

Area, $A=0.344 m^2$

Drag coefficient, $C_d=0.88$

Coefficient of rolling resistance, $\mu_r=0.007$

Friction force, $f=\mu_rmg=0.007\times 60\times 9.8=4.1 N$

Where $g=9.8 m/s^2$

Let speed of cyclist=v'

Drag force, $F_d=\frac{1}{2}\rho_{air}AC_dv^2$

Density of air, $\rho_{air}=1.225 kg/m^3$

$F_d=\frac{1}{2}\times 1.225\times 0.344\times 0.88(5)^2=4.635N$

Power,P= $(F_d+f)\times v'$

$340=(4.1+4.635) v'=8.735v'$

$v'=\frac{340}{8.735}=38.9m/s$

$v'=87.1 mph$

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3 years ago

A large plate carries a uniform charge density σ = 8. 85 × 10-9 c/m2. a pattern showing equipotential surfaces with a 5 v potent

vfiekz [6]

The potential difference comes out to be

$10 \times 10 {}^{ - 3} m$

Given:

σ = 8. 85 × 10-9 c/m2

we know,

$E = \frac{σ}{2ε0}$

$E = \frac{8.85 \times 10 {}^{ - 9} }{2ε0}$

$E = \frac{v}{d}$

given the potential difference between two equipotential surface=5v

E=∆v

∆d=∆v/E

$= \frac{5 \times 8.85 \times 10 { }^{ - 12} \times 2 }{8.85 \times 10 {}^{ - 9} }$

$Δ = 10 \times 10 {}^{ - 3} m$

Thus the potential difference is

$10 \times 10 {}^{ - 3} m$

Learn more about potential difference from here: brainly.com/question/28165869

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1 year ago

14. Which one of the following pictures shows the object that is the most dense? *

rewona [7]

Answer:

B

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2 years ago