1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
dolphi86 [110]
3 years ago
10

Astronauts on the first trip to Mars take along a pendulum that has a period on earth of 1.26 s. The period on the other planet

turns out to be 3.82 s. What is the free-fall acceleration on that other planet?
Physics
1 answer:
Angelina_Jolie [31]3 years ago
6 0

Answer:

1.066 m/s^2

Explanation:

On the earth and on Mars respectively, the time periods of the pendulums are expressed as:

Tearth = 2pi * sqrt(L/gearth)

Tmars = 2pi * sqrt(L/gMars)

Divide the two equations above:

Tearth/Tmars = sqrt(gMars/gearth)

gMars = gearth(Tearth/Tmars)^2 = (9.80m/s^2)(1.26s/3.82s)^2 = 1.066 m/s^2

You might be interested in
When are car velocity is positive and acceleration is negative, what is happening to the cars motion
DanielleElmas [232]
The motion of the car is negative
7 0
3 years ago
two barges full of salted toad guts have a collision. the red barge has a mass of 150000kg and is traveling northwest at 0.25m/s
solniwko [45]

The final velocity of the red barge in the collision elastic is 0.311 m/s when it collides with blue barge pf mass 1000000 kg.

Final velocity(v3)  of the red barge is calculated by following formula

m1×v1+ m2×v2= (m1+m2)v3

Substituting the value of m1= 150000 kg, v1= 0.25 m/s, m2= 1000000 kg, v2= 0.32 m/s

150000 × 0.25+ 1000000×0.32= (150000+1000000)×v3

37500+ 320000= 1150000×v3

357500= 1150000×v3

v3= 0.311 m/s

<h3>What is elastic collision velocity? </h3>
  • The velocity of the target particle after a head-on elastic impact in which the projectile is significantly more massive than the target will be roughly double that of the projectile, but the projectile velocity will remain virtually unaltered.

For more information on elastic collision velocity kindly visit to

brainly.com/question/29051562

#SPJ9

6 0
1 year ago
When two light waves arrive at the same place at the same time they create a?
Montano1993 [528]
Hey there,
<span>They interfere essentially like any other form of wave. 
</span>
Hope this helps :))

~Top
3 0
4 years ago
5. The atmosphere is heated both by the Sun and by the Earth's surface. Water radiates heat differently than land, so the air te
Ierofanga [76]
<span>They would feel that the water is cold.

</span> The atmosphere is heated both by the Sun and by the Earth's surface. Water radiates heat differently than land, so the air temperature over the ocean is usually different than the air temperature over land. <span>
The difference in air temperature over land compared to over water causes convection currents in the atmosphere. How would a person at the beach experience these convection currents? 
</span>They would feel that the water is cold.

NOT:
They would feel the heat of the Sun.
They would feel that the sand is hot.
<span>They would feel wind as the air moves.</span>
3 0
3 years ago
Read 2 more answers
A planet of mass m 6.75 x 1024 kg is orbiting in a circular path a star of mass M 2.75 x 1029 kg. The radius of the orbit is R 8
Umnica [9.8K]

Answer:

The orbital period of the planet is 387.62 days.

Explanation:

Given that,

Mass of planetm =6.75\times10^{24}\ kg

Mass of star m'=2.75\times10^{29}\ kg

Radius of the orbitr =8.05\times10^{7}\ km

Using centripetal and gravitational force

The centripetal force is given by

F = \dfrac{mv^2}{r}

F=m\omega^2r

We know that,

\omega=\dfrac{2\pi}{T}

F=m(\dfrac{2\pi}{T})^2r....(I)

The gravitational force is given by

F = \dfrac{mm'G}{r^2}....(II)

From equation (I) and (II)

m(\dfrac{2\pi}{T})^2r=\dfrac{mm'G}{r^2}

Where, m = mass of planet

m' = mass of star

G = gravitational constant

r = radius of the orbit

T = time period

Put the value into the formula

T^2=\dfrac{4\pi^2R^3}{m'G}

T^2=\dfrac{4\times(3.14)^2\times(8.05\times10^{7})^3}{2.75\times10^{29}\times6.67\times10^{-11}}

T=2\times3.14\times\sqrt{\dfrac{(8.05\times10^{10})^3}{2.75\times10^{29}\times6.67\times10^{-11}}}

T =3.34\times10^{7}\ s

T= 387.62\days

Hence, The orbital period of the planet is 387.62 days.

4 0
3 years ago
Read 2 more answers
Other questions:
  • A commuter train passes a passenger platform at a constant speed of 40.0 m/s. The train horn is sounded at its characteristic fr
    9·1 answer
  • What is the speed of a jet plane that flies 7200 km in 9 hours (in km/hr)?
    9·1 answer
  • A source from which organisms generally take elements is called a/an
    10·2 answers
  • You are writing a science report and want to find accurate trustworthy information. Which would be the best resource?
    13·2 answers
  • If the bus driver is the last one off the bus then who closes the door?
    8·2 answers
  • An object is released from rest at time t = 0 and falls through the air, which exerts a resistive force such that the accelerati
    14·1 answer
  • Describe the factors that determine power in your own words (not from a website)
    11·2 answers
  • If the sun did not rotate, what would happen to the global winds? why?
    12·1 answer
  • Which of the following will cause water to change to ice!
    6·2 answers
  • A very long, straight wire has charge per unit length 3.20 x 10^-10 c&gt;m. At what distance from the wire is the electric-field
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!