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inessss [21]
4 years ago
15

Listed below are sets of elements.

Physics
2 answers:
WINSTONCH [101]4 years ago
8 0
Check the attached file for the answer.

Zarrin [17]4 years ago
4 0

Answer:

b.

Explanation:

i just took it

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PLEASEEEE HELP :( it would mean a lot
Elis [28]

Answer:

Each object exerts a force on the other, and the two forces are equal and in opposite directions

Explanation:

Newton's 3rd Law Of Motion:

For every action, there is an equal and opposite reaction.

5 0
3 years ago
Read 2 more answers
A positive charge of 3.2 x 10 -5 C is located 0.85 m away from another positive charge of 7.4 x 10 -6 C. What is the electric fo
zmey [24]

by using Coulumbs Law its 2.95N.

7 0
3 years ago
What is the radius of a communications satellite’s orbital path that is in a uniform circular orbit around Earth and has a perio
pantera1 [17]
If the period of a satellite is T=24 h = 86400 s that means it is in geostationary orbit around Earth. That means that the force of gravity Fg and the centripetal force Fcp are equal:

Fg=Fcp

m*g=m*(v²/R),
 
where m is mass, v is the velocity of the satelite and R is the height of the satellite and g=G*(M/r²), where G=6.67*10^-11 m³ kg⁻¹ s⁻², M is the mass of the Earth and r is the distance from the satellite. 

Masses cancel out and we have:

G*(M/r²)=v²/R, R=r so:

G*(M/r)=v²

r=G*(M/v²), since v=ωr it means v²=ω²r² and we plug it in,

r=G*(M/ω²r²),

r³=G*(M/ω²), ω=2π/T, it means ω²=4π²/T² and we plug that in:

r³=G*(M/(4π²/T²)), and finally we take the third root to get r:

r=∛{(G*M*T²)/(4π²)}=4.226*10^7 m= 42 260 km which is the height of a geostationary satellite. 
3 0
3 years ago
A locomotive accelerates a 25-car train along a level track. Every car has a mass of 7.7 ✕ 104 kg and is subject to a friction f
MaRussiya [10]

To solve the problem it is necessary to apply the concepts related to Force of Friction and Tension between the two bodies.

In this way,

The total mass of the cars would be,

m_T = 25(7.7*10^4)Kg

m_T = 1.925*10^6Kg

Therefore the friction force at 29Km / h would be,

f=250v

f= 250*29Km/h

f = 250*29*(\frac{1000m}{1km})(\frac{1h}{3600s})

f = 2013.889N

In this way the tension exerts between first car and locomotive is,

T=m_Ta+f

T=(1.925*10^6)(0.2)+2013.889

T= 3.8701*10^5N

Therefore the tension in the coupling between the car and the locomotive is 3.87*10^5N

6 0
3 years ago
Which contribution did Kepler make to viewing the solar system?
amm1812
He proposed the sun-centered model of the solar system.
8 0
3 years ago
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