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finlep [7]
4 years ago
5

Can someone help me on this question

Physics
1 answer:
DanielleElmas [232]4 years ago
8 0
Box-1 . . .  D
Box-2 . . .  B
Box-3 . . .  A
Box-4 . . .  C
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Mester Exam 1 11 of 35
frez [133]
Please show picture of diagrams
8 0
3 years ago
The centripetal force acting on the space shuttle
tamaranim1 [39]

Answer:

(4) weight

Explanation:

The centripetal force acting on the space shuttle in orbit is given by:

F=m\frac{v^2}{r}

where

m is the mass of the shuttle

v is the tangential speed of the shuttle

r is the radius of its circular orbit

When the shuttle orbits the Earth, the centripetal force that keeps the shuttle in circular motion is given by the gravitational attraction between the shuttle and the Earth, which corresponds to the weight of the shuttle, and it is given by:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M is the Earth's mass

And this force, therefore, corresponds to the centripetal force.

7 0
3 years ago
Read 2 more answers
Two 110 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Z at –10 m/s when t
Marysya12 [62]
From the law of conservation of momentum
m1u1+ m2u2= m1v1+ m2v2
110*8+ 110*-10= 110*-10 + 110* v2
v2= 8 m/sec
8 0
3 years ago
Satellites remain in orbit around earth because
svetoff [14.1K]
Earth's gravity and the satellite's velocity keeps it so that it stays in orbit. (there is a more complicated side, too...) 
3 0
3 years ago
A cyclical heat engine, operating between temperatures of 450º C and 150º C produces 4.00 MJ of work on a heat transfer of 5.00
gogolik [260]

Answer:

(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%

Explanation:

Using the formula that relate heat and work from the thermodynamic theory as:W=Q=Q_{in}-Q_{out} solving to Q_out we get:Q_{out}=Q_{in}-W=5(MJ)-4(MJ)=1(MJ) this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: n=1-\frac{T_{Low} }{T_{High}} where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get:n=1-\frac{423.15}{723.15} =0.415=41.5%

5 0
3 years ago
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