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Mekhanik [1.2K]
3 years ago
6

A new planet is discovered in an approximately circular orbit beyond Pluto. It moves at a rate of approximately 1° per year. It'

s semi-major axis is approximately
Physics
1 answer:
Mazyrski [523]3 years ago
6 0
A new planet is discovered in an approximately circular orbit beyond Pluto. It moves at a rate of approximately 1° per year. It's semi-major axis is approximately:
answer:::::50 AU
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An automobile starter motor has an equivalent resistance of 0.055 Ï and is supplied by a 12.0 v battery which has a 0.0305 Ï int
LiRa [457]
12 V is the f.e.m. \epsilon of the battery. The potential difference that is applied to the motor is actually the fem minus the voltage drop on the internal resistance r:
\epsilon - Ir
this is equal to the voltage drop on the resistance of the motor R:
RI
so we can write:
\epsilon - Ir = RI
and using r=0.0305~\Omega and R=0.055~\Omega we can find the current I:
I= \frac{\epsilon}{R+r}=140~A
8 0
3 years ago
A particular car engine operates between temperatures of 440°C (inside the cylinders of the engine) and 20°C (the temperature of
Step2247 [10]

One of the concepts to be used to solve this problem is that of thermal efficiency, that is, that coefficient or dimensionless ratio calculated as the ratio of the energy produced and the energy supplied to the machine.

From the temperature the value is given as

\eta = 1-\frac{T_L}{T_H}

Where,

T_L = Cold focus temperature

T_H = Hot spot temperature

Our values are given as,

T_L = 20\° C = (20+273) K = 293 K

T_H = 440\° C = (440+273) K = 713 K

Replacing we have,

\eta = 1-\frac{T_L}{T_H}

\eta = 1-\frac{293}{713}

\eta = 0.589

Therefore the maximum possible efficiency the car can have is 58.9%

4 0
3 years ago
(a) when rebuilding her car's engine, a physics major must exert 300 n of force to insert a dry steel piston into a steel cylind
Vilka [71]
There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03

Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
F_f = \mu N
where \mu is the coefficient of friction, while N is the normal force. So we have:
F=\mu N
since we know that F=300 N and \mu=0.3, we can find N, the magnitude of the normal force:
N= \frac{F}{\mu}= \frac{300 N}{0.3}=1000 N

b) The problem is identical to that of the first part; however, this time the coefficienct of friction is \mu=0.03 due to the presence of the oil. Therefore, we have:
N= \frac{F}{\mu}= \frac{300 N}{0.03}=10000 N
8 0
3 years ago
Random errors can be reduced by taking repeated measurements.Error and uncertainty are interchangeable words that describe the s
fiasKO [112]

repeated mesurement can reduce the error

it is true

if you take any mesurement repeatedly and the average is taken, the error will be less

5 0
1 year ago
A man with a weight of 100 N is located
valentina_108 [34]

Answer:

Given,

  mass of man = 100 N = 10 kg

   height = h = 25m

   since the man does not move anything with his force, work done by him is zero

   work done on the man = gain in potential energy

   P.E=mgh

   P.E=10×9.8×25

  P.E=2.45KJ

Explanation:

so, potential energy gained by man is 2.45 KJ

5 0
2 years ago
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