A concave mirror forms a real image at 15.0 cm from the mirror surface along the principal axis. If the corresponding object is
1 answer:
Answer:
Focal length, f = 10 cm
Explanation:
We have,
A concave mirror forms a real image at 15.0 cm from the mirror surface along the principal axis, it means that the image distance is 15 cm
The object is placed at a distance of 30 cm, u = -30 cm
It is required to find the focal length of the mirror. Let it is f. Using mirror's formula to find it:
So, the focal length of the mirror is 10 cm.
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<span>6) On Earth, the gravity acceleration is </span><span>. The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is
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</span>
2) On Mars, the value of the gravitational acceleration is different:
3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus:
4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg:
5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as
<span>6) On Earth, the gravity acceleration is </span>
</span>
</span>
Answer:
f ’= 97.0 Hz
Explanation:
This is an exercise of the doppler effect use the frequency change due to the relative movement of the fort and the observer
in this case the source is the police cases that go to vs = 160 km / h
and the observer is vo = 120 km / h
the relationship of the doppler effect is
f ’= f₀ (v + v₀ / v- )
let's reduce the magnitude to the SI system
v_{s} = 160 km / h (1000 m / 1km) (1h / 3600s) = 44.44 m / s
v₀ = 120 km / h (1000m / 1km) (1h / 3600s) = 33.33 m / s
we substitute in the equation of the Doppler effect
f ‘= 100 (330+ 33.33 / 330-44.44)
f ’= 97.0 Hz