The focal length of given concave lens will be -26.85 cm
The height of an image to the height of an object is the ratio that is used to determine a lens' magnification. Additionally, it is provided in terms of object and image distance. It is equivalent to the object distance to image distance ratio.
Given concave lens creates a virtual image at -47.0 cm and a magnification of +1.75.
We have to find focal length
The focal length can be found out by following way:
Magnification = m = +1.75
m = hi/h
hi = -47 cm
1.75 = -47/h
h = -26.85 cm
So the focal length of given concave lens will be -26.85 cm
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Answer:
The maximum height reached by the water is 117.55 m.
Explanation:
Given;
initial velocity of the water, u = 48 m/s
at maximum height the final velocity will be zero, v = 0
the water is going upwards, i.e in the negative direction of gravity, g = -9.8 m/s².
The maximum height reached by the water is calculated as follows;
v² = u² + 2gh
where;
h is the maximum height reached by the water
0 = u² + 2gh
0 = (48)² + ( 2 x -9.8 x h)
0 = 2304 - 19.6h
19.6h = 2304
h = 2304 / 19.6
h = 117.55 m
Therefore, the maximum height reached by the water is 117.55 m.
Answer:
Norway, Sweden, Finland and Iceland
Explanation:
Sea ice is a frozen seawater that floats on the ocean surface. It is formed between the Antarctic and Arctic hemisphere. It disappears in summer but not completely. The countries that experienced sea ice in 1986 were eight (8) in number but the countries bordered by open water were in September 2017 were Norway, Iceland, Finland and Russia.
1935.5 N is the "net force" acting on a car.
<u>Explanation</u>:
Given that,
Mass of the car is 790 kg.
Velocity of the car is 7 m/s. (v)
It turned around with 20 m. (r)
We know that, Net force = m × a
Now, Net force = m × a
Net force = 790 × 2.45
Net force = 1935.5 N
Answer:
The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg
Explanation:
Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water
Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C
To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.
Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C
Latent heat of ice = L = 334000 J/kg
Specific heat capacity of water = C = 4186 J/kg.°C
Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m
Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J
543600 m = 36627.5
m = 0.0674 kg = 67.4 g of ice.
