A small 20-kg canoe is floating downriver at a speed of 2 m/s. 40 J is the canoe’s kinetic energy.
Answer: Option A
<u>Explanation:</u>
The given canoe has the mass and is being given to move at a speed. Therefore the kinetic energy of the canoe can be calculated using the following method,
Given that mass of the canoe = 20 kg and its speed =1 m/s
As we know that the Kinetic energy has the formula,
Therefore, substituting the value into the equation, we get,
= 40 J
Answer: the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm
Explanation:
Given that;
diameter of the mirror d = 1.7 m
height h = 180 km = 180 × 10³ m
wavelength λ = 500 nm = 5 × 10⁻⁹ m
Now Angular separation from the peak of the central maximum is expressed as;
sin∅= 1.22 λ / d
sin∅ = (1.22 × 5 × 10⁻⁹) / 1.7
sin∅ = 3.588 × 10⁻⁷
we know that;
sin∅ = object separation / distance from telescope
object separation =
sin∅ × distance from telescope
object separation = 3.588 × 10⁻⁷ × 180 × 10³
object separation =6.45 × 10⁻² m
then we convert to centimeter
object separation = 6.45 cm
Therefore the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm
Answer
Pressure, P = 1 atm
air density, ρ = 1.3 kg/m³
a) height of the atmosphere when the density is constant
Pressure at sea level = 1 atm = 101300 Pa
we know
P = ρ g h
h = 7951.33 m
height of the atmosphere will be equal to 7951.33 m
b) when air density decreased linearly to zero.
at x = 0 air density = 0
at x= h ρ_l = ρ_sl
assuming density is zero at x - distance
now, Pressure at depth x
integrating both side
now,
h = 15902.67 m
height of the atmosphere is equal to 15902.67 m.
High school???
No way
It's work.
Answer:
5.3 cm
Explanation:
This question is an illustration of real and apparent distance.
From the question, we have the following given parameters
Real Distance, R = 8.0cm
Refractive Index, μ = 1.5
Required
Determine the apparent distance (A)
The relationship between R, A and μ is:
μ = R/A
i.e.
Refractive Index = Real Distance ÷ Apparent Distance
Substitute values in the above formula
1.5 = 8/A
Multiply both sides by A
1.5 * A = A * 8/A
1.5A = 8
Divide both side by 1.5
1.5A/1.5 = 8/1.5
A = 8/1.5
A = 5.3cm
Hence, the letters would appear at a distance of 5.3cm
