According to motion in straight line t1≠t2
A biker travels d1 meters in t1 seconds at v1 m/s for the first leg and d2 meters in t2 seconds at v2 m/s for the second leg. It's possible that t1t2 if his average speed is equal to the average of v1 and v2.
An object is said to be in motion if its position in relation to its surroundings changes over time. It is a shift in an object's position over time. The only type of motion that exists is motion in a straight line.
A reference system is constantly used to describe a particle's motion. An arbitrary origin point is used to create a reference system, and a coordinate system is imagined to be connected to it. The reference system for a specific problem is the coordinate system that has been selected for it. For the majority of the problems, we typically select an earth-based coordinate system as the reference system.
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> Non-zero numbers (like 1,2,3,4...) are always significant
> A zero sandwiched between two non-zero numbers is always significant
> Trailing zeros in a decimal (not whole number like million) are always significant.
<span>0,020170 = 2.0170 × 10^-2
5 sig-figs
</span>
The answer is apparent weight is zero.
You are still accelerating downwards at 9.8m/s^2 (if you are on Earth).
You still are being affected by the Earth's gravity.
Not all because of the previous two statements.
Not none because apparent weight is zero as you are falling.
The work done by a constant force in a rectilinear motion is given by:
where F is the magnitude of the force, d is the distance and θ is the angle between the force and the displacement vector.
In this case we have two forces then we need to add the work done by each of them; for the first force we have a magnitude of 17 N, a displacement of 12 m and and angle of 0° (since both the displacement and the force point right); for the second force we have a magnitude of 36 N, a displacement of 12 m and an angle of 30°. Plugging these values we have that the total work is:
Therefore, the total work done is 578.123 J and the answer is option E
Answer:
R = 1.2295 10⁵ m
Explanation:
After reading your problem they give us the diameter of the lens d = 4.50 cm = 0.0450 m, therefore if we use the Rayleigh criterion for the resolution in the diffraction phenomenon, we have that the minimum separation occurs in the first minimum of diffraction of one of the bodies m = 1 coincides with the central maximum of the other body
θ = 1.22 λ / D
where the constant 1.22 leaves the resolution in polar coordinates and D is the lens aperture
how angles are measured in radians
θ = y / R
where y is the separation of the two bodies (bulbs) y = 2 m and R the distance from the bulbs to the lens
R =
let's calculate
R =
R = 1.2295 10⁵ m