Give an example in which there are clear distinctions among distance traveled, displacement, and magnitude of displacement. spec
ifically identify each quantity in your example.
1 answer:
The example I've given is shown as an illustration in the picture. Suppose a jogger covers 2 km as he heads to the east. Then, he heads up north for another 1.5 km. Now, if you are to find the distance, you simply have to take the sum.
Distance = 2 + 1.5 = 3.5 km
If you are to find the displacement, you find the distance from the starting point to the end point, represented by the red arrow. Using the pythagorean theorem,
Displacement = √(2 km)² + (1.5 km)² = 2.5 km
If you are to find the magnitude of displacement, this is equivalent to distance. So, it is equal to 3.5 km. The displacement is a vector quantity, which presents the magnitude and the direction. The magnitude of the displacement does not take into account the direction of motion, just the total distance covered.
Distance = 2 + 1.5 = 3.5 km
If you are to find the displacement, you find the distance from the starting point to the end point, represented by the red arrow. Using the pythagorean theorem,
Displacement = √(2 km)² + (1.5 km)² = 2.5 km
If you are to find the magnitude of displacement, this is equivalent to distance. So, it is equal to 3.5 km. The displacement is a vector quantity, which presents the magnitude and the direction. The magnitude of the displacement does not take into account the direction of motion, just the total distance covered.
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Answer:
(A) Force on the object will be 58.85 N
(B) Apparent weight will be equal to 57.21 N
Explanation:
We have given mass of Neptune m = 1026 kg
Radius R = 25000 Km
Time period = 16 hour
We know that 1 hour = 3600 sec
So 16 hour = 16×86400 = 57600 sec
Acceleration due to gravity at north pole of Neptune
(A) Mass of the object m = 5.5 kg
So gravitational force on the object
(B) Velocity will be equal to
Apparent weight of the object will be equal to
(a) The electric field strength between two parallel conducting plates does not exceed the breakdown strength for air ()
(b) The plates can be close together to 1.7 mm with this applied voltage
<u>Explanation:</u>
Given data:
Dielectric strength of air =
Distance between the plates = 2.00 mm =
Potential difference, V =
We need to find
a) whether the electric field strength between two parallel conducting plates exceed the breakdown strength for air or not
b) the minimum distance at which the plates can be close together with this applied voltage.
The voltage difference (V) between two points would be equal to the product of electric field (E) and distance separation (d). The equation form is and apply all given value,
From the above, concluding that The electric field strength between two parallel conducting plates () does not exceed the breakdown strength for air (
)
b) To find how close together can the plates be with this applied voltage:
The formula would be,
Apply all known values, we get