You are given three pieces of wire that have different shapes (dimensions). You connect each piece of wire separately to a batte
1 answer:
Answer:
Explanation:
The resistance of a wire is given by:
where
is the resistivity of the material
L is the length of the wire
A is the cross-sectional area of the wire
1) The first wire has length L and cross-sectional area A. So, its resistance is:
2) The second wire has length twice the first one: 2L, and same thickness, A. So its resistance is
3) The third wire has length L (as the first one), but twice cross sectional area, 2A. So, its resistance is
By comparing the three expressions, we find
So, this is the ranking of the wire from most current (least resistance) to least current (most resistance).
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Answer:
(a). The speed is 115.28 m/s.
(b). The time is 94 min.
(c). The International Space Station travels 43202.400 km.
(d). Speed is scalar quantity.
Velocity is vector quantity.
Explanation:
Given that,
Top speed = 415 km/h
1 miles = 1609 m
(a). We need to calculate the speed in m/s
Using conversion of km/h to m/s
(b). We need to calculate the distance from London to Edinburgh in km
Using conversion for distance
Distance in meter,
We need to calculate the time
Using formula of time
Put the value into the formula
(c). Speed v' =7.66 km/s
v'=7660 m/s
In the time it takes the car to travel from London to Edinburgh,
We need to calculate the distance
Using formula of distance
Put the value into the formula
(d). Speed :
Speed is equal to the distance divided by time.
It is scalar quantity.
Velocity :
Velocity is equal to the displacement divided by time.
It is vector quantity.
Hence, (a). The speed is 115.28 m/s.
(b). The time is 94 min.
(c). The International Space Station travels 43202.400 km.
(d). Speed is scalar quantity.
Velocity is vector quantity.
Refer to the diagram shown below.
After 5 minutes (300 seconds):
The man travels north by (2 ft/s)*(300 s) = 600 ft
The woman, located at q, 500 east of p, begins walking south at 4 ft/s.
The distance separating them is
d₁ = √(600² + 500²) = 781.025 ft
After 20 minutes:
The man has traveled for 20 minutes (1200 s).
The woman has traveled for 15 minutes (900 s).
The man has moved (2 ft/s)*(1200 s) = 2400 ft north of p.
The woman has moved (4 ft/s)*(900 s) = 3600 ft south of q.
The distance separating them is
d₂ = √(6000² + 500²) = 6020.8 ft
The separation from d₁ to d₂ occurs in 15 minutes (900s).
Therefore the rate of separation is
Rate = (d₂ - d₁ ft)/(900 s) = (6020.8 - 781.025)/900 = 5.822 ft/s
or
Rate = (5.822 ft/s)*(60 s/min) = 349.32 ft/min
Answer: 349.32 ft/min (or 5.82 ft/s)
After 5 minutes (300 seconds):
The man travels north by (2 ft/s)*(300 s) = 600 ft
The woman, located at q, 500 east of p, begins walking south at 4 ft/s.
The distance separating them is
d₁ = √(600² + 500²) = 781.025 ft
After 20 minutes:
The man has traveled for 20 minutes (1200 s).
The woman has traveled for 15 minutes (900 s).
The man has moved (2 ft/s)*(1200 s) = 2400 ft north of p.
The woman has moved (4 ft/s)*(900 s) = 3600 ft south of q.
The distance separating them is
d₂ = √(6000² + 500²) = 6020.8 ft
The separation from d₁ to d₂ occurs in 15 minutes (900s).
Therefore the rate of separation is
Rate = (d₂ - d₁ ft)/(900 s) = (6020.8 - 781.025)/900 = 5.822 ft/s
or
Rate = (5.822 ft/s)*(60 s/min) = 349.32 ft/min
Answer: 349.32 ft/min (or 5.82 ft/s)