Answer:
the time Joshua travels 1 mile is 12.5 min
Explanation:
Let's start by finding the distance traveled on each lap,
Let's reduce everything to the SI system
R = 400 m
d = 1 mile (1609 m / 1 mile) = 1609 m
L = 2 pi R
L = 2 pi 400
L = 2513 m
Let us form a rule of proportions if 2 turns of Julian is 3 turns Joshua, for 1 turn of Joshua how many turns Julian took
lap Julian = 2/3 turn Joshua
Let's calculate what distance is the same for both of them since they are on the same track
1 lap = 2513 m
d. Julian = 2/3 2513 m
d Julian = 1675 m distance Joshua
Let us form the last rule of three or proportions if 1609 m you travel in 12 min how long it takes to travel 1675 m
t Julian = 1675/1609 12
t = 12.5 s
Since this is the distance Joshua travels, this is the time Joshua travels 1 mile
Answer:
38 cm from q1(right)
Explanation:
Given, q1 = 3q2 , r = 60cm = 0.6 m
Let that point be situated at a distance of 'x' m from q1.
Electric field must be same from both sides to be in equilibrium(where EF is 0).
=> k q1/x² = k q2/(0.6 - x)²
=> q1(0.6 - x)² = q2(x)²
=> 3q2(0.6 - x)² = q2(x)²
=> 3(0.6 - x)² = x²
=> √3(0.6 - x) = ± x
=> 0.6√3 = x(1 + √3)
=> 1.03/2.73 = x
≈ 0.38 m = 38 cm = x
Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2
Explanation:
(1) Given mass = 0.125 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2
(2) Given mass = 0.250 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2
(3) Given mass = 0.375 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2
(4) Given mass = 0.500 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2