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:
The change of the volume of the device during this cooling is 
Explanation:
Given that,
Mass of oxygen = 10 g
Pressure = 20 kPa
Initial temperature = 110°C
Final temperature = 0°C
We need to calculate the change of the volume of the device during this cooling
Using formula of change volume
Put the value into the formula
Hence, The change of the volume of the device during this cooling is
Answer:
A: 4
B: 7
C. 3
Source:
Trust me bro
(Don’t act put this I jus need to answer questions sorry)<\3
Answer:
aaksj
Explanation:
a) the capacitance is given of a plate capacitor is given by:
C = \epsilon_0*(A/d)
Where \epsilon_0 is a constant that represents the insulator between the plates (in this case air, \epsilon_0 = 8.84*10^(-12) F/m), A is the plate's area and d is the distance between the plates. So we have:
The plates are squares so their area is given by:
A = L^2 = 0.19^2 = 0.0361 m^2
C = 8.84*10^(-12)*(0.0361/0.0077) = 8.84*10^(-12) * 4.6883 = 41.444*10^(-12) F
b) The charge on the plates is given by the product of the capacitance by the voltage applied to it:
Q = C*V = 41.444*10^(-12)*120 = 4973.361 * 10^(-12) C = 4.973 * 10^(-9) C
c) The electric field on a capacitor is given by:
E = Q/(A*\epsilon_0) = [4.973*10^(-9)]/[0.0361*8.84*10^(-12)]
E = [4.973*10^(-9)]/[0.3191*10^(-12)] = 15.58*10^(3) V/m
d) The energy stored on the capacitor is given by:
W = 0.5*(C*V^2) = 0.5*[41.444*10^(-12) * (120)^2] = 298396.8*10^(-12) = 0.298 * 10 ^6 J
Answer:
Time interval;Δt ≈ 37 seconds
Explanation:
We are given;
Angular deceleration;α = -1.6 rad/s²
Initial angular velocity;ω_i = 59 rad/s
Final angular velocity;ω_f = 0 rad/s
Now, the formula to calculate the acceleration would be gotten from;
α = Change in angular velocity/time interval
Thus; α = Δω/Δt = (ω_f - ω_i)/Δt
So, α = (ω_f - ω_i)/Δt
Making Δt the subject, we have;
Δt = (ω_f - ω_i)/α
Plugging in the relevant values to obtain;
Δt = (0 - 59)/(-1.6)
Δt = -59/-1.6
Δt = 36.875 seconds ≈ 37 seconds
