Frequency:
Cardio : Five or more
HIIT: 5 or 4
Strength: 2-3 days on non-consecutive days a week
Intensity:
How hard you work during the session
For cardio : workout at your target heart rate.
For HIIT: varying intensity from high to low or rest ( no slow pace)
For strength: you should lift enough weight to complete the needed number of reps and sets.
Time:
For cardio 30- 60 min
For HIIT 45 min (short periods of bursts followed by rest)
For strength depends on the amount of weight.
Type:
Either cardio, HIIT or strength training (resistance training)
NOTE: This info is given by an unqualified brofessor, who uses his hard gained bro-science in the expense of scientific facts to solve others broblems.
Answer:
+5.4×10⁻⁷ C
Explanation:
Electric potential: This can be defined as the work done in bringing a unit charge from infinity to that point against the action of the field. The S.I unit of potential is volt (V)
The formula for potential is
V = kq/r............................ Equation 1
Where V = electric potential, k = proportionality constant, q = charge, r = distance.
making q the subject of the equation,
q = Vr/k............................ Equation 2
Given: V = 490 V, r = 10 m,
Constant: k = 9×10⁹ Nm²/C²
Substitute into equation 2
q = 490(10)/(9×10⁹)
q = 5.4×10⁻⁷ C
q = +5.4×10⁻⁷ C
Hence the charge is +5.4×10⁻⁷ C
First, we calculate for the weight of the object by multiplying the given mass by the acceleration due to gravity which is equal to 9.8 m/s²
Weight = (14 kg)(9.8 m/s²)
Weight = 137.2 N
The component of the weight that is along the surface of the inclined plane is equal to this weight times the sine of the given angle.
Weight = (137.2 N)(sin 52°)
weight = 108.1 N
Answer:
2.69 m/s
Explanation:
Hi!
First lets find the position of the train as a function of time as seen by the passenger when he arrives to the train station. For this state, the train is at a position x0 given by:
x0 = (1/2)(0.42m/s^2)*(6.4s)^2 = 8.6016 m
So, the position as a function of time is:
xT(t)=(1/2)(0.42m/s^2)t^2 + x0 = (1/2)(0.42m/s^2)t^2 + 8.6016 m
Now, if the passanger is moving at a constant velocity of V, his position as a fucntion of time is given by:
xP(t)=V*t
In order for the passenger to catch the train
xP(t)=xT(t)
(1/2)(0.42m/s^2)t^2 + 8.6016 m = V*t
To solve this equation for t we make use of the quadratic formula, which has real solutions whenever its determinat is grater than zero:
0≤ b^2-4*a*c = V^2 - 4 * ((1/2)(0.42m/s^2)) * 8.6016 m =V^2 - 7.22534(m/s)^2
This equation give us the minimum velocity the passenger must have in order to catch the train:
V^2 - 7.22534(m/s)^2 = 0
V^2 = 7.22534(m/s)^2
V = 2.6879 m/s
1 kilometre is equal to 1000m
and 4.1 minutes is equal to 246 seconds
thus 1000/246 = 4.065 m/s
and the direction is towards the west
