Answer:
b) 68,9 km/h a) picture
Explanation:
In this problem, since velocity is expressed in km/h and time in minutes, we have to convert either time to hours or velocity to km/min. It is easier to use hours.
Using this formula we pass time to hours:
Now we can plot speed vs time (image 1). The problem says that the driver uses constant speed, so all lines have to be horizontal.
Using the values of the speed we calculate the distance in each interval
Using these values and the fact that she was having lunch in the third one (therefore stayed in the same position), we plot position vs time, using initial position zero (image 2, distance is in km, not meters).
Finally, we compute the average speed with the distance over time:
Answer:
v₀ = 13.9 10³ m / s
Explanation:
Let's analyze this exercise we can use the basic kinematics relationships to love the initial velocity and the acceleration we can look for from Newton's second law where force is gravitational attraction.
F = m a
G m M / x² = m dv / dt = m dv/dx dx/dt
G M / x² = dv/dx v
GM dx / x² = v dv
We integrate
v² / 2 = GM (-1 / x)
We evaluate between the lower limits where x = Re = 6.37 10⁶m and the velocity v = vo and the upper limit x = 2.50 10⁸m with a velocity of v = 8.50 10³ m/s
½ ((8.5 10³)² - v₀²) = GM (-1 /(2.50 10⁸) + 1 / (6.37 10⁶))
72.25 10⁶ - v₀² = 2 G M (+0.4 10⁻⁸ - 1.57 10⁻⁷)
72.25 10⁶ - v₀² = 2 6.63 10⁻¹¹ 5.98 10²⁴ (-15.3 10⁻⁸)
72.25 10⁶ - v₀² = -1.213 10⁸
v₀² = 72.25 10⁶ + 1,213 10⁸
v₀² = 193.6 10⁶
v₀ = 13.9 10³ m / s
Answer:
50.4 N
Explanation:
Q1 = Q
Q2 = 4 Q
Distance = d
The force is given by
.... (1)
Now,
Q3 = 2 Q
Q4 = 7 Q
distance = d/3
.... (2)
Divide equation (2) by equation (1), we get
F' / 1.60 = 126 / 4
F' = 50.4 N
Thus, the force is 50.4 N.
