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: 75V
Explanation:
Given that,
total resistance (Rtotal) = 150Ω
Current (I) = 0.5A
Change in electric potential (V) = ?
Recall that potential difference is the product of amount of current and the amount of resistance in the circuit. And its unit is volts.
So, apply the formula V = I x Rtotal
V = 0.5A x 150Ω
V = 75V
Thus, the change in electric potential across the circuit is 75 Volts
Explanation:
Given:
v₀ = 250 mph
v = 0 mph
t = 25 s
Find: a
v = at + v₀
(0 mph) = a (25 s) + (250 mph)
a = -10 mph/s
Answer:
4.47 km
Explanation:
If we draw the path of the van then we get a shape with two exposed points A and D. If we draw a line from point D perpendicular to BA we get point E. This gives us a right angled triangle ADE.
From Pythagoras theorem
AD² = AE² + ED²
Hence, the van is 4.47 km from its initial point
1) push down on the end of the lever, and 2) 3/4 of the way from the fulcrum
