Complete question is;
A ski jumper travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 24 m/s. The landing incline below her falls off with a slope of θ = 59◦ . The acceleration of gravity is 9.8 m/s².
What is the magnitude of the relative angle φ with which the ski jumper hits the slope? Answer in units of ◦
Answer:
14.08°
Explanation:
The time covered will be given by the formula;
t = (2V_x•tan θ)/g
t = (2 × 24 × tan 59)/9.8
t = 8.152 s
Now, the slope of the flight path at the point of impact will be given by the formula;
tan α = V_y/V_x
We are given V_x = 24 m/s
V_y will be gotten from the formula;
v = gt
Thus;
V_y = gt
V_y = 9.8 × (8.152) = 78.89 m/s
Thus;
tan α = 78.89/24
tan α = 3.2871
α = tan^(-1) 3.2871
α = 73.08°
Thus ;
Relative angle φ = α - θ = 73.08 - 59 = 14.08°
Answer:
i(t) = (E/R)[1 - exp(-Rt/L)]
Explanation:
E−vR−vL=0
E− iR− Ldi/dt = 0
E− iR = Ldi/dt
Separating te variables,
dt/L = di/(E - iR)
Let x = E - iR, so dx = -Rdi and di = -dx/R substituting for x and di we have
dt/L = -dx/Rx
-Rdt/L = dx/x
interating both sides, we have
∫-Rdt/L = ∫dx/x
-Rt/L + C = ㏑x
x = exp(-Rt/L + C)
x = exp(-Rt/L)exp(C) A = exp(C) we have
x = Aexp(-Rt/L) Substituting x = E - iR we have
E - iR = Aexp(-Rt/L) when t = 0, i(0) = 0. So
E - i(0)R = Aexp(-R×0/L)
E - 0 = Aexp(0) = A × 1
E = A
So,
E - i(t)R = Eexp(-Rt/L)
i(t)R = E - Eexp(-Rt/L)
i(t)R = E(1 - exp(-Rt/L))
i(t) = (E/R)(1 - exp(-Rt/L))
Answer:
The amount of charge the space shuttle collects is -1.224nC
Explanation:
The magnitude of Electric potential is given as;
V = kq/r
where;
V is the electric potential in volts
k is coulomb's constant
r is the radius of the sphere or distance moved by the charge
given; V = -1.1 V, k = 8.99 x 10⁹ Nm²/C², r = 10m
Substituting this values in the above equation, we estimate the amount of charge space shuttle collects.
q = (V*r)/k
q = (-1.1 *10)/(8.99 x 10⁹ )
q = -1.224 X 10⁻⁹ C
q = -1.224nC
Therefore, the amount of charge the space shuttle collects is -1.224nC
Answer:
The distance of separation is 
Explanation:
The mass of the each ball is 
The negative charge on each ball is 
Now we are told that the lower ball is restrained from moving this implies that the net force acting on it is zero
Hence the gravitational force acting on the lower ball is equivalent to the electrostatic force i.e
=> 
here k the the coulomb's constant with a value 
So
![0.01 * 9.8 = \frac{ 9*10^9 *[1*10^{-6} * 1*10^{-6}]}{d}](https://tex.z-dn.net/?f=0.01%20%2A%209.8%20%20%3D%20%20%5Cfrac%7B%209%2A10%5E9%20%2A%5B1%2A10%5E%7B-6%7D%20%2A%201%2A10%5E%7B-6%7D%5D%7D%7Bd%7D)
The atmospheric P is greater than the P in the flask, since
the Hg level is lacking down lower on the side open to the atmosphere.
43.4 cm x (10 mm / 1 cm) = 435 mm
the density of Hg is 13.6 / 0.791 = 17.2 times better than the liquid in the
manometer. This means that 1 mmHg = 17.2 mm of manometer liquid.
435 mm manometer liquid x (1 mm Hg / 17.2 mm manometer liquid) = 25.3 mm
Hg
The pressure in the flask is 755 - 25.3 = 729.7 mmHg.
729.7 mmHg x (1 atm / 760 mmHg ) = 0.960 atm.
