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:
(2) −1 e
Explanation:
A quark is the lightest elementary particles which form hadron such as proton and neutron. A quark has fractional charge.
Up, charm and top quarks have
charge where as down, strange and bottom quarks have
charge.
The antiparticle of up quark is antiup quark and has charge
charge.
The antiparticle of down quark is antidown quark and has charge
charge.
An antibaryon is composed of two anti-up quark and one anti-down quark.
Net charge of the anti-baryon is:
Thus, antibaryon has -1e charge.
Answer:1. Roche limit
2.hydrogen
3.atmosphere
4.mercury
5.venus
6.when an object passes the Roche limit, the strength of gravity on the object increases. If the density of the planet is higher, then the object can break up farther away from the planet. If the density is lower, then the Roche limit is located closer to the planet
7.Farther our in the solar system, beyond the frost line, hydrogen was at a low enough temperature that it could condense. This allowed hydrogen to accumulate under gravity, eventually forming the Jovian planets
Explanation:
Complete Question
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.
I = 1.2 A at time 5 secs.
Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.
Answer:
The charge is 
Explanation:
From the question we are told that
The diameter of the wire is 
The radius of the wire is 
The resistivity of aluminum is 
The electric field change is mathematically defied as

Generally the charge is mathematically represented as

Where A is the area which is mathematically represented as

So

Therefore

substituting values
![Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5Cint%5Climits%5E%7Bt%7D_%7B0%7D%20%7B%20%5B%200.0004t%5E2%20-%200.0001t%20%2B0.0004%5D%20%7D%20%5C%2C%20dt)
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%20t%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
From the question we are told that t = 5 sec
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%205%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
![Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004%285%29%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20%285%29%5E2%7D%7B2%7D%20%2B0.0004%285%29%5D%20%7D)
