Answer:
The electric field due to charge at distance r is
Explanation:
Given that,
Initial position of particle
Final position of particle
We need to calculate the magnitude of the position vector
Using formula of position vector
Put the value into the formula
We need to calculate the unit vector along electric field direction
Using formula of unit vector
Put the value into the formula
We need to calculate the electric field due to charge at distance r
Using formula of electric field
Put the value into the formula
Hence, The electric field due to charge at distance r is
Answer:
A) 23 joules
B)8
Explanation: trust me
vi = 18.5 m/s
vf = 46.1 m/s
t = 2.47 s
Find:
d = ??
a = ??
a = (Delta v)/t
a = (46.1 m/s - 18.5 m/s)/(2.47 s)
a = 11.2 m/s2
d = vi*t + 0.5*a*t2
d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s2)*(2.47 s)2
d = 45.7 m + 34.1 m
d = 79.8 m
(Note: the d can also be calculated using the equation vf2 = vi2 + 2*a*d)
Question seems to be missing. Found it on google:
a) How long is the ski jumper airborne?
b) Where does the ski jumper land on the incline?
a) 4.15 s
We start by noticing that:
- The horizontal motion of the skier is a uniform motion, with constant velocity
and the distance covered along the horizontal direction in a time t is
- The vertical motion of the skier is a uniformly accelerated motion, with initial velocity and constant acceleration (where we take the downward direction as positive direction). Therefore, the vertical distance covered in a time t is
The time t at which the skier lands is the time at which the skier reaches the incline, whose slope is
below the horizontal
This happens when:
Substituting and solving for t, we find:
b) 143.6 m
Here we want to find the distance covered along the slope of the incline, so we need to find the horizontal and vertical components of the displacement first:
The distance covered along the slope is just the magnitude of the resultant displacement, so we can use Pythagorean's theorem:
Answer:
851 m
Explanation:
Let's say the distance to one mountain is x, and the distance to the other mountain is y, such that x < y.
First, the wave moves a distance x, reflects off the nearer mountain, and travels another distance x back to the boy. This is the first echo. The time is:
t₁ = 2x / c
Meanwhile, the wave also moves a distance y in the opposite direction, reflects off the further mountain, and travels another distance y back to the boy. This is the second echo. The time is:
t₂ = 2y / c
Finally, each wave reflects off the opposite mountain and travels a total distance of 2x + 2y. This is the third echo. The time is:
t₃ = (2x + 2y) / c
Which means:
t₃ = t₁ + t₂
We are given the differences between the times:
t₂ − t₁ = 1.92 s
t₃ − t₂ = 1.47 s
Since t₃ = t₁ + t₂:
t₁ + t₂ − t₂ = 1.47 s
t₁ = 1.47 s
Which means:
t₂ − 1.47 s = 1.92 s
t₂ = 3.39 s
And:
t₃ = 1.47 s + 3.39 s
t₃ = 4.86 s
Therefore:
4.86 = (2x + 2y) / c
4.86c = 2x + 2y
2.43c = x + y
Given c = 350 m/s:
x + y = (2.43 s) (350 m/s)
x + y = 850.5 m
Rounding to three significant figures, the distance between the mountains is 851 m.