Answer:
Airplane speed relative to the ground is 260 km/h and θ = 22.6º direction from north to east
Explanation:
This is a problem of vector composition, a very practical method is to decompose the vectors with respect to an xy reference system, perform the sum of each component and then with the Pythagorean theorem and trigonometry find the result.
Let's take the north direction with the Y axis and the east direction as the X axis
Vy = 240 km / h airplane
Vx = 100 Km / h wind
a) See the annex
Analytical calculation of the magnitude of the speed and direction of the aircraft
V² = Vx² + Vy²
V = √ (240² + 100²)
V = 260 km/h
Airplane speed relative to the ground is 260 km/h
Tan θ = Vy / Vx
tan θ = 100/240
θ = 22.6º
Direction from north to eastb
b) What direction should the pilot have so that the resulting northbound
Vo = 240 km/h airplane
Vox = Vo cos θ
Voy = Vo sin θ
Vx = 100 km / h wind
To travel north the speeds the x axis (East) must add zero
Vx -Vox = 0
Vx = Vox = Vo cos θ
100 = 240 cos θ
θ = cos⁻¹ (100/240)
θ = 65.7º
North to West Direction
The speed in that case would be
V² = Vx² + Vy²
To go north we must find Vy
Vy² = V² - Vx²
Vy = √( 240² - 100²)
Vy = 218.2 km / h
Her estimate would not be greater, so it would be near the actual estimate. the numbers 70 and 10 are smaller than the numbers 73 and 12, so it is not possible for 70 x 10 to be more than 73 x 12 is. i hope this helps!!
Answer:
<em>0.25</em>
Explanation:
According to newtons law of motion
\sum F_x = ma
F_f = ma
nR = ma
nmg = ma
ng = a
n = a/g
g is the acceleration due to gravity
Given
a = 2.42m/s²
g = 9.8m/s²
Substitute into the formula;
n = 2.42/9.8
n = 0.25
<em>Hence the coefficient of kinetic friction is 0.25</em>
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Answer:
332m/s
Explanation:
We know the formula for velocity is v=d/t, but this particular question is asking about an echo. The problem tells us the distance to the bird from the bat, which is 19.4. To find the echo, you need to find the distance from the bat to the bird and back.
So
19.4 x 2 = 38.8
And then plug into the equation
v = 38.8 / 0.117
= 332m/s
0.030 hours
First part of run
d = 1.5 km
s = 85 km/h
t = d / s = 1.5/85
t1 = 0.017647 hours
Second part of run
d = 0.80 km
s = 67 km/h
t = d / s = 0.80/67
t2 = 0.011940 hours
t1 + t2 = 0.017647 + 0.011940
= 0.030 ℎours