Answer:
35.7 m
Explanation:
Let
We have to find the distance between Joe's and Karl'e tent.
Substitute the values then we get
Because vertical component of B lie in IV quadrant and y-inIV quadrant is negative.
By triangle addition of vector
Hence, the distance between Joe's and Karl's tent=35.7 m
Answers:
a) -171.402 m/s
b) 17.49 s
c) 1700.99 m
Explanation:
We can solve this problem with the following equations:
(1)
(2)
(3)
Where:
is the bomb's final jeight
is the bomb'e initial height
is the bomb's initial vertical velocity, since the airplane was moving horizontally
is the time
is the acceleration due gravity
is the bomb's range
is the bomb's initial horizontal velocity
is the bomb's fina velocity
Knowing this, let's begin with the answers:
<h3>b) Time</h3>With the conditions given above, equation (1) is now written as:
(4)
Isolating :
(5)
(6)
(7)
<h3>a) Final velocity</h3>
Since , equation (3) is written as:
(8)
(9)
(10) The negative sign ony indicates the direction is downwards
<h3>c) Range</h3>
Substituting (7) in (2):
(11)
(12)
Answer:
a) v = 0.7071 v₀, b) v= v₀, c) v = 0.577 v₀, d) v = 1.41 v₀, e) v = 0.447 v₀
Explanation:
The speed of a wave along an eta string given by the expression
v =
where T is the tension of the string and μ is linear density
a) the mass of the cable is double
m = 2m₀
let's find the new linear density
μ = m / l
iinitial density
μ₀ = m₀ / l
final density
μ = 2m₀ / lo
μ = 2 μ₀
we substitute in the equation for the velocity
initial v₀ =
with the new dough
v =
v = 1 /√2 \sqrt{ \frac{T_o}{ \mu_o} }
v = 1 /√2 v₀
v = 0.7071 v₀
b) we double the length of the cable
If the cable also increases its mass, the relationship is maintained
μ = μ₀
in this case the speed does not change
c) the cable l = l₀ and m = 3m₀
we look for the density
μ = 3m₀ / l₀
μ = 3 m₀/l₀
μ = 3 μ₀
v =
v = 1 /√3 v₀
v = 0.577 v₀
d) l = 2l₀
μ = m₀ / 2l₀
μ = μ₀/ 2
v =
v = √2 v₀
v = 1.41 v₀
e) m = 10m₀ and l = 2l₀
we look for the density
μ = 10 m₀/2l₀
μ = 5 μ₀
we look for speed
v =
v = 1 /√5 v₀
v = 0.447 v₀