Ok so use trigonometry to work out the vertical component of velocity.
sin(25) =opp/hyp
rearrange to:
30*sin(25) which equals 12.67ms^-1
now use SUVAT to get the time of flight from the vertical component,
V=U+at
Where V is velocity, U is the initial velocity, a is acceleration due to gravity or g. and t is the time.
rearranges to t= (V+u)/a
plug in some numbers and do some maths and we get 2.583s
this is the total air time of the golf ball.
now we can use Pythagoras to get the horizontal component of velocity.
30^2-12.67^2= 739.29
sqrt739.29 = 27.19ms^-1
and finally speed = distance/time
so--- 27.19ms^-1*2.583s= 70.24m
The ball makes it to the green, and the air time is 2.58s
Answer:
244mm
Explanation:
I₁ = 3.35A
I₂ = 6.99A
μ₀ = 4π*10^-7
force per unit length (F/L) = 6.03*10⁻⁵N/m
B = (μ₀ I₁ I₂ )/ 2πr .........equation i
B = F / L ..........equation ii
equating equation i & ii,
F / L = (μ₀ I₁ I₂ )/ 2πr
Note F/L = B = F
F = (μ₀ I₁ I₂ ) / 2πr
2πr*F = (μ₀ I₁ I₂ )
r = (μ₀ I₁ I₂ ) / 2πF
r = (4π*10⁻⁷ * 3.35 * 6.99) / 2π * 6.03*10⁻⁵
r = 1.4713*10⁻⁵ / 6.03*10⁻⁵
r = 0.244m = 244mm
The distance between the wires is 244m
Answer:
1/3 the distance from the fulcrum
Explanation:
On a balanced seesaw, the torques around the fulcrum calculated on one side and on another side must be equal. This means that:
where
W1 is the weight of the boy
d1 is its distance from the fulcrum
W2 is the weight of his partner
d2 is the distance of the partner from the fulcrum
In this problem, we know that the boy is three times as heavy as his partner, so
If we substitute this into the equation, we find:
and by simplifying:
which means that the boy sits at 1/3 the distance from the fulcrum.
