Answer:
θ ≈ 65°
Explanation:
From the question given above, the following data were obtained:
Maximum height (H) = 8 m
Range (R) = 15 m
Initial velocity (u) = v
Angle θ =?
H = u²Sine²θ / 2g
8 = v²Sine²θ / 2g
Cross multiply
8 × 2g = v²Sine²θ
16g = v²Sine²θ
Divide both side by Sine²θ
v² = 16g / Sine²θ.... (1)
R = u²Sine2θ / g
15 = v²Sine2θ / g
Cross multiply
15 × g = v²Sine2θ
15g = v²Sine2θ
Divide both side by Sine2θ
v² = 15g / Sine 2θ... (2)
Summary:
v² = 16g / Sine²θ.... (1)
v² = 15g / Sine 2θ... (2)
Equate equation 1 and 2
16g / Sine²θ = 15g / Sine 2θ
16 / Sine²θ = 15 / Sine 2θ
Recall:
Sine²θ = SineθSineθ
Sine 2θ = 2SineθCosθ
16 / Sine²θ = 15 / Sine 2θ
16 / SineθSineθ = 15 / 2SineθCosθ
16 / Sineθ = 15 / 2Cosθ
Cross multiply
15 × Sineθ = 16 × 2Cosθ
15Sineθ = 32Cosθ
Divide both side by Cosθ
15Sineθ / Cosθ = 32
Divide both side by 15
Sineθ / Cosθ = 32/15
Recall:
Sineθ / Cosθ = Tanθ
Sineθ / Cosθ = 32/15
Tanθ = 32/15
Tanθ = 2.1333
Take the inverse of Tan
θ = Tan¯¹ 2.1333
θ ≈ 65°
Answer:
Ans 171.875 m(according to the units)
Explanation:
As we know,
to find the distance, we need to use this eqn..
speed = distance/ time
So, time is 1.92 s (as given)
the speed is 330 m/s ( in air speed)
so to find it is easy to find by using this
eqn..
speed =distance/time
330 =distance/1.92
330/1.92 =distance
171.875 = distance
Answer:
mparing this value with the weight of the vehicle we see W₂> 1.6 10⁴ N
therefore the woman can lift the car
Explanation:
For this exercise on Pascal's principle, we use that the pressure in every incompressible liquid is the same
P₁ = P₂
the press is defined by
P = F / A
for the woman's side
the area of a circle is
A = π D₁² / 4
P₁ = W (4 /π d₁²)
for the car side
A₂ =π d₂²
P₂ = W₂ (4 /π d₂²)
we substitute in the first equation
W (4 /π d₁²) = W₂ 4 /π d₂²)
W / d₁² = W₂ / D₂²
W₂ = (d₂ / d₁)² W
the weight of the woman is
W = mg
we calculate
W₂ = (24/3)₂ 50 9.8
W₂ = 3,136 10⁴ N
when comparing this value with the weight of the vehicle we see
W₂> 1.6 10⁴ N
therefore the woman can lift the car
Answer:
1.424 μC
Explanation:
I'm assuming here, that the charged ball is suspended by the string. If the string also is deflected by the angle α, then the forces acting on it would be: mg (acting downwards),
tension T (acting along the string - to the pivot point), and
F (electric force – acting along the line connecting the charges).
We then have something like this
x: T•sin α = F,
y: T•cosα = mg.
Dividing the first one by the second one we have
T•sin α/ T•cosα = F/mg, ultimately,
tan α = F/mg.
Since we already know that
q1=q2=q, and
r=2•L•sinα,
k=9•10^9 N•m²/C²
Remember,
F =k•q1•q2/r², if we substitute for r, we have
F = k•q²/(2•L•sinα)².
tan α = F/mg =
= k•q²/(2•L•sinα)² •mg.
q = (2•L•sinα) • √(m•g•tanα/k)=
=(2•0.5•0.486) • √(0.0142•9.8•0.557/9•10^9) =
q = 0.486 • √(8.61•10^-12)
q = 0.486 • 2.93•10^-6
q = 1.424•10^-6 C
q = 1.424 μC.