The angle of incline of the hill above the horizontal is 8.81°.
Since the hill has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction.
<h3>Tangent of the angle of the incline of the hill,</h3>
The tangent of the angle of the incline of the hill, Ф is
tanФ = vertical rise/horizontal distance = grade of hill
Now, the vertical rise = 15.5 m and the horizontal distance = 100.0 m
So, substituting the values of the variables into the equation, we have
tanФ = vertical rise/horizontal run
tanФ = 15.5 m/100.0 m
tanФ = 0.155
<h3>Angle of incline of the hill</h3>
Taking inverse tan of both sides, we have
Ф = tan⁻¹(0.155)
Ф = 8.81°
So, the angle of incline of the hill above the horizontal is 8.81°.
Learn more about angle of incline of a hill here:
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From the average speed you can fix an equation:
Average speed = distance / time
You know the average speed = 65.1 kg / h, then
65.1 = distance / total time,
where total time is the time traveling plus 22.0 minutes
Call t the time treavelling and pass 22 minutes to hours:
65.1 = distance / [t + 22/60] ==> distance = [t + 22/60]*65.1
From the constant speed, you can fix a second equation
Constant speed = distance / time traveling
94.5 = distance / t ==> distance = 94.5 * t
The distance is the same in both equations, then you have:
[t +22/60] * 65.1 = 94.5 t
Now you can solve for t.
65.1t + 22*65.1/60 = 94.5t
94.5t - 65.1t = 22*65.1/60
29.4t = 23.87
t = 23.87 / 29.4
t = 0.812 hours
distance = 94.5 km/h * 0.812 h = 76.7 km
Answers: 1) 0.81 hours, 2) 76.7 km
Answer:magnitude -5; angle 160°
Explanation:
Vector A is described as having magnitude 5 and angle -20°.
To get an equivalent vector, we either leave the magnitude at 5 and add 360° to the angle, or we reverse the magnitude to -5 and add 180° to the angle.
5 @ -20° = 5 @ 340°
5 @ -20° = -5 @ 160°
The third one is the answer.
Answer:
Explanation:
Let T be the tension .
Applying newton's second law on the downward movement of the bucket
mg - T = ma
On the drum , a torque of TR will be acting which will create an angular acceleration of α in it . If I be the moment of inertia of the drum
TR = Iα
TR = Ia/ R
T = Ia/ R²
Replacing this value of T in the other equation
mg - T = ma
mg - Ia/ R² = ma
mg = Ia/ R² +ma
a ( I/ R² +m)= mg
a = mg / ( I/ R² +m)
mg - T = ma
mg - ma = T
mg - m x mg / ( I/ R² +m) = T
mg - m²g / ( I/ R² +m ) = T
mg - mg / ( 1 + I / m R² ) = T
b ) T = Ia/ R²
I = TR² / a
c ) Moment of inertia of hollow cylinder
I = 1/2 M ( R² - R² / 4 )
= 3/4 x 1/2 MR²
= 3/8 MR²
I / R² = 3/8 M
a = mg / ( I/ R² +m)
a = mg / ( 3/8 M + m )
T = Ia/ R²
= 3/8 MR² x mg / ( 3/8 M + m ) x 1 /R²
= 
Answer:
(a) A = 0.0800 m, λ = 20.9 m, f = 11.9 Hz
(b) 250 m/s
(c) 1250 N
(d) Positive x-direction
(e) 6.00 m/s
(f) 0.0365 m
Explanation:
(a) The standard form of the wave is:
y = A cos ((2πf) t ± (2π/λ) x)
where A is the amplitude, f is the frequency, and λ is the wavelength.
If the x term has a positive coefficient, the wave moves to the left.
If the x term has a negative coefficient, the wave moves to the right.
Therefore:
A = 0.0800 m
2π/λ = 0.300 m⁻¹
λ = 20.9 m
2πf = 75.0 rad/s
f = 11.9 Hz
(b) Velocity is wavelength times frequency.
v = λf
v = (20.9 m) (11.9 Hz)
v = 250 m/s
(c) The tension is:
T = v²ρ
where ρ is the mass per unit length.
T = (250 m/s)² (0.0200 kg/m)
T = 1250 N
(d) The x term has a negative coefficient, so the wave moves to the right (positive x-direction).
(e) The maximum transverse speed is Aω.
(0.0800 m) (75.0 rad/s)
6.00 m/s
(f) Plug in the values and find y.
y = (0.0800 m) cos((75.0 rad/s) (2.00 s) − (0.300 m⁻¹) (1.00 m))
y = 0.0365 m
