Answer:
Vd = 1.597 ×10⁻⁴ m/s
Explanation:
Given: A = 3.90×10⁻⁶ m², I = 6.00 A, ρ = 2.70 g/cm³
To find:
Drift Velocity Vd=?
Solution:
the formula is Vd = I/nqA (n is the number of charge per unit volume)
n = No. of electron in a mole ( Avogadro's No.) / Volume
Volume = Molar mass / density ( molar mass of Al =27 g)
V = 27 g / 2.70 g/cm³ = 10 cm³ = 1 × 10 ⁻⁵ m³
n= (6.02 × 10 ²³) / (1 × 10 ⁻⁵ m³)
n= 6.02 × 10 ²⁸
Now
Vd = (6A) / ( 6.02 × 10 ²⁸ × 1.6 × 10⁻¹⁹ C × 3.9×10⁻⁶ m²)
Vd = 1.597 ×10⁻⁴ m/s
The scale in N, reading if the elevator moves upward at a constant speed of 1.5 m/s^2 is 862.5 N.
weight of man = 75kg
speed of elevator, a = 1.5 






So, the scale reading in the elevator is greater than his 862.5 N weight. This indicates that the person is being propelled upward by the scale, which it must do in order to do so, with a force larger than his weight. According to what you experience in quickly accelerating or slowly moving elevators, it is obvious that the faster the elevator acceleration, the greater the scale reading.
Speed can be defines as the pace at which the position of an object changes in any direction. Since speed simply has a direction and no magnitude, it is a scalar quantity.
Learn more about speed here:-
brainly.com/question/19127881
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Answer:
the planes wings are lifting at an angle to gravity so the plane isn't lifting as much against gravity when it banks. some of the wing lift is going into turning the plane. :) so it needs more lift to bank and stay up
Explanation:
Answer:
the final angular velocity of the platform with its load is 1.0356 rad/s
Explanation:
Given that;
mass of circular platform m = 97.1 kg
Initial angular velocity of platform ω₀ = 1.63 rad/s
mass of banana
= 8.97 kg
at distance r = 4/5 { radius of platform }
mass of monkey
= 22.1 kg
at edge = R
R = 1.73 m
now since there is No external Torque
Angular momentum will be conserved, so;
mR²/2 × ω₀ = [ mR²/2 +
(
R)² +
R² ]w
m/2 × ω₀ = [ m/2 +
(
)² +
]w
we substitute
w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1
w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )
w = 48.55 × [ 1.63 / ( 76.3908 ) ]
w = 48.55 × 0.02133
w = 1.0356 rad/s
Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s
Answer:
ω₂=1.20
Explanation:
Given that
mass of the turn table ,M= 15 kg
mass of the ice ,m= 9 kg
radius ,r= 25 cm
Initial angular speed ,ω₁ = 0.75 rad/s
Initial mass moment of inertia



Final mass moment of inertia



Lets take final speed of the turn table after ice evaporated =ω₂ rad/s
Now by conservation angular momentum
I₁ ω₁ =ω₂ I₂

ω₂=1.20