Answer:
Explanation:
a ) speed of passenger = circumference / time
= 2π R / Time
= 2 x 3.14 x 50 / 60
= 5.23 m /s
b )
centrifugal force = m v² /R
= (882 /9.8 ) x 5.23² / 50
= 77.47 N
Apparent weight at the highest point
real weight - centrifugal force
= 882 - 77.47
= 804.53 N
Apparent weight at the lowest point
real weight + centrifugal force
= 882 +77.47
= 959.47 N
c ) if the passenger’s apparent weight at the highest point were zero
centrifugal force = weight
mv² /R = mg
v² = gR
= 9.8 X 50
v = 22.13 m /s
d )
apparent weight
mg - mv² / R
= 882 - (882 / 9.8 )x 22.13²/50
= 882 + 882
= 1764 N
=
Answer:
250 m/s
Explanation:
The mass of the bullet, m₁ = 100 g = 0.1 kg
The mass of the gun, m₂ = 5 kg
The backward velocity of the gun, v₂ = -5 m/s
Given that the momentum is conserved, we have;
The total initial momentum = The total final momentum
The gun and the bullet are at rest, therefore, we have;
The initial momentum = 0
The total final momentum = m₁·v₁ + m₂·v₂
Where;
v₁ = The forward velocity of the bullet
Therefore, we get;
m₁·v₁ + m₂·v₂ = 0
0.1 kg × v₁ + 5 kg × (-5 m/s) = 0
0.1 kg × v₁ = 5 kg × 5 m/s
v₁ = (5 kg × 5 m/s)/(0.1 kg) = 250 m/s
The forward velocity of the bullet, v₁ = 250 m/s
Sometimes, you can play sport for a longer time than you should. You should always stop and take a little break and always drink water. Also, if something starts to hurt, stop, let someone responsible know and take a break.
Answer:
a) the spring will stretch 60.19 mm
with the same box attached as it accelerates upwards
b) spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²
Explanation:
Given that;
Gravitational acceleration g = 9.81 m/s²
Mass m = 5 kg
Extension of the spring X = 50 mm = 0.05 m
Spring constant k = ?
we know that;
mg = kX
5 × 9.81 = k(0.05)
k = 981 N/m
a)
Given that; Acceleration of the elevator a = 2 m/s² upwards
Extension of the spring in this situation = X1
Force exerted by the spring = F
we know that;
ma = F - mg
ma = kX1 - mg
we substitute
5 × 2 = 981 × X1 - (5 ×9.81 )
X1 = 0.06019 m
X1 = 60.19 mm
Therefore the spring will stretch 60.19 mm
with the same box attached as it accelerates upwards
B)
Acceleration of the elevator = a
The spring is relaxed i.e, it is not exerting any force on the box.
Only the weight force of the box is exerted on the box.
ma = mg
a = g
a = 9.81 m/s² downwards.
Therefore spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²
5 distinct angles from the vertical axis can the orbital angular momentum vector l make for an electron
The characteristic of any rotating object determined by the product of the moment of inertia and the angular velocity is known as angular momentum. It is a characteristic of rotating bodies determined by the sum of their moment of inertia and angular velocity.
The image of a person sitting in a rotating chair while clutching a spinning bicycle wheel is another well-known illustration of the conservation of angular momentum. The bicycle wheel is then turned over, spinning in the opposite direction. The definition of angular momentum (L), with some simplification, is the object's distance from the rotation axis multiplied by the linear momentum: L = r*p or L = mvr.
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