In a closed system, the loss of momentum of one object is same as________ the gain in momentum of another object
according to law of conservation of momentum, total momentum before and after collision in a closed system in absence of any net external force, remains conserved . that is
total momentum before collision = total momentum after collision
P₁ + P₂ = P'₁ + P'₂
where P₁ and P₂ are momentum before collision for object 1 and object 2 respectively.
P'₁ - P₁ = - (P'₂ - P₂)
so clearly gain in momentum of one object is same as the loss of momentum of other object
The problem statement is simply asking us to convert units. We convert from units of ft^3 to units of m^3. To do this, we need a conversion factor. For this case, we use 1 m is equal to 3.28084 ft. We do as follows:
5.0 ft^3 ( 1 m / 3.28084 ft )^3 = 0.1416 m^3
Answer:
a) v₃ = 19.54 km, b) 70.2º north-west
Explanation:
This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition
vector 1 moves 26 km northeast
let's use trigonometry to find its components
cos 45 = x₁ / V₁
sin 45 = y₁ / V₁
x₁ = v₁ cos 45
y₁ = v₁ sin 45
x₁ = 26 cos 45
y₁ = 26 sin 45
x₁ = 18.38 km
y₁ = 18.38 km
Vector 2 moves 45 km north
y₂ = 45 km
Unknown 3 vector
x3 =?
y3 =?
Vector Resulting 70 km north of the starting point
R_y = 70 km
we make the sum on each axis
X axis
Rₓ = x₁ + x₃
x₃ = Rₓ -x₁
x₃ = 0 - 18.38
x₃ = -18.38 km
Y Axis
R_y = y₁ + y₂ + y₃
y₃ = R_y - y₁ -y₂
y₃ = 70 -18.38 - 45
y₃ = 6.62 km
the vector of the third leg of the journey is
v₃ = (-18.38 i ^ +6.62 j^ ) km
let's use the Pythagorean theorem to find the length
v₃ = √ (18.38² + 6.62²)
v₃ = 19.54 km
to find the angle let's use trigonometry
tan θ = y₃ / x₃
θ = tan⁻¹ (y₃ / x₃)
θ = tan⁻¹ (6.62 / (- 18.38))
θ = -19.8º
with respect to the x axis, if we measure this angle from the positive side of the x axis it is
θ’= 180 -19.8
θ’= 160.19º
I mean the address is
θ’’ = 90-19.8
θ = 70.2º
70.2º north-west
Answer:
To increase kinetic friction, the amount of fine water droplets sprayed before the game is limited.
To reduce kinetic friction. increase the amount of fine water droplets during pregame preparation and sweeping in front of the curling stones.
Explanation:
In curling sports, since the ice sheets are flat, the friction on the stone would be too high and the large smooth stone would not travel half as far. Thus controlling the amount of fine water droplets sprayed before the game is limited pregame is necessary to increase friction.
On the other hand, reducing ice kinetic friction involves two ways. The first way is adding bumps to the ice which is known as pebbling. Fine water droplets are sprayed onto the flat ice surface. These droplets freeze into small "pebbles", which the curling stones "ride" on as they slide down the ice. This increases contact pressure which lowers the friction of the stone with the ice. As a result, the stones travel farther, and curl less.
The second way to reduce the kinetic friction is sweeping in front of the large smooth stone. The sweeping action quickly heats and melts the pebbles on the ice leaving a film of water. This film reduces the friction between the stone and ice.
Answer:
a)106.48 x 10⁵ kg.m²
b)144.97 x 10⁵ kgm² s⁻¹
Explanation:
a)Given
m = 5500 kg
l = 44 m
Moment of inertia of one blade
= 1/3 x m l²
where m is mass of the blade
l is length of each blade.
Putting all the required values, moment of inertia of one blade will be
= 1/3 x 5500 x 44²
= 35.49 x 10⁵ kg.m²
Moment of inertia of 3 blades
= 3 x 35.49 x 10⁵ kg.m²
= 106.48 x 10⁵ kg.m²
b) Angular momentum 'L' is given by
L = x ω
where,
= moment of inertia of turbine i.e 106.48 x 10⁵ kg.m²
ω=angular velocity =2π f
f is frequency of rotation of blade i.e 13 rpm
f = 13 rpm=>= 13 / 60 revolution per second
ω = 2π f => 2π x 13 / 60 rad / s
L= x ω =>106.48 x 10⁵ x 2π x 13 / 60
= 144.97 x 10⁵ kgm² s⁻¹
