Answer:
a) θ = 2500 radians
b) α = 200 rad/s²
Explanation:
Using equations of motion,
θ = (w - w₀)t/2
θ = angle turned through = ?
w = final angular velocity = 1420 rad/s
w₀ = initial angular velocity = 420
t = time taken = 5s
θ = (1420 - 420) × 5/2 = 2500 rads
Again,
w = w₀ + αt
α = angular accelaration = ?
1420 = 420 + 5α
α = 1000/5 = 200 rad/s²
Answer:
The car stops in 7.78s and does not spare the child.
Explanation:
In order to know if the car stops before the distance to the child, you take into account the following equation:
(1)
vo: initial speed of the car = 45km/h
a: deceleration of the car = 2 m/s^2
t: time
xo: initial distance to the child = 25m
x: final distance to the child = 0m
It is necessary that the solution of the equation (1) for time t are real.
You first convert the initial speed to m/s, then replace the values of the parameters and solve the quadratic polynomial for t:


You take the first value t1 because it has physical meaning.
The solution for t is real, then, the car stops in 7.78s and does not spare the child.
<span>For a point mass the moment of inertia is just
the mass times the square of perpendicular distance to the rotation axis, I =
mr^2. That point mass relationship becomes the basis for all other moments of
inertia since any object can be built up from a collection of point masses. So the
I = (1.2 kg)(0.66m/2)^2 = 0.1307 kg m2</span>
Answer:
(a) 61.25 N
(b) 6.25 kg
(c) 6.25 Kg
Explanation:
Weight on moon = 10 N
Acceleration due to gravity on moon = 1.6 m/s^2
Acceleration due to gravity on earth = 9.8 m/s^2
Let m be the mass of the package.
(a) Weight on earth = mass x acceleration due to gravity on earth
Weight on earth = 6.25 x 9.8 = 61.25 N
(b) Weight on moon = mass x acceleration due to gravity on moon
10 = m x 1.6
m = 6.25 kg
(c) Mass of the package remains same as mass does not change, so the mass of package on earth is 6.25 kg.
Answer:
Momentum of block B after collision =
Explanation:
Given
Before collision:
Momentum of block A =
= 
Momentum of block B =
= 
After collision:
Momentum of block A =
= 
Applying law of conservation of momentum to find momentum of block B after collision
.

Plugging in the given values and simplifying.


Adding 200 to both sides.


∴ 
Momentum of block B after collision =