Answer:
Explanation:
The mass of the block is 0.5kg
m = 0.5kg.
The spring constant is 50N/m
k =50N/m.
When the spring is stretch to 0.3m
e=0.3m
The spring oscillates from -0.3 to 0.3m
Therefore, amplitude is A=0.3m
Magnitude of acceleration and the direction of the force
The angular frequency (ω) is given as
ω = √(k/m)
ω = √(50/0.5)
ω = √100
ω = 10rad/s
The acceleration of a SHM is given as
a = -ω²A
a = -10²×0.3
a = -30m/s²
Since we need the magnitude of the acceleration,
Then, a = 30m/s²
To know the direction of net force let apply newtons second law
ΣFnet = ma
Fnet = 0.5 × -30
Fnet = -15N
Fnet = -15•i N
The net force is directed to the negative direction of the x -axis
Answer:
1) t = 3.45 s, 2) x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s
,
5) θ = -40.2º
Explanation:
This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.
1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff
y = y₀ +
t - ½ g t²
When leaving the cliff the speed is horizontal v_{oy}= 0 and at the bottom of the cliff y = 0
0 = y₀ - ½ g t2
t = √ 2y₀ / g
t = √ (2 60 / 9.8)
t = 3.45 s
2) The horizontal distance traveled
x = v₀ₓ t
x = 40 3.45
x = 138 m
3) The vertical velocity at the point of impact
v_{y} = I go - g t
v_{y} = 0 - 9.8 3.45
v_{y} = -33.81 m /s
the negative sign indicates that the speed is down
4) the resulting velocity at this point
v = √ (vₓ² + v_{y}²)
v = √ (40² + 33.8²)
v = 52.37 m / s
5) angle of impact
tan θ = v_{y} / vx
θ = tan⁻¹ v_{y} / vx
θ = tan⁻¹ (-33.81 / 40)
θ = -40.2º
6) sin (-40.2) = -0.6455
7) tan (-40.2) = -0.845
8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis
Answer:
Explanation:
Given
Speed of ball 
Plane is inclined at an angle 
To win the Game we need to hit the target at
away
Launch angle of ball 
Motion of ball can be considered in two planes i.e. Vertical to the plane and horizontal to the plane
So Net acceleration in vertical plane is 
Range of Projectile is given by

for 





so ball must be launched at an angle of 
Incomplete question as we have not told to find what.So the complete question is here
A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of mass m, moves with velocity (-60 m/s)i and a second piece, also of mass m, moves with velocity (-60 m/s)j. The third piece has mass 3m.Just after the explosion, what are the (a) magnitude and (b) direction of the velocity of the third piece?
Answer:

Explanation:
Given data
The vessel at rest
Piece one,of mass m,moves with velocity=(-60 m/s)i
Piece two,of mass m,moves with velocity=(-60 m/s)j
Piece three,of mass 3m
As the linear momentum is conserved in this system,Because the system is closed and no external force acting on it
So momentum is given as

As the vessel at rest so the initial momentum is zero
So

Answer:
θ = 45º
Explanation:
The light that falls on the second polarized is polarized, therefore it is governed by the law of Maluz
I = I₀ cos² θ
in the problem they ask us
I = ½ I₀
let's look for the angles
½ I₀ = I₀ cos² θ
cos θ = √ ½ = 0.707
θ = cos 0.707
θ = 45º