Complete question is;
A ski jumper travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 24 m/s. The landing incline below her falls off with a slope of θ = 59◦ . The acceleration of gravity is 9.8 m/s².
What is the magnitude of the relative angle φ with which the ski jumper hits the slope? Answer in units of ◦
Answer:
14.08°
Explanation:
The time covered will be given by the formula;
t = (2V_x•tan θ)/g
t = (2 × 24 × tan 59)/9.8
t = 8.152 s
Now, the slope of the flight path at the point of impact will be given by the formula;
tan α = V_y/V_x
We are given V_x = 24 m/s
V_y will be gotten from the formula;
v = gt
Thus;
V_y = gt
V_y = 9.8 × (8.152) = 78.89 m/s
Thus;
tan α = 78.89/24
tan α = 3.2871
α = tan^(-1) 3.2871
α = 73.08°
Thus ;
Relative angle φ = α - θ = 73.08 - 59 = 14.08°
Answer:
57.6Joules
Explanation:
Rotational kinetic energy of a body can be determined using the expression
Rotational kinetic energy = 1/2Iω²where;
I is the moment of inertia around axis of rotation. = 5kgm/s²
ω is the angular velocity = ?
Note that torque (T) = I¶ where;
¶ is the angular acceleration.
I is the moment of inertia
¶ = T/I
¶ = 3.0/5.0
¶ = 0.6rad/s²
Angular acceleration (¶) = ∆ω/∆t
∆ω = ¶∆t
ω = 0.6×8
ω = 4.8rad/s
Therefore, rotational kinetic energy = 1/2×5×4.8²
= 5×4.8×2.4
= 57.6Joules
Explanation:
Whole system will accelerate under the action of applied force. The box will experience the force against the friction and when this force exceeds then the box will move. so
Ff = μs×m1×g
m1×a = μs×m1×g
a = μs×g
The applied force is given by
F = (m1 + m2)×a so
F = μs×g×(m1+m2)
Answer:
Spring constant, k = 24.1 N/m
Explanation:
Given that,
Weight of the object, W = 2.45 N
Time period of oscillation of simple harmonic motion, T = 0.64 s
To find,
Spring constant of the spring.
Solution,
In case of simple harmonic motion, the time period of oscillation is given by :

m is the mass of object


m = 0.25 kg


k = 24.09 N/m
or
k = 24.11 N/m
So, the spring constant of the spring is 24.1 N/m.
Answer:
75 m
Explanation:
The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.
The horizontal component of the velocity of the projectile is

and it is constant during the motion;
the total time of flight is
t = 5 s
Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:
