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:
Explanation:
Mass of the cable car, m = 5800 kg
It goes 260 m up a hill, along a slope of 
Therefore vertical elevation of the car = 
Now, when you get into the cable car, it's velocity is zero, that is, initial kinetic energy is zero (since K.E. = 
). Similarly as the car reaches the top, it halts and hence final kinetic energy is zero.
Therefore the only possible change in the cable car system is the change in it's gravitational potential energy.
Hence, total change in energy = mgh = 
where, g = acceleration due to gravity
h = height/vertical elevation
<span>First, we use the kinetic energy equation to create a formula:
Ka = 2Kb
1/2(ma*Va^2) = 2(1/2(mb*Vb^2))
The 1/2 of the right gets cancelled by the 2 left of the bracket so:
1/2(ma*Va^2) = mb*Vb^2 (1)
By the definiton of momentum we can say:
ma*Va = mb*Vb
And with some algebra:
Vb = (ma*Va)/mb (2)
Substituting (2) into (1), we have:
1/2(ma*Va^2) = mb*((ma*Va)/mb)^2
Then:
1/2(ma*Va^2) = mb*(ma^2*Va^2)/mb^2
We cancel the Va^2 in both sides and cancel the mb at the numerator, leving the denominator of the right side with exponent 1:
1/2(ma) = (ma^2)/mb
Cancel the ma of the left, leaving the right one with exponent 1:
1/2 = ma/mb
And finally we have that:
mb/2 = ma
mb = 2ma</span>
Answer:
The gravitational force exerted on the object is 75 N (answer D)
Explanation:
Hi there!
The gravitational force is calculated as follows:
F = m · g
Where:
F = force of gravity.
m = mass of the object.
g = acceleration due to gravity (unknown).
For a falling object moving in a straight line, its height at a given time can be calculated using the following equation:
y = y0 + v0 · t + 1/2 · a · t²
Where:
y = position at time t.
y0 = initial position.
v0 = initial velocity.
t = time.
g = acceleration due to gravity.
Let´s place the origin of the frame of reference at the point where the object is released so that y0 = 0. Let´s also consider the downward direction as negative.
Then, after 2 seconds, the height of the object will be -30 m:
y = y0 + v0 · t + 1/2 · g · t²
-30 m = 0 m + 0 m/s · 2 s + 1/2 · g · (2 s)²
-30 m = 1/2 · g · 4 s²
-30 m = 2 s ² · g
-30 m/2 s² = g
g = -15 m/s²
Then, the magnitude of the gravitational force will be:
F = m · g
F = 5 kg · 15 m/s²
F = 75 N
The gravitational force exerted on the object is 75 N (answer D)
Have a nice day!
Answer:
16,18,22
Or
1,3,7
Explanation:
The detailed explanation is contained in the image attached. The lengths are found using Pythagoras theorem and the two lengths reflects the two values of x yielded by the quadratic equation
