<u>Answer:</u>
<em>Newtons II law: </em>
<em> </em>It is defined as<em> "the net force acting on the object is a product of mass and acceleration of the body"</em> . Also it defines that the <em>"acceleration of an object is dependent on net force and mass of the body".</em>
Let us assume that,a string is attached to the cart, which passes over a pulley along the track. At another end of the string a weight is attached which hangs over the pulley. The hanging weight provides tension in the spring, and it helps in accelerating the cart. We assume that the string is massless and no friction between pulley and the string.
Whenever the hanging weight moves downwards, the cart will accelerate to right side.
<em>For the hanging weight/mass</em>
When hanging weight of mass is m₁ and accelerate due to gravitational force g.
Therefore we can write F = m₁ .g
and the tension acts in upward direction T (negetive)
Now, Fnet = m₁ .g - T
= m₁.a
So From Newtons II law<em> F = m.a</em>
Answer:(a)891.64 N
(b)0.7
Explanation:
Mass of crate 
Crate slows down in 
initial speed 
inclination 
From Work-Energy Principle
Work done by all the Forces is equal to change in Kinetic Energy
change in kinetic energy
(b)Coefficient of sliding friction
and 
Answer:
The bomb will remain in air for <u>17.5 s</u> before hitting the ground.
Explanation:
Given:
Initial vertical height is, 
Initial horizontal velocity is, 
Initial vertical velocity is, 
Let the time taken by the bomb to reach the ground be 't'.
So, consider the equation of motion of the bomb in the vertical direction.
The displacement of the bomb vertically is 
Acceleration in the vertical direction is due to gravity, 
Therefore, the displacement of the bomb is given as:
So, the bomb will remain in air for 17.5 s before hitting the ground.
For this case, what we can do is use the Pythagorean theorem to find the magnitude of the displacement of the car.
We have then
From here, we clear the value of d.
We have then:
Rewriting:
Answer:
The magnitude of the car's displacement is:
d = 20 miles
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°
