A 3.90 kg block is in equilibrium on an incline of 31.0◦. The acceleration of gravity is 9.81 m/s2 . What is Fn of the incline o
1 answer:
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Answer:
False statement = There must be a non-zero net force acting on the object.
Explanation:
An object is moving at a constant speed along a straight line. If the speed is constant then its velocity must be constant. We know that the rate of change of velocity is called acceleration of the object i.e.
a = 0
⇒ The acceleration of the object is zero.
The product of force and acceleration gives the magnitude of force acting on the object i.e.
F = m a = 0
⇒ The net force acting on the object must be zero.
So, the option (a) is not true. This is because the force acting on the object is zero. First option contradicts the fact.
Answer:
∅ = 89.44°
Explanation:
In situations like this air resistance are usually been neglected thereby making g= 9.81 m/
Bring out the given parameters from the question:
Initial Velocity () = 1000 m/s
Target distance (d) = 2000 m
Target height (h) = 800 m
Projection angle ∅ = ?
Horizontal distance = tcos ∅ .......................... Equation 1
where = velocity in the X - direction
t = Time taken
Vertical Distance = y = t -
g
................... Equation 2
Where = Velocity in the Y- direction
t = Time taken
=
sin∅
Making time (t) subject of the formula in Equation 1
t = d/(cos ∅)
t = =
=
...................Equation 3
substituting equation 3 into equation 2
Vertical Distance = d =
-
g
Vertical Distance = h = sin∅ -
g
Vertical Distance = h = dtan∅ - g
Applying geometry
=
+ 1
Vertical Distance = h = d tan∅ - 2 g ( + 1)
substituting the given parameters
800 = 2000 tan ∅ - 2 (9.81)( + 1)
800 = 2000 tan ∅ - 19.6( + 1) Equation 4
Replacing tan ∅ = Q .....................Equation 5
In order to get a quadratic equation that can be easily solve.
800 = 2000 Q - 19.6 + 19.6
Rearranging 19.6 - 2000 Q + 780.4 = 0
= 101.6291
= 0.411
Inserting the value of Q Into Equation 5
tan ∅ = 101.63 or tan ∅ = 0.4114
Taking the Tan inverse of each value of Q
∅ = 89.44° ∅ = 22.37°