What is the frequency (in Hz) at which he bounces, given that his mass plus the mass of his equipment is 84.0 kg?
1 answer:
Complete question:
The length of nylon rope from which a mountain climber is suspended has an effective force constant of 1.40 ×10⁴ N/m.
What is the frequency (in Hz) at which he bounces, given that his mass plus the mass of his equipment is 84.0 kg?
Answer:
The frequency (in Hz) at which he bounce is 2.054 Hz
Explanation:
Given;
effective force constant, K = 1.40 ×10⁴ N/m.
The total mass = his mass plus the mass of his equipment, m = 84 kg
The frequency (in Hz) at which he bounce is given by;
Therefore, the frequency (in Hz) at which he bounce is 2.054 Hz
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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°
Answer:
- The impulse exerted by the ball is 7.2 kg.m/s
- The average force exerted on the ball by the bat is 600 N.
Explanation:
Given;
mass of the baseball, m = 0.144 kg
velocity of the baseball, v₁ = 20 m/s
velocity of the batter, v₂ = -30 m/s (opposite direction to the ball's speed)
The impulse exerted by the ball is calculated as follows;
J = ΔP = mv₁ - mv₂
ΔP = m(v₁ - v₂)
ΔP = 0.144 [20 - (-30)]
ΔP = 0.144 ( 20 + 30 )
ΔP = 0.144 (50)
ΔP = 7.2 kg.m/s
The average force exerted on the ball by the bat is calculated as;