A person who weighs 69kg rides with an acceleration of +1.45 m/s2. What is the force exerted by the legs on the person's upper b
1 answer:
Answer:
The force exerted by the legs on the person's upper body 32.106 N.
Given:
Mass = 69 kg
Acceleration = 1.45
To find:
Force exerted by legs on the person's upper body = ?
Formula used:
Force = mass × acceleration
Solution:
Mass of legs = mass of body
Mass of legs = 22.08 kg
According to Newton's second law of motion,
Force = mass × acceleration
Force = 22.08 × 1.45
Force = 32.106 N
The force exerted by the legs on the person's upper body 32.106 N.
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Answer:
a) θ₁ = 0.487º , b) t = 0.400 s , x = 11.73 ft
Explanation:
For this exercise let's use the projectile launch relationships.
The initial height is I = 5 ft and the final height y = 4 ft
y = y₀ + t - ½ g t²
The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft
x = v₀ₓ t
t = x / v₀ₓ
We replace
y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²
v_{oy} = v₀ sin θ
v₀ₓ = vo cos θ
y –y₀ = x tan θ - ½ g x² / v₀² cos² θ
5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)
1 = 120 tan θ - 0.0213 sec² θ
Let's use the trigonometry relationship
Sec² θ = 1 - tan² θ
1 = 120 tan θ - 0.0213 (1 –tan²θ)
0.0213 tan²θ + 120 tanθ -1.0213 = 0
We change variables
u = tan θ
u² + 5633.8 u - 48.03 = 0
We solve the second degree equation
u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2
u = [- 5633.8 ± 5633.82] / 2
u₁ = 0.0085
u₂= -5633.81
u = tan θ
θ = tan⁻¹ u
For u₁
θ₁ = tan⁻¹ 0.0085
θ₁ = 0.487º
For u₂
θ₂ = -89.99º
The launch angle must be 0.487º
b) let's look for the time it takes for the arrow to arrive
x = v₀ₓ t
t = x / v₀ cos θ
t = 120 / (300 cos 0.487)
t = 0.400 s
The deer must be at a distance of
v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s
x = v t
x = 29.33 0.4
x = 11.73 ft
Answer:
Magnetic field will be ZERO at the given position
Explanation:
As we know that the magnetic field due to moving charge is given as
so here we know that for the direction of magnetic field we will use
so we have
so magnetic field must be ZERO
So whenever charge is moving along the same direction where the position vector is given then magnetic field will be zero
Answer: a. Mass per unit length =0.0245kg/m
b. Tension =2.45x10^-8N
C. Tension = 2.45 x10^-8N
Fundamental frequency =200Hz
Explanation:
