Muscular strength is a measure of how much force you can exert in one repetition. Muscular endurance refers to the ability to perform a specific muscular action for a prolonged period of time.
Answer:
The value is 
Explanation:
From the question we are told that
The velocity is ![a = 120 \ km/h = [tex]\frac{120 * 1000}{3600} = 33.3 \ m/s](https://tex.z-dn.net/?f=a%20%3D%20%20120%20%5C%20%20km%2Fh%20%3D%20%20%5Btex%5D%5Cfrac%7B120%20%2A%20%201000%7D%7B3600%7D%20%3D%20%2033.3%20%5C%20%20m%2Fs)
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The time taken is 
The time taken for contact is 
The velocity of the of the car after contact is ![v_c = 71.0 \ km/h = [tex]\frac{71 *1000}{3600} = 19.7 2 \ m/s](https://tex.z-dn.net/?f=v_c%20%3D%20%2071.0%20%5C%20%20km%2Fh%20%3D%20%5Btex%5D%5Cfrac%7B71%20%2A1000%7D%7B3600%7D%20%3D%20%2019.7%202%20%5C%20%20m%2Fs)
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From the equation of kinematics we have that
Here u = 0 \ m/s since the car is initially motionless
=> 
=>
Answer:
a) T = 1,467 s
, b) A = 0.495 m
, c) v = 4.97 10⁻² m / s
Explanation:
The simple harmonic movement is described by the expression
x = A cos (wt + Ф)
Where the angular velocity is
w = √ k / m
a) Ask the period
Angular velocity, frequency and period are related
w = 2π f = 2π / T
T = 2π / w
T = 2pi √ m / k
T = 2π √ (1.2 / 22)
T = 1,467 s
f = 1 / T
f = 0.68 Hz
b) ask the amplitude
The mechanical energy of a harmonic oscillator
E = ½ k A²
A = √2 E / k
A = √ (2 2.7 / 22)
A = 0.495 m
c) the mass changes to 8.0 kg
As released from rest Ф = 0, the equation remains
x = A cos wt
w = √ (22/8)
w = 1,658
x = 3.0 cos (1,658 t)
Speed is
v = dx / dt
v = -A w sin wt
The speed is maximum when without wt = ±1
v = Aw
v = 0.03 1,658
v = 4.97 10⁻² m / s
Answer:
U = √Rg/sin2θ
Explanation:
Using the formula for "range" in projectile motion to derive the average speed before the ball hits the ground.
Range is the distance covered by the body in the horizontal direction from the point of launch to the point of landing.
According to the range formula,
R = U²sin2θ/g
Cross multiplying we have;
Rg = U²sin2θ
Dividing both sides by sin2θ, we have;
U² = Rg/sin2θ
Taking the square root of both sides we have;
√U² = √Rg/sin2θ
U = √Rg/sin2θ
Therefore, his average speed if he is to meet the ball just before it hits the ground is √Rg/sin2θ
