Answer: X = 52,314.12 N
Explanation: Let X be the force the feet of the athlete exerts on the floor.
According to newton's third law of motion the floor gives an upward reaction based on the weight of the athlete and the barbell which is known as the normal reaction ( based on the mass of the athlete and the barbell)
Mass of athlete = 87kg, mass of barbell = 600/ hence total normal reaction from the floor = 87* 61.22/ 9.8 *9.8 = 52,200N.
The athlete lifts the barbell from rest thus making it initial velocity u=0, distance covered = S = 0.65m and the time taken = 1.3s
The acceleration of the barbell is gotten by using the equation of constant acceleration motion
S= ut + 1/2at²
But u = 0
S = 1/2at²
0.65 = 1/2 *a (1.3)²
0.65 = 1.69 * a/2
0.65 * 2 = 1.69 * a
a = 0.65 * 2/ 1.69
a = 0.77m/s²
According to newton's second law of motion
Resultant force = mass * acceleration
And resultant force in this case is
X - 52,200 = (87 + 61.22) * 0.77
X - 52,200 = 148.22 * 0.77
X - 52, 200 = 114.132
X = 114.132 + 52,200
X = 52,314.12 N
Answer:
5.01 J
Explanation:
Info given:
mass (m) = 0.0780kg
height (h) = 5.36m
velocity (v) = 4.84 m/s
gravity (g) = 9.81m/s^2
1. First, solve for Kinetic energy (KE)
KE = 1/2mv^2
1/2(0.0780kg)(4.84m/s)^2 = 0.91 J
so KE = 0.91 J
2. Next, solve for Potential energy (PE)
PE = mgh
(0.0780kg)(9.81m/s^2)(5.36m) = 4.10 J
so PE = 4.10 J
3. Mechanical Energy , E = KE + PE
Plug in values for KE and PE
KE + PE = 0.91J + 4.10 J = 5.01 J
Answer:
Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.
Explanation:
Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.
Answer:
The charge on the third object is − 21.7nC
Explanation:
From Gauss's Law
Φ = Q/ε₀
where;
Φ is the total electric flux through the shell = − 533 N⋅m²/C
Q is the total charge Q in the shell = ?
ε₀ is the permittivity of free space = 8.85 x 10⁻¹²
From this equation; Φ = Q/ε₀
Q = Φ * ε₀ = − 533 * 8.85 x 10⁻¹²
Q = −4.7 X 10⁻⁹ C = -4.7nC
Q = q₁ + q₂ + q₃
− 4.7nC = − 14.0 nC + 31.0 nC + q₃
− 4.7nC − 17nC = q₃
− 21.7nC = q₃
Therefore, the charge on the third object is − 21.7nC
