Answer:
Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N
Explanation:
Total force required = Mass x Acceleration,
F = ma
Here we need to consider the system as combine, total mass need to be considered.
Total mass, a = m₁+m₂+m₃ = 584 + 838 + 322 = 1744 kg
We need to accelerate the group of rocks from the road at 0.250 m/s²
That is acceleration, a = 0.250 m/s²
Force required, F = ma = 1744 x 0.25 = 436 N
Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N
Answer:
7.8 N
Explanation:
Applying,
I = Ft................. Eqaution 1
Where I = Impulse on the egg test dummy, F = Force on the egg test dummy during the time interval, t = time interval
make F the subject of the equation
F = I/t.................. Equation 2
From the question,
Given: T = -0.39 N.s, t = 0.050 s
Substitute these vales into equation 2
F = -0.39/0.050
F = -7.8 N
Hence the force that act on the egg test dummy is 7.8 N
Answer:
The time taken to rotate the sphere one time is, t = 22 s
Explanation:
Given data,
The mass of the sphere, m = 8200 kg
The radius of the sphere, r = 90 cm
= .9 m
The force applied by the girl, F = 75 N
The moment of inertia of the sphere is,
I = 2/5 mr²
= (2/5) 8200 x (.9)²
= 2657 kg·m²
The torque,
τ = I α
75 x 0.9 = 2657 x α
α = 0.0254 rad/s²
The angular displacement,
θ = ½αt²
2π = ½ x 0.0254 rad/s² x t²
t = 22 s
Hence, the time taken to rotate the sphere one time is, t = 22 s
The correct answer is actually D. because rapidly is an adverb of time
I hope I've helped!
Vi = 4 m/s
vf = 22 m/s
t = 3s
a = ?
vf = vi + a * t
vf - vi = a * t
a = (vf - vi) / t
a = (22 - 4) / 3
a = 6 m / s^2
