Answer:
a) W = 46.8 J and b) v = 3.84 m/s
Explanation:
The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy
W = ΔK = 
-K₀
a) work is the scalar product of force by distance
W = F . d
Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.
W = F d cos θ
W = 39.0 1.20 cos 0
W = 46.8 J
b) zero initial kinetic language because the package is stopped
W -
= -K₀
W - fr d= ½ m v² - 0
W - μ N d = ½ m v
on the horizontal surface using Newton's second law
N-W = 0
N = W = mg
W - μ mg d = ½ m v
v² = (W -μ mg d) 2/m
v = √(W -μ mg d) 2/m
v = √[(46.8 - 0.30 4.30 9.8 1.20) 2/4.3
]
v = √(31.63 2/4.3)
v = 3.84 m/s
Answer:
a) 
, b) 
Explanation:
The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:
Where 
is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:
Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:
a) t = 50 s.
b) t = 100 s.
Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:
Since 
, then the angular velocity is equal to zero. Therefore:
Answer:
Time of flight A is greatest
Explanation:
Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.
So
H = u₁² sin²θ₁ /2g
H = u₂² sin²θ₂ /2g
H = u₃² sin²θ₃ /2g
On the basis of these equation we can write
u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃
For maximum range we can write
D = u₁² sin2θ₁ /g
1.5 D = u₂² sin2θ₂ / g
2 D =u₃² sin2θ₃ / g
1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁
1.5 = u₂ cosθ₂ /u₁ cosθ₁ ( since , u₁ sinθ₁ =u₂ sinθ₂ )
u₂ cosθ₂ >u₁ cosθ₁
u₂ sinθ₂ < u₁ sinθ₁
2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g
Time of flight B < Time of flight A
Similarly we can prove
Time of flight C < Time of flight B
Hence Time of flight A is greatest .
Answer:
B. - 0.328
Explanation
Potential Energy:<em> This is the energy of a body due to position.</em>
<em>The S.I unit of potential energy is Joules (J).</em>
<em>It can be expressed mathematically as</em>
<em>Ep = mgh........................... Equation 1</em>
<em>Where Ep = potential energy, m = mass of the coin, h = height, g = acceleration due to gravity,</em>
<em>Given: m = 2.74 g = 0.00274 kg, h = 12.2 m, g = 9.8 m/s²</em>
Substituting these values into equation 1
Ep = 0.00274×12.2×9.8
Ep = 0.328 J.
Note: Since the potential energy at the surface is zero, the potential Energy with respect to the surface = -0.328 J
The right option is B. - 0.328
<em />
Answer:
E = -4000 N / C
Explanation:
The potential and electric field are related
V = - E s
E = - V / s
we reduce the magnitudes to the SI system
s = 4 mm (1 m / 1000 mm) = 0.004 m
we calculate
E = - 16 /0.004
E = -4000 N / C
