The work done is the product between the intensity of the force applied F, the amount of the displacement d of the book and the cosine of the angle
between the direction of the force and the direction of the displacement:
In our problem, the student is lifting the book, so he is applying a force directed upward, and the book is moving upward, so F and d are parallel and therefore the angle is zero, so
Therefore, the work done is
Answer:
The force exerted by the rope on her arms is 273.7 N = 0.274 kN
Explanation:
Step 1: Data given
Mass of the ice skater = 55.6 kg
Velocity = 1.73 m/s
She then moves in a circle of radius 0.608 m around the pole.
Step 2:
Force exterted by the horizontal rope is the centripetal force acting on theice skater:
Fc = M*ac
⇒ with ac = v²/r
Fc = M * v²/r
Fc = 55.6 * 1.73²/0.608
Fc =273.69 N
The force exerted by the rope on her arms is 273.7 N = 0.274 kN
The force decreases due to less frictional resistance. it becomes easier to keep it moving as it is already moving, one of the Newton's laws
Answer:
φ = -7.16 × 10⁴ Nm²/C
Explanation:
q = -3.80 × 10⁻⁶ C
ε₀ = 8.85 10⁻¹² C²/Nm²
By Gauss's law
φ = q/ε₀
as the cube has 6 faces so for one side the flux will
φ =1/6 ( q/ε₀)
φ = 1/6 ( -3.80 × 10⁻⁶ C / 8.85 10⁻¹² C²/Nm²)
φ = -71,563.088 Nm²/C
φ = -7.16 × 10⁴ Nm²/C
Explanation:
(a) What is the maximum height the arrow will attain?
Given:
v₀ᵧ = 54 sin 27.0° m/s = 24.5 m/s
vᵧ = 0 m/s
aᵧ = -9.8 m/s²
Find: Δy
vᵧ² = v₀ᵧ² + 2aᵧΔy
(0 m/s)² = (24.5 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 30.7 m
(b) The target is at the height from which the arrow was shot. How far away is it?
Given:
Δy = 0 m
v₀ᵧ = 54 sin 27.0° m/s = 24.5 m/s
aᵧ = -9.8 m/s²
Find: t
Δy = v₀ᵧ t + ½ aᵧt²
0 m = (24.5 m/s) t + ½ (-9.8 m/s²) t²
t = 5.00 s
Given:
v₀ₓ = 54 cos 27.0° m/s = 48.1 m/s
aₓ = 0 m/s²
t = 5.00 s
Find: Δx
Δx = v₀ₓ t + ½ aₓt²
Δx = (48.1 m/s) (5.00 s) + ½ (0 m/s²) (5.00 s)²
Δx = 241 m