1. False
2. Positive
I’m just joking I don’t know the answersss
F = m . g = 76.5 x 9..8 = 749.7
Net Force = 3225 - 749.7 = 2475.3
F = m.a
2475.3 = 76.5 a
a = 32.35
V = at + v1
V = at + 0
V = 32.35 x 0.15
V = 4.8525
Hope this helps
Answer:
Same reading.
Explanation:
Assume that after the string breaks the ball falls through the liquid with constant speed. If the mass of the bucket and the liquid is 1.20 kg, and the mass of the ball is 0.150 kg,
A.) Before the string break, the total weight = weight of the can + weight of the water.
According to Archimedes' Principle which state that: “A body immersed in a liquid loses weight by an amount equal to the weight of the liquid displaced.” Archimedes principle also states that: “When a body is immersed in a liquid, an upward thrust, equal to the weight of the liquid displaced, acts on it
B.) After the string break.
The scale will have the same reading as before the string break.
Answer:
always same
Explanation:
velocity and speed are same upto some extend but velocity is vector while speed is scalar quantity
Answer:
θ = 12.95º
Explanation:
For this exercise it is best to separate the process into two parts, one where they collide and another where the system moves altar the maximum height
Let's start by finding the speed of the bar plus clay ball system, using amount of momentum
The mass of the bar (M = 0.080 kg) and the mass of the clay ball (m = 0.015 kg) with speed (v₀ = 2.0 m / s)
Initial before the crash
p₀ = m v₀
Final after the crash before starting the movement
= (m + M) v
p₀ = 
m v₀ = (m + M) v
v = v₀ m / (m + M)
v = 2.0 0.015 / (0.015 +0.080)
v = 0.316 m / s
With this speed the clay plus bar system comes out, let's use the concept of conservation of mechanical energy
Lower
Em₀ = K = ½ (m + M) v²
Higher
= U = (m + M) g y
Em₀ = 
½ (m + M) v² = (m + M) g y
y = ½ v² / g
y = ½ 0.316² / 9.8
y = 0.00509 m
Let's look for the angle the height from the pivot point is
L = 0.40 / 2 = 0.20 cm
The distance that went up is
y = L - L cos θ
cos θ = (L-y) / L
θ = cos⁻¹ (L-y) / L
θ = cos⁻¹-1 ((0.20 - 0.00509) /0.20)
θ = 12.95º