Answer:
(a) 0.613 m
(b) 0.385 m
(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s
v = 3.68 m/s², θ = 72.6° below the horizontal
Explanation:
(a) Take down to be positive.
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 0.350 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²
Δy = 0.613 m
(b) Given in the x direction:
v₀ = 1.10 m/s
a = 0 m/s²
t = 0.350 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²
Δx = 0.385 m
(c) Find: vₓ and vᵧ
vₓ = aₓt + v₀ₓ
vₓ = (0 m/s²) (0.350 s) + 1.10 m/s
vₓ = 1.10 m/s
vᵧ = aᵧt + v₀ᵧ
vᵧ = (10 m/s²) (0.350 s) + 0 m/s
vᵧ = 3.50 m/s
The magnitude is:
v² = vₓ² + vᵧ²
v = 3.68 m/s²
The direction is:
θ = atan(vᵧ / vₓ)
θ = 72.6° below the horizontal
Answer:
b. 9.5°C
Explanation:
= Mass of ice = 50 g
= Initial temperature of water and Aluminum = 30°C
= Latent heat of fusion = 
= Mass of water = 200 g
= Specific heat of water = 4186 J/kg⋅°C
= Mass of Aluminum = 80 g
= Specific heat of Aluminum = 900 J/kg⋅°C
The equation of the system's heat exchange is given by

The final equilibrium temperature is 9.50022°C
Answer:
a) v₁fin = 3.7059 m/s (→)
b) v₂fin = 1.0588 m/s (→)
Explanation:
a) Given
m₁ = 0.5 Kg
L = 70 cm = 0.7 m
v₁in = 0 m/s ⇒ Kin = 0 J
v₁fin = ?
h<em>in </em>= L = 0.7 m
h<em>fin </em>= 0 m ⇒ U<em>fin</em> = 0 J
The speed of the ball before the collision can be obtained as follows
Einitial = Efinal
⇒ Kin + Uin = Kfin + Ufin
⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0
⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))
⇒ v₁fin = 3.7059 m/s (→)
b) Given
m₁ = 0.5 Kg
m₂ = 3.0 Kg
v₁ = 3.7059 m/s (→)
v₂ = 0 m/s
v₂fin = ?
The speed of the block just after the collision can be obtained using the equation
v₂fin = 2*m₁*v₁ / (m₁ + m₂)
⇒ v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)
⇒ v₂fin = 1.0588 m/s (→)
Answer:
215955.06 m/s^2
Explanation:
length of barrel, s = 0.89 m
initial velocity of the bullet, u = 0 m/s
Final velocity of the bullet, v = 620 m/s
Let a be the acceleration of the bullet in the barrel
Use third equation of motion, we get


a = 215955.06 m/s^2
Thus, the acceleration of the bullet inside the barrel is 215955.06 m/s^2.
If the cross-section of a wire of fixed length is doubled, the resistance of that wire change into doubled.We know that <span>the total </span>length<span> of the wires will </span>affect<span> the amount of </span>resistance. <span> The longer the wire, the more </span>resistance<span> that there will be so the answer is doubled.</span>