Weight is a force caused by gravity. The weight of an object is the gravitational force between the object and Earth.
Answer:
114.44 J
Explanation:
From Hook's Law,
F = ke................. Equation 1
Where F = Force required to stretch the spring, k = spring constant, e = extension.
make k the subject of the equation
k = F/e.............. Equation 2
Given: F = 10 lb = (10×4.45) N = 44.5 N, e = 4 in = (4×0.254) = 1.016 m.
Substitute into equation 2
k = 44.5/1.016
k = 43.799 N/m
Work done in stretching the 9 in beyond its natural length
W = 1/2ke²................. Equation 3
Given: e = 9 in = (9×0.254) = 2.286 m, k = 43.799 N/m
Substitute into equation 3
W = 1/2×43.799×2.286²
W = 114.44 J
Answer:
(a) W = 1329.5 J = 1.33 KJ
(b) ΔU = 24.27 KJ
Explanation:
(a)
Work done by the gas can be found by the following formula:
where,
W = Work = ?
P = constant pressure = (0.991 atm)(
) = 100413 Pa
ΔV = Change in Volume = 18.7 L - 5.46 L = (13.24 L)(
) = 0.01324 m³
Therefore,
W = (100413 Pa)(0.01324 m³)
<u>W = 1329.5 J = 1.33 KJ</u>
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(b)
Using the first law of thermodynamics:
ΔU = ΔQ - W (negative W for the work done by the system)
where,
ΔU = change in internal energy of the gas = ?
ΔQ = heat added to the system = 25.6 KJ
Therefore,
ΔU = 25.6 KJ - 1.33 KJ
<u>ΔU = 24.27 KJ</u>
<u>Momentum</u>
- a vector quantity; has both magnitude and direction
- has the same direction as object's velocity
- can be represented by components x & y.
Find linebacker momentum given m₁ = 120kg, v₁ = 8.6 m/s north
P₁ = m₁v₁
P₁ = (120)(8.6)
[ P₁ = 1032 kg·m/s ] = y-component, linebacker momentum
Find halfback momentum given m₂ = 75kg, v₂ = 7.4 m/s east
P₂ = m₂v₂
P₂ = (75)(7.4)
[ P₂ = 555 kg·m/s ] = x-component, halfback momentum
Find total momentum using x and y components.
P = √(P₁)² + (P₂)²
P = √(1032)² + (555)²
[[ P = 1171.77 kg·m/s ]] = magnitude
!! Finally, to find the magnitude of velocity, take the divide magnitude of momentum by the total mass of the players.
P = mv
P = (m₁ + m₂)v
1171.77 = (120 + 75)v <em>[solve for v]</em>
<em />v = 1171.77/195
v = 6.0091 ≈ 6.0 m/s
If asked to find direction, take inverse tan of x and y components.
tanθ = (y/x)
θ = tan⁻¹(1032/555)
[ θ = 61.73° north of east. ]
The magnitude of the velocity at which the two players move together immediately after the collision is approximately 6.0 m/s.
Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants (
= 0) zero, so let's use the equation
y = t -1/2 g t²
y= - ½ g t²
t = √2y/g
b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity
x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
vₓ = vox = constant
= - gt
= 0 - g √2y/g
= - √2gy
tan θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
d) The projectile is higher than the cliff because it is a horizontal launch
