Answer:
F = 69.3 N
Explanation:
For this exercise we use Newton's second law, remembering that the static friction force increases up to a maximum value given by
fr = μ N
We define a reference system parallel to the floor
block B ( lower)
Y axis
N - W₁-W₂ = 0
N = W₂ + W₂
N = (M + m) g
X axis
F -fr = M a
for block A (upper)
X axis
fr = m a (2)
so that the blocks do not slide, the acceleration in both must be the same.
Let's solve the system by adding the two equations
F = (M + m) a (3)
a =
the friction force has the formula
fr = μ N
fr = μ (M + m) g
let's calculate
fr = 0.34 (2.0 + 0.250) 9.8
fr = 7.7 N
we substitute in equation 2
fr = m a
a = fr / m
a = 7.7 / 0.250
a = 30.8 m / s²
we substitute in equation 3
F = (2.0 + 0.250) 30.8
F = 69.3 N
<h2>Apartment Explosion Reported </h2>
The apartment’s explosion, reportedly caused by a gas leak, produced a violent release of gas and heat. The heat increased the temperature of the air in the room, which means an increase in the air's molecular kinetic energy.
When heat is provided then temperature increases and the molecules of substances move rapidly by increase of kinetic energy (K.E) temperature increases. It is understood that heat increases temperature.
Answer:
E = 1.25×10¹³ N/m²
Explanation:
Young's modulus is defined as:
E = stress / strain
E = (F / A) / (dL / L)
E = (F L) / (A dL)
Given:
F = 10 kg × 9.8 m/s² = 98 N
L = 1 m
dL = 10⁻⁵ m
A = π/4 (0.001 m)² = 7.85×10⁻⁷ m²
Solve:
E = (98 N × 1 m) / (7.85×10⁻⁷ m² × 10⁻⁵ m)
E = 1.25×10¹³ N/m²
Round as needed.
Answer:
a) 14.2 atm
b) 4.46 atm
c) 1.06 atm
Explanation:
For an ideal gas,
PV = nRT
P = pressure of the gas
V = volume occupied by the gas
n = number of moles of the gas
R = molar gas constant = 0.08206 L.atm/mol.K
T = temperature of the gas in Kelvin
a) For HF,
P =?, V = 2.5L, n = 1.35 moles, T = 320K
P = 1.35 × 0.08206 × 320/2.5
P = 14.2 atm
b) For NO₂
P =?, V = 4.75L, n = 0.86 moles, T = 300K
P = 0.86 × 0.08206 × 300/4.75
P = 4.46 atm
c) For CO₂
P =?, V = 5.5 × 10⁴ mL = 55L, n = 2.15 moles, T = 57°C = 330K
P = 2.15 × 0.08206 × 330/55
P = 1.06 atm
