Answer:
Explanation:
Given
Force P is acting upward
C is vertical contact Force
W is the weight of the crate
As P is unable to move the Block therefore Normal reaction keeps on acting on block
thus we can say that
P-W+C=0
P=W-C
The partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm
<u>Explanation:</u>
H₂ + Br₂ ⇒ 2HBr
PH₂ = 0.782atm
PBr₂ = 0.493atm
Kp = (PHBr)²/ (PH₂) (PBr₂) = 1.4 X 10⁻²¹
At equilibrium:
Let 2x = pressure of HBr
PH₂ = 0.782 -x
PBr₂ = 0.493 - x
Kp = (2x)^2 / (0.782-x)(0.493-x)
Now, because Kp is very small, x will be very small compared to 0.782 and 0.493.
Then,
Kp = 1.4X10⁻²¹ = (4x²) / (0.782)(0.493)
x = 1.2X10⁻¹¹
PHBr = 2x = 2.4 X 10⁻¹¹ atm
Therefore, the partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm
The acceleration produced in a body is always in the direction of the resultant force acting on the body. Therefore, we may determine the horizontal acceleration using the horizontal force applied. To do this, we may apply the mathematical form of Newton's second law:
Force = mass * acceleration
acceleration = force / mass
Substituting the values,
a = 100 / 0.15
a = 666.7 m/s²
The acceleration of the hockey puck is 670 m/s²
Answer:
b ≈ 64 Kg/s
Explanation:
Given
Fd = −bv
m = 2.5 kg
y = 6.0 cm = 0.06 m
g = 9.81 m/s²
The object in the pan comes to rest in the minimum time without overshoot. this means that damping is critical (b² = 4*k*m).
m is given and we find k from the equilibrium extension of 6.0 cm (0.06 m):
∑Fy = 0 (↑)
k*y - W = 0 ⇒ k*y - m*g = 0 ⇒ k = m*g / y
⇒ k = (2.5 kg)*(9.81 m/s²) / (0.06 m)
⇒ k = 408.75 N/m
Hence, if
b² = 4*k*m ⇒ b = √(4*k*m) = 2*√(k*m)
⇒ b = 2*√(k*m) = 2*√(408.75 N/m*2.5 kg)
⇒ b = 63.9335 Kg/s ≈ 64 Kg/s
Answer:
Explanation:
Voltage, V = 12 V
Charge, q = 2 micro coulomb = 2 x 10^-6 C
Work = energy
W = 0.5 x q x V
W = 0.5 x 2 x 10^-6 x 12
W = 12 x 10^-6 J