Answer:
A) The free body diagrams for both the load of bricks and the counterweight are attached.
B) a = 2.96 m/s²
Explanation:
A)
The free body diagrams for both the load of bricks and the counterweight are attached.
B)
The acceleration of upward acceleration of the load of bricks is given by the following formula:
a = g(m₁ - m₂)/(m₁ + m₂)
where,
a = upward acceleration of load of bricks = ?
g = 9.8 m/s²
m₁ = heavier mass = mass of counterweight = 28 kg
m₂ = lighter mass = mass of load of bricks = 15 kg
Therefore, using these values in equation, we get:
a = (9.8 m/s²)(28 kg - 15 kg)/(28 kg + 15 kg)
<u>a = 2.96 m/s²</u>
Answer:
T = 273 + (-50) = 273 – 50 = 223 K
R = 188.82 J / kg K for CO2
Density (Martian Atmosphere) = P / RT = 900 / 188.92 x 223 = 900 / 42129.16 = 0.0213 kg /
T = 273 +18 = 291 K, R = 287 J / kg k (for air) P = 101.6 k Pa = 101600 Pa
Density (Earth Atmosphere) = P / RT = 101600 / 287 x 291 = 1.216 kg /
Answer:
<h3>The answer is 4.29 m/s²</h3>Explanation:
The acceleration of an object given it's velocity and time taken acting on it can be found by using the formula
From the question
velocity = 30 m/s
time = 7 s
We have
We have the final answer as
<h3>4.29 m/s²</h3>Hope this helps you