Answer:
6.53 m/s²
Explanation:
Let m₁ = 5 kg and m₂ = 10 kg. The figure is attached and free body diagrams of the objects are also attached.
Both objects (m₁ and m₂) have the same magnitude of acceleration(a). Let g be the acceleration due to gravity = 9.8 m/s². Hence:
T = m₁a (1)
m₂g - T = m₂a (2)
substituting T = m₁a in equation 2:
m₂g - m₁a = m₂a
m₂a + m₁a = m₂g
a(m₁ + m₂) = m₂g
a = m₂g / (m₁ + m₂)
a = (10 kg * 9.8 m/s²) / (10 kg + 5 kg) = 6.53 m/s²
Both objects have an acceleration of 6.53 m/s²
Hello!
Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.
Data:
Hooke represented mathematically his theory with the equation:
F = K * Δx
On what:
F (elastic force) = 2 N
K (elastic constant) = 4 N/cm
Δx (deformation or elongation of the elastic medium or distance from a spring) = ?
Solving:




simplify by 2


Answer:
B.) 1/2 cm
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I Hope this helps, greetings ... Dexteright02! =)
Answer:
Object appears to move forward at 1 cm/sec, then the velocity drops to zero for 3 sec and then moves forward at 2 cm/sec (11 - 3) / (10 - 6) = 2 cm/sec
A cation
Cations are positive ions that have lost an electron
Answer:
Babies often struggle to go to sleep sometimes warm milk and sooting sounds help babies sleep
Explanation: