J=joules, c=specific heat, q= energy, and the Tf and Ti are the final and initial temperatures cause I couldn't find a delta sign.
Answer:
we could use the formula, v=u+at,
65=25+a (10), a=4 , since the motion is declerating we have a=-4 m/s2
Answer:
The answer depends on what object you are dropping. Are you dropping a balloon or a car? (I'm joking 'bout that one.) If the mass of the object is very little, then it might drop slower. If the mass is bigger, then it might drop faster.
Good luck!
Explanation:
Answer:
V₁ = √ (gy / 3)
Explanation:
For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2
Starting point
Em₀ = U₁ + U₂
Em₀ = m₁ g y₁ + m₂ g y₂
Let's place the reference system at the point where the mass m1 is
y₁ = 0
y₂ = y
Em₀ = m₂ g y = 2 m₁ g y
End point, at height yf = y / 2
= K₁ + U₁ + K₂ + U₂
= ½ m₁ v₁² + ½ m₂ v₂² + m₁ g + m₂ g
Since the masses are joined by a rope, they must have the same speed
= ½ (m₁ + m₂) v₁² + (m₁ + m₂) g
= ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g
How energy is conserved
Em₀ =
2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g
2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2
3/2 v₁² = 2 g y -3/2 g y
3/2 v₁² = ½ g y
V₁ = √ (gy / 3)
Answer:
The time taken is
Explanation:
From the question we are told that
The length of steel the wire is
The length of the copper wire is
The diameter of the wire is
The tension is
The time taken by the transverse wave to travel the length of the two wire is mathematically represented as
Where is the time taken to transverse the steel wire which is mathematically represented as
here is the density of steel with a value
So
And
is the time taken to transverse the copper wire which is mathematically represented as
here is the density of steel with a value
So
So