Answer:
s(t) = (t1 - t2)* [((1/t1 - 3) + (1/t0 - 3))/2]
Explanation:
We will assume that v(t) is in units of m/s and t (time) is in seconds.
v(t)=1/t−3
At time 0 (initial) the equation tells us the particle has a velocity of
v(t)=1/t−3
v(0)=1/(0)−3
v(0) = - 3 m/s
The particle is moving from right to left (the negative sign) at a rate of 3 m/s.
The position of the particle would be the average velocity times the time traveled.
Distance = Velocity x Time (with Velocity being the average between times t0 and t1)
We'll use s(t) for displacement for time t.
s(t) = v*t
We need the average velocity for the time period t0 to t1.
Let t0 and t1 be the initial and final times in which the measurement takes place.
At time t0 the velocity is = 1/t0 - 3
At time t1 the velocity is = 1/t1 - 3
The displacement is the average velocity between the two points, t0 and t1. This can be written as:
[(1/t1 - 3) + (1/t0 - 3)/\]/2
Displacement: s(t) = (t1 - t2)* [((1/t1 - 3) + (1/t0 - 3))/2]
Answer:
35
Explanation:
total magnification = eyepiece lens x objective lens
TM = 10X x 25X
TM = 250X
Answer:
1.Cp₁ = 1.2 J/g.⁰C
Explanation:
For new material:
m₁ = 25 g
T₁ = 80⁰C
specific heat of water = Cp₁
For water :
m₂ = 100 g
T₂ = 20⁰C
The final temperature T=24⁰C
We know that specific heat of water Cp₂ = 4.187 kJ/kg.K
The heat lost new material = Heat gain by Water
m₁ Cp₁ ( T₁ - T ) = m₂ Cp₂ (T- T₂)
25 x Cp₁ (80- 24 ) = 100 x 4.817 (24 - 20 )
Cp₁ x 56 = 4 x 4.187 x 4
Cp₁ = 1.19 kJ/kg.K
Cp₁ = 1.2 J/g.⁰C
Answer:
Explanation:
In this problem we have three important moments; the instant in which the ball is released (1), the instant in which the ball starts to fly freely (2) and the instant in which has its maximum height (3). From the conservation of mechanical energy, the total energy in each moment has to be the same. In (1), it is only elastic potential energy; in (2) and (3) are both gravitational potential energy and kinetic energy. Writing this and substituting by known values, we obtain:
Since we only care about the velocity , we can keep only the second and third parts of the equation and solve:
So, the speed of the ball just after the launch is 17.3m/s.