At time t = 0, a 2150-kg rocket in outer space fires an engine that exerts an increasing force on it in the +x-direction. this force obeys the equation f = at^2, where t is time,
and has a magnitude of 781.25 n when r = 1.25 s. (a) find the si value of the constant a, including its units. (b) what impulse does the engine exert on the rocket during the i .50-s
interval starting 2.00 s after the engine is fired? (c) by how much does the rocket? s velocity change during this interval?

a)  A = 500 N/s²

b)  I = 5812.50 N-s

c)  Δv = 2.7034 m/s

Explanation:

Given info

m = 2150 kg

F(t) = At²

F(1.25 s) = 781.25 N

a) A = ?

We use the equation

F(t) = At²  ⇒     781.25 N = A*(1.25 s)²

⇒ A = F(t) / t² = (781.25 N) / (1.25 s)²

⇒ A = 500 N/s²

b) I = ?  if   2.00 s ≤ t ≤ 3.50 s

we apply the equation

I = ∫F(t) dt = ∫At² dt = A ∫t² dt = (500/3)*t³ + C

Since the limits of integration are 2 and 3.5,  we obtain

I = (500/3)*((3.5)³-(2)³) = 5812.50 N-s

c) Δv = ?

we can apply the equation

I = m*Δv   ⇒   Δv = I / m

⇒   Δv = 5812.50 N-s / 2150 kg

⇒   Δv = 2.7034 m/s

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

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