The engine of a locomotive exerts a constant force of 8.0 105 n to accelerate a train to 82 km/h. determine the time (in min) taken for the train of mass 1.7 107 kg to reach this speed from rest. min

It will take the train 8.2 min to reach a speed of 82 km/h

Explanation:

Hi there!

Using Newton´s second law, we can calculate the acceleration of the train:

F = m · a

Where:

F = force exerted by the locomotive.

m = mass of the train.

a = acceleration of the train.

Solving the equation for the acceleration:

F/m = a

8.0 × 10⁵ N / 1.7 × 10⁷ kg = a

a = 4.7 × 10⁻² m/s²

Let´s convert 82 km/h to m/s:

82 km/h · (1000 m / 1 km) · (1 h /3600 s) = 23 m/s

Using the equation of velocity, we can calculate the time it takes for the train to reach a velocity of 23 m/s.

v = v0 + a · t

Where:

v = velocity at time "t".

v0 = initial velocity (in this case, v0 = 0 because the train starts from rest).

a = acceleration.

t = time.

23 m/s = 0 +  4.7 × 10⁻² m/s² · t

t = 23 m/s /  4.7 × 10⁻² m/s²

t =4.9 × 10² s

Let´s convert the seconds into minutes:

4.9 × 10² s · (1 min / 60 s) = 8.2 min

It will take the train 8.2 min to reach a speed of 82 km/h

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

luster is not a good characteristic to use by itself because luster is the property of minerals that indicates how much the surface of a mineral reflects light. the luster of a mineral is affected by the brilliance of the light used to observe the mineral surface.