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Lana71 [14]
3 years ago
14

A motorist drives north for 36.0 minutes at 96.0 km/h and then stops for 15.0 minutes. He then continues north, traveling 130 km

in 2.20 h. (a) What is his total displacement? km
(b) What is his average velocity? km/h
Physics
1 answer:
Over [174]3 years ago
8 0

Answer:

a) d=187.6km

b)v=61.5km/h

Explanation:

First we convert our minutes to hours so we work always in the same units.

36min=36min(\frac{1h}{60min})=0.6h

15min=15min(\frac{1h}{60min})=0.25h

Where we used the fact that 1 hour are 60 min, thus the multiplying factor is equal to 1 (not altering the time, just changing the units).

a) On the first part the motorist travels a distance d_1=v_1t_1=(96km/h)(0.6h)=57.6km, and on the second part he travels d_2=130km.

The total displacement is d=d_1+d_2=57.6km+130km=187.6km

b) The average velocity is the relation between the total displacement and the time taken to cover it. Our total time is t=0.6h+0.25h+2.2h=3.05h, thus we have:

v=\frac{187.6km}{3.05h}=61.5km/h

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Yakvenalex [24]

Answer:

ΔH = 249 kJ/mol

Explanation:

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2H₂O(g) → 2H₂(g) + O₂(g)    (1)

To calculate the energy change to obtain one mole of H₂(g) from one mole of H₂O(g), the coefficients of the reaction (1) must be halved:

H₂O(g) → H₂(g) + 1/2O₂(g)    (2)

The enthalpy of the reaction (2) is given by:

\Delta H = \Delta H_{r} - \Delta H_{p}  

<em>Where \Delta H_{r}: is the bond enthalpy of reactants and \Delta H_{p}: is the bond enthalpy of products.</em>

<u>For the reactants we have the next bond energies:</u>

2 x (H-O) = 2 x (467)

<u>And the bond energies for the products are:</u>

H-H + (1/2) (O=O) =  436 + (1/2)(498)

So, the enthalpy of the reaction (2) is:

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I hope it helps you!    

3 0
3 years ago
A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f
konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

(c) The maximum speed attained by the object during its motion is 0.577 m/s

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

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speed of the toy, v = 0.4 m/s

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E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

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E = ¹/₂KA²

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(c) the maximum speed attained by the object during its motion

E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s

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