Answer:
The average speed can be calculated as the quotient between the distance travelled and the time needed to travel that distance.
To go to the school, he travels 2.4 km in 0.6 hours, then here the average speed is:
s = (2.4km)/(0.6 hours) = 4 km/h
To return to his home, he travels 2.4km again, this time in only 0.4 hours, then here the average speed is:
s' = (2.4 km)/(0.4 hours) = 6 km/h.
Now, if we want the total average speed (of going and returning) we have that the total distance traveled is two times the distance between his home and school, and the total time is 0.6 hours plus 0.4 hours, then the average speed is:
S = (2*2.4 km)/(0.6 hours + 0.4 hours)
S = (4.8km)/(1 h) = 4.8 km/h
Answer:
The electron’s velocity is 0.9999 c m/s.
Explanation:
Given that,
Rest mass energy of muon = 105.7 MeV
We know the rest mass of electron = 0.511 Mev
We need to calculate the value of γ
Using formula of energy
Put the value into the formula
We need to calculate the electron’s velocity
Using formula of velocity
Put the value into the formula
Hence, The electron’s velocity is 0.9999 c m/s.
Answer:
- <u>The energy change would be 46kJ</u>
- <u>The energy would be absorbed</u>
Explanation:
The <em>energy change </em>during a chemical reation, i.e. the reaction energy, is equal to the chemical energy stored in the<em> bonds of the products </em>less the chemical energy stored in the <em>bonds of the reactants</em>.
Hence:
- <em>Energy change</em> = 478 kJ - 432kJ = 46kJ
The change is positive, this is, the chemical energy of the products is greater than the chemical energy of the reactants.
That corresponds to the second graph, where the level of the energy of the products in the graph is higher than the level of the energy of the reactants. Therefore, the conclusion is that the reaction <em>absorbed energy</em> and it is endothermic.
Answer:
A) F=-20.16×10⁹N
B) if the distance doubles, force is 4 times smaller.
Explanation:
q1=-28C
q2=5mC=0.005C
d=25cm=0.25m
Electrostatic force between charges: F=k×q1×q2/d², where k is a coefficient that has the value k=9 × 10⁹ N⋅m²⋅C^(-2) for air.
Thus:
F=9×10⁹×(-28)×0.005/0.25²
F=-20.16×10⁹N
The minus sign indicates attraction.
If distance doubles, d1=2×d, then we have 4d² at the denominator and the force is 4 times smaller.
Answer:
m = 35.98 Kg ≈ 36 Kg
Explanation:
I₀ = 125 kg·m²
R₁ = 1.50 m
ωi = 0.600 rad/s
R₂ = 0.905 m
ωf = 0.800 rad/s
m = ?
We can apply The law of conservation of angular momentum as follows:
Linitial = Lfinal
⇒ Ii*ωi = If*ωf <em>(I)</em>
where
Ii = I₀ + m*R₁² = 125 + m*(1.50)² = 125 + 2.25*m
If = I₀ + m*R₂² = 125 + m*(0.905)² = 125 + 0.819025*m
Now, we using the equation <em>(I) </em>we have
(125 + 2.25*m)*0.600 = (125 + 0.819025*m)*0.800
⇒ m = 35.98 Kg ≈ 36 Kg
