To find out scientific notation, you want to make sure that number is less than 10. So do 5.000000, you don't rally need the zeros but I just want to make my point. So use 10^x meaning ten the whatever power adds zeros like 5.000000x10^6 meaning it is increasing it by six zeros moving it out of the decimals and letting become 5,000,000.
Force equals mass times acceleration. Or:
F=ma
Plug it in:
5=10a
5/10=(10a)/10
.5m/s²=a
One of the concepts to be used to solve this problem is that of thermal efficiency, that is, that coefficient or dimensionless ratio calculated as the ratio of the energy produced and the energy supplied to the machine.
From the temperature the value is given as
Where,
T_L = Cold focus temperature
T_H = Hot spot temperature
Our values are given as,
T_L = 20\° C = (20+273) K = 293 K
T_H = 440\° C = (440+273) K = 713 K
Replacing we have,
Therefore the maximum possible efficiency the car can have is 58.9%
If the velocity of the train is v=s/t, where s is the distance and t is time, then v=400/5=80m/s. To get the vertical component of the velocity we need to multiply the velocity v with a sin(α): Vv=v*sin(α), where Vv is the vertical component of the velocity and α is the angle with the horizontal. So:
Vv=80*sin(10)=80*0.1736=13.888 m/s.
So the vertical component of the velocity of the train is Vv=13.888 m/s.
Answer:
i)-6.25m/s
ii)18 metres
iii)26.5 m/s or 95.4 km/hr
Explanation:
Firstly convert 90km/hr to m/s
90 × 1000/3600 = 25m/s
(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)
0 = (25)^2 + 2A(50)
0 = 625 + 100A....then moved the other value to one
-625 = 100A
Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)
(ii) Firstly convert 54km/hr to m/s
In which this is 54 × 1000/3600 = 15m/s
then apply the same formula as that in (i)
0 = (15)^2 + 2(-6.25)s
-225 = -12.5s
Hence the stopping distance = 18metres
(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question
0 = u^2 + 2(-6.25)(56)
u^2 = 700
Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s
In km/hr....26.5 × 3600/1000 = 95.4 km/hr
