Answer:
64.2 m/s
Explanation:
We are given that
Speed ,v=38 m/s
We have to find the maximum speed when his car reach on flat ground.
Using dimensional analysis
[tex]F_{res}\propto v^2[/tex]
If 35% acceleration reduced by F(res) at 38 m/s
Then, 100% acceleration can be reduced by F(res) at v' m/s
[tex]\frac{F_1}{F_2}=\frac{v^2}{v'^2}[/tex]
[tex]v'^2=\frac{F_2}{F_1}v^2[/tex]
[tex]v'=v\sqrt{\frac{F_2}{F_1}}[/tex]
Substitute the values
[tex]v'=38\times \sqrt{\frac{100}{35}}[/tex]
[tex]v'=64.2 m/s[/tex]
Hence, the maximum speed when his car can reach on flat ground=64.2 m/s
The maximum velocity of the car at constant acceleration on the road will be 62.2 m/s.
What is the relation between resistance and speed?
The air resistance is directly proportional to the square of the velocity of an object.
R ∝ v²
The speed of the car was reduced by 35 % at 38 m/s.
So, the speed of the car was reduced by 100% at v' m/s.
The relationship can be given by,
[tex]\dfrac {R_1}{R_2} = \dfrac {v^2}{v'^2}[/tex]
Put the values in the formula and calculate for [tex]v'[/tex],
[tex]v' = \sqrt {\dfrac {R_2 v^2}{R_1 }}[/tex]
[tex]v' = \sqrt {\dfrac { 100\times 38 ^2}{35 }}\\\\v' = 62.2 \rm \ m/s[/tex]
Therefore, the maximum velocity of the car at constant acceleration on the road will be 62.2 m/s.
Learn more about resistance and speed?
https://brainly.com/question/11574961
A 10 kg remote control plane is flying at a height of 111 m. How much
potential energy does it have?
Answer:
10.88kJ
Explanation:
Given data
mass= 10kg
heigth= 111m
Applying
PE= mgh
assume g= 9.81m/s^2
substitute
PE= 10*9.81*111
PE=10889.1 Joules
PE=10.881kJ
Hence the potential energy is 10.88kJ
You perform nine (identical) measurements of the acceleration of gravity (units of m/s2): 10.1, 9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90. The true value is 9.81. Calculate the mean value of your results to three significant digits. ________
Answer: The mean value = 9.85m/s².
Explanation:
Mean = [tex]\dfrac{\text{Sum of n observations}}{n}[/tex]
The given measurements the acceleration of gravity (units of m/s²): 10.1, 9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90.
Number of measurements =9
Sum of measurements = 88.69
Mean = [tex]\dfrac{88.69}{9}=9.85444444\approx9.85[/tex]
Hence, the mean value = 9.85m/s².
Current Attempt in Progress The atomic radii of a divalent cation and a monovalent anion are 0.52 nm and 0.125 nm, respectively. (a) Calculate the force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another). Enter your answer for part (a) in accordance to the question statement N (b) What is the force of repulsion at this same separation distance
Answer:
a) F = 1.70 10⁻⁹N, F = 1.47 10⁻⁸ N,
b) * the electronegative repulsion, from the repulsion by quantum effects
Explanation:
a) The atraicione force comes from the electric force given by Coulomb's law,
F = [tex]k \frac{ q_1 q_2}{r^2}[/tex]
divalent atoms
In this case q = 2q₀ where qo is the charge of the electron -1,6 10⁻¹⁹ C and the separation is given
F = k q² / r²
F = [tex]2 \ 10^9 \ \frac{2 (1.6 \ 10^{-19} )^2}{ (0.52 10^{-9} )^2 }[/tex]
F = 1.70 10⁻⁹N
monovalent atoms
in this case the load is q = q₀
F = 2 \ 10^9 \ \frac{ (1.6 \ 10^{-19} )^2}{ (0.125 10^{-9} )^2 }
F = 1.47 10⁻⁸ N
b) repulsive forces come from various sources
* the electronegative repulsion of positive nuclei
* the electrostatic repulsion of the electrons when it comes to bringing the electron clouds closer together
* from the repulsion of electron clouds, by quantum effects