Aaron's car is moving at speed of 30 m/s
His reaction time is given as 0.7 s
but when he is tired the reaction time is doubled
Now we need to find the distance covered by his car when he is tired during the time when he react to apply brakes
So here since during this time speed is given as constant so we can say that distance covered can be product of speed and time
So here we can use
So the car will move to 42 m during the time when he apply brakes
Answer:
Current = 10 Amperes.
Explanation:
Given the following dat;
Quantity of charge, Q = 36 kilocoulombs (KC) = 36 * 1000 = 36000C
Time = 1 hour to seconds = 60*60 = 3600 seconds
To find the current;
Quantity of charge = current * time
Substituting in the equation
36000 = current * 3600
Current = 36000/3600
Current = 10 Amperes.
Answer:
vo=5.87m/s
Explanation:
Hello! In this problem we have a uniformly varied rectilinear movement.
Taking into account the data:
α =69.2
vf = 10m / s
h=2.7m
g=9.8m/s2
We know we want to know the speed on the y axis.
We calculate vfy
vfy = 10m / s * (sen69.2) = 9.35m / s
We can use the following equation.
We clear the vo (initial speed)
vo=5.87m/s
The correct answer is 1.07m.
The area surrounding an electric charge where its impact may be felt is known as the electric field. When another charge enters the field, the presence of an electric field may be felt. The electric field will either attract or repel the charge depending on its makeup. Any electric charge has a property known as the electric field. The charge and electrical force working in the field determine the strength or intensity of the electric field.
Here, is the charge per unit length, r is the distance from the wire, and
is the free space permittivity ε_0. Electric field due to the long straight wire is,
E= λ/2πε_0r
Rearrange the equation for r.
r=λ/2πε_0E
Substitute 2.41 N/C for E,
E=1.44×10^-10C/m
λ=8.85×10^-12C^2/Nm^2
r=(1.44×10^-10C/m)/(2(3.14)(8.85×10^-12C^2/Nm^2)(2.41N/C))
r=1.07m
At a distance of 1.07 m the magnitude of electric field is 2.41 N/C.
To learn more about electric field refer the link:
brainly.com/question/12821750
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