Answer
given,
distance = 140 m
time, t = 3.6 s
moving speed = 53 m/s
a) distance = (average velocity) x time
v₀ + 53 = 77.78
v₀ = 24.78 m/s or 25 m/s
b)
a = 7.8 m/s²
using equation of motion
v₀² = v₁² + 2 a s
53² = 0²+ 2 x 7.8 x s
s = 180 m
Answer:
2.068 x 10^6 m / s
Explanation:
radius, r = 5.92 x 10^-11 m
mass of electron, m = 9.1 x 10^-31 kg
charge of electron, q = 1.6 x 10^-19 C
As the electron is revolving in a circular path, it experiences a centripetal force which is balanced by the electrostatic force between the electron and the nucleus.
centripetal force =
Electrostatic force =
where, k be the Coulombic constant, k = 9 x 10^9 Nm^2 / C^2
So, balancing both the forces we get
v = 2.068 x 10^6 m / s
Thus, the speed of the electron is give by 2.068 x 10^6 m / s.
Answer:
20cm
Explanation:
A convex lens has a positive focal length and the object placed in front of it produce both virtual and real image <em>(image distance can be negative or positive depending on the nature of the image</em>).
According to the lens equation
where;
f is the focal length of the lens
u is the object distance
v is the image distance
If the magnification is - 0.6
mag = v/u = -0.5
v = -0.5u
since v = 10cm
10 = -0.5u
u = -10/0.5
u =-20 cm
Substitute u = -20cm ( due to negative magnification)and v = 10cm into the lens formula to get the focal length f
Hence the focal length of the convex lens is 20cm
Answer:
Explanation:
First we have to find the time required for train to travel 60 meters and impact the car, this is an uniform linear motion:
The reaction time of the driver before starting to accelerate was 0.50 seconds. So, remaining time for driver is 1.5 seconds.
Now, we have to calculate the distance traveled for the driver in this 0.5 seconds before he start to accelerate. Again, is an uniform linear motion:
The driver cover 10 meters in this 0.5 seconds. So, the remaining distance to be cover in 1.5 seconds by the driver are 35 meters. We calculate the minimum acceleration required by the car in order to cross the tracks before the train arrive, Since this is an uniformly accelerated motion, we use the following equation: