Answer:
B can take 0.64 sec for the longest nap .
Explanation:
Given that,
Total distance = 350 m
Acceleration of A = 1.6 m/s²
Distance = 30 m
Acceleration of B = 2.0 m/s²
We need to calculate the time for A
Using equation of motion

Put the value in the equation



We need to calculate the time for B
Using equation of motion
Put the value in the equation



We need to calculate the time for longest nap
Using formula for difference of time



Hence, B can take 0.64 sec for the longest nap .
Answer:
t = 1.05 s
Explanation:
Given,
The distance between your vehicle and car, 100 ft
The constant speed of your vehicle, u = 95 ft/s
Since, the velocity is constant, a =0
If the car stopped suddenly, time left for you to hit the brake, t = ?
Using the second equation of motion,
S = ut + ½ at²
Substituting the given values in the equation
100 = 95 x t
t = 100/95
= 1.05 s
Hence, the time left for you to hit the brakes and stop before rear ending them, t = 1.05 s
Answer:
Explanation:
<u></u>
<u>1. Formulae:</u>
Where:
- E = kinetic energy of the particle
- λ = de-Broglie wavelength
- m = mass of the particle
- v = speed of the particle
- h = Planck constant
<u><em>2. Reasoning</em></u>
An alha particle contains 2 neutrons and 2 protons, thus its mass number is 4.
A proton has mass number 1.
Thus, the relative masses of an alpha particle and a proton are:

For the kinetic energies you find:


Thus:


From de-Broglie equation, λ = h/(mv)

1 meter = 1e9 nm
To get meters, divide nanometers by 1e9: 9.95nm / 1x10^9 = 9.95x10^-9 meters
Answer: 9.95e-9 meters
Answer:
5.43 x 10^-3 Nm
Explanation:
N = 52.5, radius, r = 5.35 cm = 0.0535 m, B = 0.455 T, I = 25.3 mA = 0.0253 A
Torque = N I A B Sin theta
Here, theta = 90 degree
Torque = 52.5 x 0.0253 x 3.14 x 0.0535 x 0.0535 x 0.455
Torque = 5.43 x 10^-3 Nm