Answer with Explanation:
Newton's laws are applicable for inertial frames of reference which is a frame which is not accelerating when seen from the observer standing on earth.
For the person as he presses the brakes his frame is a decelerating frame of reference hence he cannot apply the newtons laws of motion as they are in their original form but if he analyses the motion he has to apply a correction known as pseudo-force on the object he is analyzing. Pseudo Force has no basis in newton's laws but are a correction that needs to be applied if he wishes to analyse the motion from non inertial frame of reference
While as a person standing on earth outside the car since his frame is an inertial frame of reference he can apply newton's laws of motion without any correction.
Answer:
The resolution of an analog-to-digital converter is 24.41 mV
Explanation:
Resolution of an analog-to-digital = (analogue signal input range)/2ⁿ
where;
n is the number or length of bit, and in this question it is given as 12
Also, the analogue signal input range is 100V
Resolution of an analog-to-digital = 100V/2¹²
2¹² = 4096
Resolution of an analog-to-digital = 100V/4096
Resolution of an analog-to-digital = 0.02441 V = 24.41 mV
Therefore, the resolution of an analog-to-digital converter is 24.41 mV
Explanation:
It is given that,
Mass of the box, m = 100 kg
Left rope makes an angle of 20 degrees with the vertical, and the right rope makes an angle of 40 degrees.
From the attached figure, the x and y component of forces is given by :
Let 
and 
is the resultant in x and y direction.
As the system is balanced the net force acting on it is 0. So,
.............(1)
..................(2)
On solving equation (1) and (2) we get:
(tension on the left rope)
(tension on the right rope)
So, the tension on the right rope is 1063.36 N. Hence, this is the required solution.
Answer:
a = 2.22 [m/s^2]
Explanation:
First we have to convert from kilometers per hour to meters per second
![40 [\frac{km}{h}]*[\frac{1h}{3600s}]*[\frac{1000m}{1km}] = 11.11 [m/s]](https://tex.z-dn.net/?f=40%20%5B%5Cfrac%7Bkm%7D%7Bh%7D%5D%2A%5B%5Cfrac%7B1h%7D%7B3600s%7D%5D%2A%5B%5Cfrac%7B1000m%7D%7B1km%7D%5D%20%3D%2011.11%20%5Bm%2Fs%5D)
We have to use the following kinematics equation:
where:
Vf = final velocity = 11.11 [m/s]
Vi = initial velocity = 0
a = acceleration [m/s^2]
t = time = 5 [s]
The initial speed is taken as zero, as the car starts from zero.
11.11 = 0 + (a*5)
a = 2.22 [m/s^2]
Answer:
5.95 A
Explanation:
From the question
R = ρL/A..................... Equation 1
Where R = resistance of the tungsten wire, ρ = Resistivity of the tungsten wire, L = length, A = cross sectional area.
Given: L = 1.5 m, A = 0.8 mm² = 0.8×10⁻⁶ m, ρ = 5.60×10⁻⁸ Ω.m
Substitute these values into equation 1
R = 1.5(5.60×10⁻⁸)/0.8×10⁻⁶
R = 0.084 Ω.
Finally, using Ohm law,
V = IR
Where V = Voltage, I = current
Make I the subject of the equation
I = V/R............... Equation 2
I = 0.5/0.084
I = 5.95 A
