Answer:
4.7 m³
Explanation:
We'll use the gas law P1 • V1 / T1 = P2 • V2 / T2
* Givens :
P1 = 101 kPa , V1 = 2 m³ , T1 = 300.15 K , P2 = 40 kPa , T2 = 283.15 K
( We must always convert the temperature unit to Kelvin "K")
* What we want to find :
V2 = ?
* Solution :
101 × 2 / 300.15 = 40 × V2 / 283.15
V2 × 40 / 283.15 ≈ 0.67
V2 = 0.67 × 283.15 / 40
V2 ≈ 4.7 m³
The forces that make a passenger speed up, slow down, or
turn a curve are the same forces that have the same effect
on the driver and anybody else in the car.
-- Speeding up . . .
the back of the seat
friction between the car seat and the seat of your pants
-- Slowing down . . .
the seat belt
friction between the car seat and the seat of your pants
-- Turning away from a straight line . . .
the seat belt
friction between the car seat and the seat of your pants
the door, or whatever or whomever you're leaning against
Complete question:
Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength at the midpoint between the two rings ?
Answer:
The electric field strength at the mid-point between the two rings is zero.
Explanation:
Given;
diameter of each ring, d = 10 cm = 0.1 m
distance between the rings, r = 21.0 cm = 0.21 m
charge of each ring, q = 40 nC = 40 x 10⁻⁹ C
let the midpoint between the two rings = x
The electric field strength at the midpoint between the two rings is given as;

Therefore, the electric field strength at the mid-point between the two rings is zero.
Answer:
120 m
Explanation:
Given:
wavelength 'λ' = 2.4m
pulse width 'τ'= 100T ('T' is the time of one oscillation)
The below inequality express the range of distances to an object that radar can detect
τc/2 < x < Tc/2 ---->eq(1)
Where, τc/2 is the shortest distance
First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'
f = c/λ (c= speed of light i.e 3 x
m/s)
f= 3 x
/ 2.4
f=1.25 x
hz.
As, T= 1/f
time of one oscillation T= 1/1.25 x
T= 8 x
s
It was given that pulse width 'τ'= 100T
τ= 100 x 8 x
=> 800 x
s
From eq(1), we can conclude that the shortest distance to an object that this radar can detect:
= τc/2 => (800 x
x 3 x
)/2
=120m