Answer:
c. about 1/10 as great.
Explanation:
While jumping form a certain height when we bend our knees upon reaching the ground such that the time taken to come to complete rest is increased by 10 times then the impact force gets reduced to one-tenth of the initial value when we would not do so.
This is in accordance with the Newton's second law of motion which states that the rate of change in velocity is directly proportional to the force applied on the body.
Mathematically:
since mass is constant
when 
then,
the body will experience the tenth part of the maximum force.
where:
represents the rate of change in dependent quantity with respect to time
momentum
mass of the person jumping
velocity of the body while hitting the ground.
Answer:
Angular speed ω=3771.4 rad/min
Revolution=5921 rpm
Explanation:
Given data
To find
Angular speed ω
Revolution per minute N
Solution
First we need to convert the speed of truck to inches per mile
as
1 mile=63360 inches
1 hour=60 minutes
so
Now to solve for angular speed ω by substituting the speed v and radius r in below equation
To solve for N(revolutions per minute) by substituting the angular speed ω in the following equation
Answer:
d = 1.954 Km
Explanation:
given,
total distance, D = 2.5 Km
in stretch A to B =
speed = 99 Km/h = 99 x 0.278 = 27.22 m/s time =t
in stretch B to C
time = 3.4 s
In stretch C to D
speed = 48 Km/h = 48 x 0.278 = 13.34 m/s time =t
we know,
distance = speed x time
distance of BC
using equation of motion
v = u + a t
27.22 = 13.34 - a x 3.4
a = 4.08 m/s²
uniform deceleration is equal to 4.08 m/s²
distance traveled in BC
s = 68.94 m
3000 = 27.5 t + 68.94 + 13.33 t
40.83 t = 2931.06
t = 71.79 s
distance travel in AB
distance = s x t
d = 27.22 x 71.79
d = 1954 m
d = 1.954 Km
distance between A and B is equal to 1.954 Km.
Answer:
Thermometer
Explanation:
Hence the name Thermo meaning heat.
Answer:
(a) 3.82 x 10⁷ m/s
(b) 4.5 MV/m
Explanation:
(a)
ΔV = change in the electric potential as the proton moves = 7.60 x 10⁶ Volts
q = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C
v = speed gained by the proton
m = mass of proton = 1.67 x 10⁻²⁷ kg
Using conservation of energy
Kinetic energy gained by proton = Electric potential energy
(0.5) m v² = q ΔV
inserting the values
(0.5) (1.67 x 10⁻²⁷) v² = (1.6 x 10⁻¹⁹) (7.60 x 10⁶)
v = 3.82 x 10⁷ m/s
(b)
d = distance over which the potential change = 1.70 m
Electric field is given as
E = ΔV/d
E = 7.60 x 10⁶/1.70
E = 4.5 x 10⁶ V/m
E = 4.5 MV/m
