Answer:
539N
Explanation:
Parameters given:
Mass of person = 66kg
Mass of parachute = 4kg
Downward acceleration = 2.1 m/s²
The total force acting on the system is the sum of the force of gravity and the force due to air acting opposite gravity:
F = mg - F(a)
Where g = acceleration due to gravity
Force is given as:
F = ma
=> ma = mg - F(a)
=> F(a) = mg - ma
F(a) = m(g - a)
m is the mass of the entire sustem
m = 66 + 4 = 70kg
=> F(a) = 70(9.8 - 2.1)
F(a) = 70 * 7.7
F(a) = 539N
The upward force on the open parachute from the air is 539N
Responder:
Explicación:
Para que podamos calcular el tiempo que le tomará al esquiador permanecer en el aire, encontraremos el tiempo de vuelo como se muestra;
T = 2Usin theta / 2
theta = 90 grados
U = 25 m / s
T = 25sin90 / 2 (9,8)
T = 25 / 19,62
T = 1,27 segundos
Por lo tanto, los cielos usarán 1.27 segundos en el aire.
La distancia horizontal es el rango;
Rango R = U√2H / g
R = 25√2 (80) /9,8
R = 25√160 / 9,8
R = 25 * √16,326
R = 25 * 4.04
R = 101,02 m
Por tanto, la distancia horizontal recorrida por el esquiador es 101,02 m
Answer:
The value is 
Explanation:
From the question we are told that
The distance of separation is 
The current on the one wire is 
The current on the second wire is 
Generally the magnitude of the field exerted between the current carrying wire is
Here 
is the magnetic field due to the first wire which is mathematically represented as
Here 
is the distance to the half way point of the separation and the value is
is the magnetic field due to the first wire which is mathematically represented as
Here 
is the distance to the half way point of the separation and the value is
This means that 
So
=> 
=> 
=>
<span>ADP has 2 phosphate groups, and when another phosphate group is added it becomes ATP. </span><span />
Answer:
Explanation:
Given
mass of sled =26 kg
coefficient of static friction 
coefficient of kinetic friction 
In order to move sled from rest we need to provide a force greater than static friction which is given by
After Moving Sled kinetic friction comes in to play which is less than static friction
therefore minimum force to keep moving sledge at constant velocity is 18.34 N
