Answer:
1.) Magnitude = 5596 N
2.) Direction = 60 degrees
Explanation: You are given that the breakdown vehicle A is exerting a force of 4000 N at angle 45 degree to the vertical and breakdown vehicle B is exerting a force of 2000 N
Let us resolve the two forces into X and Y component
Sum of the forces in the X - component will be 4000 × cos 45 = 2828.43 N
Sum of the forces in the Y - component will be 2000 + ( 4000 × sin 45 )
= 2000 + 2828.43
= 4828.43 N
The resultant force R will be
R = sqrt ( X^2 + Y^2 )
Substitutes the forces at X component and Y component into the formula
R = sqrt ( 2828.43^2 + 4828.43^2 )
R = sqrt ( 31313752.53 )
R = 5595.87 N
The direction will be
Tan Ø = Y/X
Substitute Y and X into the formula
Tan Ø = 4828.43 / 2828.43
Tan Ø = 1.707106
Ø = tan^-1( 1.707106 )
Ø = 59.64 degree
Therefore, approximately, the magnitude and direction of the resultant force on the truck are 5596 N and 60 degree respectively.
Answer:
a) 2.5 m/s. (In the opposite direction to the direction in which she threw the boot).
b) The centre of mass is still at the starting point for both bodies.
c) It'll take Sally 12 s to reach the shore which is 30 m from her starting point.
Explanation:
Linear momentum is conserved.
(mass of boot) × (velocity of boot) + (mass of sally) × (velocity of Sally) = 0
5×30 + 60 × v = 0
v = (-150/60) = -2.5 m/s. (Minus inicates that motion is in the opposite direction to the direction in which she threw the boot).
b) At time t = 10 s,
Sally has travelled 25 m and the boot has travelled 300 m.
Taking the starting point for both bodies as the origin, and Sally's direction as the positive direction.
Centre of mass = [(60)(25) + (5)(-300)]/(60+5)
= 0 m.
The centre of mass is still at the starting point for both bodies.
c) The shore is 30 m away.
Speed = (Distance)/(time)
Time = (Distance)/(speed) = (30/2.5)
Time = 12 s
Hope this Helps!!!
Answer:
Time period for first satellites 24.46 days and for second satellites 37.67 days
Explanation:
Given :
Distance of first satellites 
m
Distance of second satellites 
m
Distance of charon 
m
Time period of charon 
days
From the kepler's third law,
Square of the time period is proportional to the cube of the semi major axis.
For first satellites,
days
For second satellites,
days
Therefore, time period for first satellites = 24.46 days and for second satellites 37.67 days
Answer:
Given that
V= 0.06 m³
Cv= 2.5 R= 5/2 R
T₁=500 K
P₁=1 bar
Heat addition = 15000 J
We know that heat addition at constant volume process ( rigid vessel ) given as
Q = n Cv ΔT
We know that
P V = n R T
n=PV/RT
n= (100 x 0.06)(500 x 8.314)
n=1.443 mol
So
Q = n Cv ΔT
15000 = 1.433 x 2.5 x 8.314 ( T₂-500)
T₂=1000.12 K
We know that at constant volume process
P₂/P₁=T₂/T₁
P₂/1 = 1000.21/500
P₂= 2 bar
Entropy change given as
Cp-Cv= R
Cp=7/2 R
Now by putting the values
a)ΔS= 20.79 J/K
b)
If the process is adiabatic it means that heat transfer is zero.
So
ΔS= 20.79 J/K
We know that
Process is adiabatic
Answer:
B
Explanation:
Work done can be said to be positive if the applied force has a component to be in the direction of the displacement and when the angle between the applied force and displacement is positive.
In 1 and 2 work done is positive
