Answer:
maximumforce is F = mg
Explanation:
For this case we must use Newton's second law,
Σ F = m a
bold indicate vectors, so we will write it in its components x and y
X axis
Fₓ = maₓ
Axis y
Fy - W = m a
Now let's examine our case, with indicate that the bird is level, the force of the wings can have a measured angle with respect to the x axis, where the vertical component is responsible for the lift, let's use trigonometry to find the components
Cos θ = Fₓ / F
Fₓ = F cos θ
sin θ = Fy / F
Fy = F sin θ
Let's replace and calculate
F sin θ -w = m a
As the bird indicates that leveling at the same height, so the vertical acceleration is zero (ay = 0)
F sin θ = w = mg
The maximum value of this equation occurs when the sin=1, in this case
F = mg
Answer:
33.68 N
Explanation:
Data
W= 32J
d- 0.95m
F= ?
W=Fd
They are asking for the magnitude which is the force, so you need to solve for force.
F=W/d
= 32J/ 0.95m
= 33.68 N
Answer: a) 456.66 s ; b) 564.3 m
Explanation: The time spend to cover any distance a constant velocity is given by:
v= distance/time so t=distance/v
The slower student time is: t=780m/0.9 m/s= 866.66 s
For the faster students t=780 m/1,9 m/s= 410.52 s
Therefore the time difference is 866.66-410.52= 456.14 s
In order to calculate the distance that faster student should walk
to arrive 5,5 m before that slower student, we consider the follow expressions:
distance =vslower*time1
distance= vfaster*time 2
The time difference is 5.5 m that is equal to 330 s
replacing in the above expression we have
time 1= 627 s
time2 = 297 s
The distance traveled is 564,3 m
Answer:
88.3
Explanation:
Emf in a rotating coil is given by rate of change of flux:
E= dФ/dt=(NABcos∅)/ dt
N: number of turns in the coil= 80
A: area of the coil= 0.25×0.40= 0.1
B: magnetic field strength= 1.1
Ф: angle of rotation= 90- 37= 53
dt= 0.06s
E= (80 × 0.4× 0.25×1.10 × cos53)/0.06= 88.3V
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.
