Answer:
The normal force N1 exerted by the floor is
The normal force N2 exerted by the wall is
The frictional force exerted by the wall is
Explanation:
From the question we are told that
The length of the ladder is
The weight of the ladder is
The distance of the ladder position on the wall from the floor is
The mass of the person is
Applying Pythagoras theorem
The length of the position the ladder on the ground from the base of the wall is
substituting values
In order the for the ladder not to shift from the ground the sum of the moment about the position of the ladder on the ground must be equal to zero this is mathematically represented as
Now the force exerted by the floor on the ladder is mathematically represented as
substituting values
Now the horizontal forces acting on the ladder are and they are in opposite direction so
Answer:
The velocity is 19.39 m/s
Solution:
As per the question:
Mass, m = 75 kg
Radius, R = 19.2 m
Now,
When the mass is at the top position in the loop, then the necessary centrifugal force is to keep the mass on the path is provided by the gravitational force acting downwards.
where
v = velocity
g = acceleration due to gravity
Answer:
The plane is 2353.7 mi from the starting position.
Explanation:
Please, see the attached figure for a graphic representation of the problem.
We have 2 displacement vectors "a" and "b" and a vector "c" that is the sum of vectors "a" plus "b" (c = a + b). The module of "c" will be the distance of the plane from the starting point.
vector a = (xa, ya)
vector b = (xb, yb)
where “xa” and “xb” are the horizontal components of the vectors “a” and “b” respectively and “ya” and “yb” are the vertical components of each vector.
Then, the vector c = a + b will be:
c = (xa + xb, ya + yb)
The module of a vector is calculated using the following expression for a vector “v”:
module of v =
Then, the module of c will be:
module of c = = distance from starting point
Then, we have to find the components of vectors “a” and “b”
The distance traveled during the first 1.5 hours of the trip is the module of the vector “a”. Then:
module of a = = distance traveled during the first 1.5 hours.
The distance can be calculated using the equation of the position of an object moving in a straight line at constant speed:
x = x0 + v * t
where
x = position at time t
x0 = initial position
v = speed
t = time
Considering x0 as the starting point (x0 = 0)
x = 675 mi/h * 1.5 h = 1012.5 mi
Then:
module of a = = 1012. 5 mi
Since the plane moves only on the horizontal (see figure), the "y" component of the vector, "ya", will be 0.
Then:
(1012.5 mi)² = xa²
xa = 1012. 5mi
a = (1012.5 mi, 0)
In the same way, we have fo find the components of the vector “b”. The module of “b” will be the distance traveled during this part of the flight:
module of b = = x = x0 + v * t
Considering x0 as the point at which the plane turns (x0 = 0)
x = 675 mi / h * 2 h = 1350 mi
Using trigonometry, we can calculate xb and yb (see figure):
sin angle = opposite / hypotenuse
cos angle = adjacent / hypotenuse
In this case:
opposite = yb
adjacent = xb
hypotenuse = module of “b”
Then:
sin 10° = yb / module of “b”
sin 10° * module of “b” = yb
In the same way:
cos 10° * module of “b” = xb
Since module of “b” = 1350 mi
xb = 1329.5 mi
yb = 234.4 mi
b = (1329.5 mi, 234.4 mi)
The vector c = a+b can now be calculated:
c = (xa + xb, ya + yb)
c =(1012.5 mi + 1329.5 mi, 0 mi + 234.4 mi) = (2342 mi, 234.4 mi)
The module of c will be:
module of c = = 2353.7 mi
The plane is 2353.7 mi from the starting position.
Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ = / vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m