You pull downward with a force of 27.3 N on a rope that passes over a disk-shaped pulley of mass 1.43 kg and radius 0.0792 m. Th
e other end of the rope is attached to a 0.700-kg mass. Calculate the tension in the rope on both sides of the pulley. Enter tension for the part of the rope that you are pulling on first. Then enter the tension for the part of the rope with the mass.
1 answer:
Answer:
The tension in the left side string = 17.21 N
The tension in the right side string = F = 27.3 N
Explanation:
Given that
F= 27.3 N
M= 1.43 kg ,r= 0.0792 m
Moment of inertia of disk ,I = 0.5 m r²
I = 0.5 x 1.43 x 0.0792² = 0.0044 kg.m²
m= 0.7 kg
Lets take linear acceleration of system is a m/s²
Lets take tension in left side string = T
From Newtons law
T- mg = ma
T- 0.7 x 10 = 0.7 a ----------1
(F - T) r = I α
α = Angular acceleration of disk
a = α r
(F - T) r = I α
(F - T) r² = I a
( 27.3 - T) x 0.0792² = 0.0044 a --------2
Form equation 1 and 2
a= 1.42 T - 10 m/s²
a = 1.42 ( 27.3 - T) m/s²
1.42 T - 10 = 38.9 - 1.42 T
T=17.21 N
The tension in the right side string = F = 27.3 N
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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:
f = 2 Hz
Explanation:
The frequency of a wave is defined as the no. of waves passing per unit of time. Therefore, the frequency of a wave can be calculated by the following formula:
where,
f = frequency of the wave = ?
t = time passed = 1 s
n = no. of waves passing in time t = 2
Therefore,
<u>f = 2 Hz</u>