Answer:
Because 'distance per second' is a velocity, not an acceleration.
Explanation:
Because 'distance per second' is a velocity, not an acceleration. For example, at 1 m/s an object is travelling a distance of 1 metre every second. But a rate of acceleration is a steady increase in velocity. So at 1 m/s^2, an object's velocity is increasing by 1 m/s every second.
Answer:
The acceleration of the collar is 10 m/s²
Explanation:
Given;
mass of the collar, m = 1 kg
applied force on the bar, F = 10 N
The acceleration of the collar can be calculated by applying Newton's second law of motion;
F = ma
where;
F is the applied force
m is mass of the object
a is the acceleration
a = F / m
a = 10 / 1
a = 10 m/s²
Therefore, the acceleration of the collar is 10 m/s²
Answer:
A) 5.2 x 10³ N
B) 8.8 x 10³ N
Explanation:
Part A)
= weight of the craft in downward direction = tension force in the cable when stationary = 7000 N
= Tension force in upward direction
= Drag force in upward direction = 1800 N
Force equation for the motion of craft is given as
-
-
= 0
7000 - 1800 -
= 0
= 5200 N
= 5.2 x 10³ N
Part B)
= weight of the craft in downward direction = tension force in the cable when stationary = 7000 N
= Tension force in upward direction
= Drag force in downward direction = 1800 N
Force equation for the motion of craft is given as
-
-
= 0
- 7000 - 1800 = 0
= 8800 N
= 8.8 x 10³ N
The best choice would be letter C hope this helps
Answer:
a) w = 7.27 * 10^-5 rad/s
b) v1 = 463.1 m/s
c) v1 = 440.433 m/s
Explanation:
Given:-
- The radius of the earth, R = 6.37 * 10 ^6 m
- The time period for 1 revolution T = 24 hrs
Find:
What is the earth's angular speed?
What is the speed of a point on the equator?
What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?
Solution:
- The angular speed w of the earth can be related with the Time period T of the earth revolution by:
w = 2π / T
w = 2π / 24*3600
w = 7.27 * 10^-5 rad/s
- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.
v1 = R*w
v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)
v1 = 463.1 m/s
- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.
π/2 ........... s
x ............ 1/5 s
x = π/2*5 = 18°
- The radius of the earth R' at point where θ = 18° from the equator is:
R' = R*cos(18)
R' = (6.37 * 10 ^6)*cos(18)
R' = 6058230.0088 m
- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.
v2 = R'*w
v2 = (6058230.0088)*(7.27 * 10^-5)
v2 = 440.433 m/s