on the globe , the largest continent on earth , asia lies in between eastern and northern hemisphere. it is also known that country china comes in asian continent. hence we can say that china also lies in between eastern and northern hemisphere.
hence the two hemispheres are northern and eastern.
C) The weight of the object on planet A will be less than the weight of the object on planet B.
This happened because: when gravitational pull of a planet is weak, an object will weigh less as a result of less gravity forcing it down to the ground.
Answer:
(a) 98 N
(b) 158 N
(c) 38 N
Explanation:
<h2>
Part (a)</h2>
When the acceleration is 0 m/s², the net force on the mass is 0 N. Therefore, the tension force is equal to the weight force due to Newton's Second Law:
- ∑F_y = T - w = ma_y
- ∑F_y = T - w = m(0 m/s²)
- ∑F_y = T - w = 0
- ∑F_y = T = w
Since the tension in the cable and the weight of the mass are equal to each other, we can solve for the weight force of the mass by using:
- w = mg
- w = (10 kg)(9.8 m/s²)
- w = 98 N
Since T = w, we can say that T = 98 N.
<h2>Part (b)</h2>
Let's set the upwards direction to be positive and the downwards direction to be negative. We can use Newton's Second Law to solve for the tension in the cable if the acceleration is 6 m/s² upward:
- ∑F_y = T - w = ma_y
- ∑F_y = T - mg = m(6 m/s²)
- ∑F_y = T - mg = 6m
Plug the known values into the equation and solve for T.
- T - mg = 6m
- T - (10 kg)(9.8 m/s²) = 6(10 kg)
- T - 98 = 60
- T = 158 N
The tension in the cable if the acceleration is +6 m/s² is 158 N.
<h2>
Part (c)</h2>
The process is the same, but this time acceleration is -6 m/s².
- ∑F_y = T - w = ma_y
- ∑F_y = T - mg = m(-6 m/s²)
- ∑F_y = T - mg = -6m
Plug known values into the equation and solve for T.
- T - mg = -6m
- T - (10 kg)(9.8 m/s²) = -6(10 kg)
- T - 98 = -60
- T = 38 N
The tension in the cable if the acceleration is -6 m/s² is 38 N.
Answer:
With any object at any time the free fall acceleration is 10 m/s2
Explanation:
That is because when you throw a ball into the air it decreases by about 10 m/s each second, on the way down it increases its speed by about 10 m/s each second. So the the anwser is 10 m/s2.
This question is not complete.
The complete question is as follows:
One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?
Explanation:
a. Using the expression;
T = 2π√R/g
where R = radius of the space = diameter/2
R = 800/2 = 400m
g= acceleration due to gravity = 9.8m/s^2
1/T = number of revolutions per second
T = 2π√R/g
T = 2 x 3.14 x √400/9.8
T = 6.28 x 6.39 = 40.13
1/T = 1/40.13 = 0.025 x 60 = 1.5 revolution/minute