The resultant force on the object is
∑ <em>F</em> = 〈0, 8〉 N + 〈6, 0〉 N = 〈6, 8〉 N
which has a magnitude of
<em>F</em> = √((6 N)² + (8 N)²) = √(100 N²) = 10 N
By Newton's second law, the acceleration has magnitude <em>a</em> such that
<em>F</em> = <em>m a</em>
10 N = (2 kg) <em>a</em>
<em>a</em> = (10 N) / (2 kg)
<em>a</em> = 5 m/s²
so the answer is B.
Because its just enough to where its not out f the gravitational pull and not close enough to be pulled back to earth. Hope it helps<span />
Answer:
3.62m/s and 2.83m/s
Explanation:
Apply conservation of momentum
For vertical component,
Pfy = Piy
m* Vof (sin38) - m*Vgf (sin52) = 0
Divide through by m
Vof(sin38) - Vgf(sin52) = 0
Vof(sin38) = Vgf(sin52)
Vof (sin38/sin52) = Vgf
0.7813Vof = Vgf
For horizontal component
Pxf= Pxi
m* Vof (cos38) - m*Vgf (cos52) = m*4.6
Divide through by m
Vof(cos38) + Vgf(cos52) = 4.6
Recall that
0.7813Vof = Vgf
Vof(cos38) + 0.7813 Vof(cos52) = 4.6
0.7880Vof + 0.4810Vof = 4.
1.269Vof = 4.6
Vof = 4.6/1.269
Vof = 3.62m/s
Recall that
0.7813Vof = Vgf
Vgf = 0.7813 * 3.62
Vgf = 2.83m/s
Answer:
v = 18.84 m/s
Explanation:
Given that,
The length of the string, r = 1.5 m (it will act as radius)
The rubber stopper makes 120 complete circles every minute.
Since, 1 minute = 60 seconds
It means, its frequency is 2 circles every second.
Let we need to find the average speed of the rubber stopper. It can be calculated as follows :

d is distance,
and 1/T = f (frequency)

So, the average speed of the rubber stopper is 18.84 m/s.
Answer:
15009
Explanation:
PE = mgh
PE = 61.2(9.81)(10 * 2.50)
PE = 15009.3