Newton's second law tells you:
Sum of forces on an object = ma
Here, the forces acting on the bundle are the tension in the string and the force of gravity, these two must combine to yield the acceleration of the bundle.
So we have:
T-mg = ma
or T=m(g+a)
We know m=8.7kg, we need to find a from the information
starting from rest, an accelerating object covers distance according to:
<span>dist = 1/2 at^2 </span>
to cover 1m in 1.8s, we have:
a=2d/t^2 = 2x1/1.8^2 = 0.62 m/s/s
Thus, the tension in the string is:
<span>T = m(g+a)
= 8.7</span>kg(9.8m/s/s+0.62m/s/s)
<span>
<span>T = 90.654 N
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Answer:
25.59 m/s²
Explanation:
Using the formula for the force of static friction:
--- (1)
where;
static friction force
coefficient of static friction
N = normal force
Also, recall that:
F = mass × acceleration
Similarly, N = mg
here, due to min. acceleration of the car;
From equation (1)
However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.
Thus,
where;
and g = 9.8 m/s²
Answer:
0.16joules
Explanation:
Using the relation for The gravitational potential energy
E= Mgh
Where,
E= Potential energy
h = Vertical Height
M = mass
g = Gravitational Field Strength
To find the vertical component of angle of launch Where the angle is 22°
h= sin theta
So E = mghsintheta
= 0.18 x 0.98 x 0.253 sin22
=0.16joules
Explanation:
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