A ray diagram is shown.
2 answers:
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Newton's second law states that the resultant of the forces applied to an object is equal to the product between the object's mass and its acceleration:
where in our problem, m is the mass the (child+cart) and a is the acceleration of the system.
We are only concerned about what it happens on the horizontal axis, so there are two forces acting on the cart+child system: the force F of the man pushing it, and the frictional force
acting in the opposite direction. So Newton's second law can be rewritten as
or
since the frictional force is 15 N and we want to achieve an acceleration of
, we can substitute these values to find what is the force the man needs:
where in our problem, m is the mass the (child+cart) and a is the acceleration of the system.
We are only concerned about what it happens on the horizontal axis, so there are two forces acting on the cart+child system: the force F of the man pushing it, and the frictional force
or
since the frictional force is 15 N and we want to achieve an acceleration of
Explanation:
Distance travelled = Area under the line
= ut + ½ (v-u)t
Acceleration (a) = (v-u)/t and so (v-u) = at
Therefore,
Distance travelled (s) = ut + ½ (v-u)t = ut + ½ (at)t = ut + ½ at²
Thus,proved.
Answer:
The minimum coefficient of friction is 0.544
Solution:
As per the question:
Radius of the curve, R = 48 m
Speed of the car, v = 16 m/s
To calculate the minimum coefficient of static friction:
The centrifugal force on the box is in the outward direction and is given by:
where
= coefficient of static friction
The net force on the box is zero, since, the box is stationary and is given by:
Answer:Orbital period =21.22hrs
Explanation:
given that
mass of earth M = 5.97 x 10^24 kg
radius of a satellite's orbit, R= earth's radius + height of the satellite
6.38X 10^6 + 3.25 X10^7 m =3.89 X 10^7m
Speed of satellite, v=
where G = 6.673 x 10-11 N m2/kg2
V= \sqrt (6.673x10^-11 x 5.97x10^ 24)/(3.89 X 10^ 7m)
V =10,241082.2
v= 3,200.2m/s
a) Orbital period
=
V=
T= 2 r/ V
= 2 X 3.142 X 3.89 X 10^7m/ 3,200.2m/s
=76,385.1 s
60 sec= 1min
60mins = 1hr
76,385.1s =hr
76,385.1/3600=21.22hrs