Answer:
a) 4.31 m/s²
b) 215.5 m
Explanation:
a) According to Newton's first law of motion
The net force applied to particular mass produced acceleration, a, according to
F = ma
F = 140 N
m = 32.5 kg
a = ?
140 = 32.5 × a
a = 140/32.5 = 4.31 m/s²
b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s
u = initial velocity of the probe = 0 m/s (since it was initially at rest)
a = 4.31 m/s²
t = 10 s
s = distance travelled = ?
s = ut + at²/2
s = 0 + (4.31×10²)/2 = 215.5 m
Answer:
<em>His angular velocity will increase.</em>
Explanation:
According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.
The angular momentum of a system = 
'ω'
where
' is the initial rotational inertia
ω' is the initial angular velocity
the rotational inertia = 
where m is the mass of the system
and r' is the initial radius of rotation
Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.
we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to .
From
'ω' = ω
since is now reduced, ω will be greater than ω'
therefore, the angular velocity increases.
Answer:
40 N
Explanation:
F=ma where F is the applied force, m is the mass of object and a is the acceleration.
Since there is no friction, substituting 20 Kg for m and 2 m/s squared for a then we obtain
F=20*2=40 N
Answer:
The difference in the decibel corresponses to a constant difference in the loudness perceived.
The refore the sound intensity from the orchestra is like 100 times that of the violin.
Explanation:
Answer: T= 715 N
Explanation:
The only external force (neglecting gravity) acting on the swinging mass, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:
T = mv² / r
At the moment that the mass be released, it wil continue moving in a straight line at the same tangential speed that it had just an instant before, which is the same speed included in the centripetal force expression.
So the kinetic energy will be the following:
K = 1/2 m v² = 15. 0 J
Solving for v², and replacing in the expression for T:
T = 1.9 Kg (3.97)² m²/s² / 0.042 m = 715 N
