Answer:
The magnitude of the rate of change of the child's momentum is 794.11 N.
Explanation:
Given that,
Mass of child = 27 kg
Speed of child in horizontal = 10 m/s
Length = 3.40 m
There is a rate of change of the perpendicular component of momentum.
Centripetal force acts always towards the center.
We need to calculate the magnitude of the rate of change of the child's momentum
Using formula of momentum
Put the value into the formula
Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.
Answer:
<em>0.45 mm</em>
Explanation:
The complete question is
a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?
A) 0.45 mm
B) 0.63 mm
C.) 0.68 mm
D) 0.91 mm
Current in the fuse is 1.0 A
Current density of the fuse when it melts is 620 A/cm^2
Area of the wire in the fuse = I/ρ
Where I is the current through the fuse
ρ is the current density of the fuse
Area = 1/620 = 1.613 x 10^-3 cm^2
We know that 10000 cm^2 = 1 m^2, therefore,
1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2
Recall that this area of this wire is gotten as
A = 
where d is the diameter of the wire
1.613 x 10^-7 = 
6.448 x 10^-7 = 3.142 x 
=
d = 4.5 x 10^-4 m = <em>0.45 mm</em>
Answer:
C. Both reach the bottom at the same time.
Explanation:
For a rolling object down an inclined plane , the acceleration is given below
a = g sinθ / (1 + k² / r² )
θ is angle of inclination , k is radius of gyration , r is radius of the cylinder
For cylindrical object
k² / r² = 1/2
acceleration = g sinθ /( 1 + 1/2 )
= 2 g sinθ / 3
Since it does not depend upon either mass or radius , acceleration of both the cylinder will be equal . Hence they will reach the bottom simultaneously.
Answer:
<h2>a)
Nathan's acceleration is 5 m/s²
</h2><h2>b)
Nathan's displacement during this time interval is 15.625 m</h2><h2>
c) Nathan's average velocity during this time interval is 6.25 m/s</h2>
Explanation:
a) We have equation of motion v = u + at
Initial velocity, u = 0 m/s
Final velocity, v = 12.5 m/s
Time, t = 2.5 s
Substituting
v = u + at
1.25 = 0 + a x 2.5
a = 5 m/s²
Nathan's acceleration is 5 m/s²
b) We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 5 m/s²
Time, t = 2.5 s
Substituting
s = ut + 0.5 at²
s = 0 x 2.5 + 0.5 x 5 x 2.5²
s = 15.625 m
Nathan's displacement during this time interval is 15.625 m
c) Displacement = 15.625 m
Time = 2.5 s
We have
Displacement = Time x Average velocity
15.625 = 2.5 x Average velocity
Average velocity = 6.25 m/s
Nathan's average velocity during this time interval is 6.25 m/s
Answer:
The correct answer is A The distance is greater in the first hour because her speed is faster.
Explanation:
During the first hour, Anna is driving at a speed of 50 km/h. During the second hour, she is only driving at a speed of 30 km/h. The faster she goes, the farther she will go.
Hope this helps,
♥<em>A.W.E.</em><u><em>S.W.A.N.</em></u>♥
