Frequency= speed/ wavelength
=0.5m/s divided 0.1.m
=5.0 Hz
The answer would be letter D.
Answer:
Explanation:
The cannonball goes a horizontal distance of 275 m . It travels a vertical distance of 100 m
Time taken to cover vertical distance = t ,
Initial velocity u = 0
distance s = 100 m
acceleration a = 9.8 m /s²
s = ut + 1/2 g t²
100 = .5 x 9.8 x t²
t = 4.51 s
During this time it travels horizontally also uniformly so
horizontal velocity Vx = horizontal displacement / time
= 275 / 4.51 = 60.97 m /s
Vertical velocity Vy
Vy = u + gt
= 0 + 9.8 x 4.51
= 44.2 m /s
Resultant velocity
V = √ ( 44.2² + 60.97² )
= √ ( 1953.64 + 3717.34 )
= 75.3 m /s
Angle with horizontal Ф
TanФ = Vy / Vx
= 44.2 / 60.97
= .725
Ф = 36⁰ .
Answer:
The answer is "a, c and b"
Explanation:
- Its total block power is equal to the amount of potential energy and kinetic energy.
- Because the original block expansion in all situations will be the same, its potential power in all cases is the same.
- Because the block in the first case has no initial speed, the block has zero film energy.
- For both the second example, it also has the
velocity, but the kinetic energy is higher among the three because its potential and kinetic energy are higher. - While over the last case the kinetic speed is greater and lower than in the first case, the total energy is also higher than the first lower than that of the second.
- The greater the amplitude was its greater the total energy, therefore lower the second, during the first case the higher the amplitude.
For a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s is mathematically given as
F= 618.9 N
<h3>What is the centripetal
force?</h3>
Generally, the equation for the angular speed is mathematically given as
w = v/R
Therefore
w= 4.7/1.8
w= 2.611 rad/s
Where total momentum
Tm= 642.96 + 272.32
Tm= 915.28
and total inertia
Ti= 184 + 246.24
Ti= 430.24
In conclusion, centripetal force
F= mrw^2
F = m*R*w2^2
F = 76*1.8*2.127^2
F= 618.9 N
Read more about mass
brainly.com/question/15959704
CQ
Flag
a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s in the counter clockwise direction when viewed from above a person with mass m=76 kg and velocity v=4.7 m/s runs on a path tangent to the merry go round once at the merry go round the person jumps on and holds on to the rim of the merry go round angular speed of the merry go round after the person jumps on 2.127 rad/s Once the merry go round travels at this new angular speed with what force does the person need to hold on?
