The answer is D. <span> There would be a decrease in the population of marine organisms.
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Remember Newton's second law: F=ma
to get the force in newtons, mass should be in kg and acceleration in m/s^2
conveniently, we don't need to convert units
we just need to multiply the two to get the force
65* 0.3 = 19.5 kg m/s^2 or N
if significant digit is an issue, the least number if sig figs is 1 so the answer would be 20 N
Answer:
Explanation:
Let c be the circumference and r be the radius
c = 2πr , r = c / 2π , area A = π r² = π (c/2π )² = (1/4π) x c²
flux (ψ) = BA = 1 X 1/4π X c²
dψ/dt = 1/4π x 2c dc/dt =1/2π x c x dc/dt
at t = 8 s
c = 161 - 13 x 8 = 57 cm , dc/dt = 13 cm/s
e = dψ/dt = (1 / 2π )x 57 x 13 x 10⁻⁴ = 118 x 10⁻⁴ V.
Yes potential increases while kinetic decreases
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
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CQ
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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?
