A force of 8,480 N is applied to a cart to accelerate it at a rate of 26.5 m/s2. What is the mass of the cart?
1 answer:
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Answer:
so the transverse displacement is 0.089963 m
Explanation:
Given data
equation y(x,t)=Acos(kx−ωt)
speed v = 9.00 m/s
amplitude A = 9.00 × 10^−2 m
wavelength λ = 0.480 m
to find out
the transverse displacement
solution
we know
v = angular frequency / wave number
and
wave number = 2 / λ = 2
/ 0.480 = 13.0899
angular frequency = v k
angular frequency = 9.00 × 13.0899
angular frequency = 117.8097 rad/sec = 118 rad/sec
so
equation y(x,t)=Acos(kx−ωt)
y(x,t)=9.00 × 10^−2 cos(13.0899 x−118t)
when x =0 and and t = 0
maximum y(x,t)= 9.00 × 10^−2 cos(13.0899 (0) − 118 (0))
maximum y(x,t)= 9.00 × 10^−2 m
and when x = x = 1.59 m and t = 0.150 s
y(x,t)=9.00 × 10^−2 cos(13.0899 (1.59) −118(0.150) )
y(x,t)=9.00 × 10^−2 × (0.99959)
y(x,t) = 0.089963 m
so the transverse displacement is 0.089963 m
Answer:
the correct one is 2. the equipotential lines must be closer together where the field has more intensity
Explanation:
The equipotential line concept is a line or surface where a test charge can move without doing work, therefore the potential in this line is constant and they are perpendicular to the electric field lines.
In this exercise we have a charge and a series of equipotential lines, if this is a point charge the lines are circles around the charge, where the potential is given by
V = k q / r
also the electric field and the electuary potential are related
E =
therefore the equipotential lines must be closer together where the field has more intensity
When checking the answers, the correct one is 2