<span>The answer is (2) 0.50 Hz. The frequency (f) of oscillation is the number of oscillations (n) per time (t) in seconds: f = n/t. A duck floating on a lake oscillates up and down 5.0 times (n = 5.0) during a 10.-second interval (t = 10.0 s). So, the frequency of duck's oscillations is: f = 5.0/10.0 s = 0.50 1/s = 0.50 Hz.Hope I helped! :) Cheers!</span>
Answer:

Explanation:
given,
F = 14.1 i + 0 j + 5.1 k
displacement = 6 m
Assuming block is moving in x- direction
we know,
dW = F dx


![W = F[x]_0^6](https://tex.z-dn.net/?f=W%20%3D%20F%5Bx%5D_0%5E6)


hence, work done by the force is equal to 
Answer:
455,000 Pa
Explanation:
PV = nRT
If n is constant:
PV / T = PV / T
(101,325 Pa) (718 mL) / (273 K) = P (175 mL) / (26 + 273) K
P = 455,000 Pa
Answer:
Decreases
Explanation:
"Effort" usually refers to the applied force. An inclined plane decreases the force required while increasing the distance that the force is exerted over. So even though there's less force needed, the amount of work stays the same.
Answer:
A. Ein = 8.05*10^-4 V/m
B. Clockwise sense
Explanation:
A. the magnitude of the electric field induced in the ring is obtaind by using the following formula:
(1)
Ein: induced electric field
ds: differential of a path of the ring
ФB: magnetic flux in the ring
The Ein vector is parallel to ds in the complete ring. Furthermore, the area of the ring is constant, hence, you have in the equation (1):
(2)
dB/dt = -0.280T/s (it is decreasing)
A: area of the ring = π(r/2)^2= (π/4) r^2
r: radius of the ring = 4.60/2 = 2.30 cm
Then, you replace the values of all variables in the equation (2):

hence, the induced electric field is 8.05*10^-4 V/m
B. The induced current in the ring produced a magnetic field that is opposite to the magnetic field of the magnet. The, in this case you have that the induced current is in a clockwise sense.