This question is incomplete; here is the complete question:
Marco is conducting an experiment. He knows the wave that he is working with has a wavelength of 32.4 cm. If he measures the frequency as 3 hertz, which statement about the wave is accurate?
A. The wave has traveled 32.4 cm in 3 seconds.
B. The wave has traveled 32.4 cm in 9 seconds.
C. The wave has traveled 97.2 cm in 3 seconds.
D. The wave has traveled 97.2 cm in 1 second.
The answer to this question is D. The wave has traveled 97.2 cm in 1 second.
Explanation:
The frequency of a wave, which is in this case 3 hertz, represents the number of waves that go through a point during 1 second. According to this, if the frequency of the wave is 3 hertz this means in 1 second there were 3 waves. Moreover, if you multiply the wavelength (32.4cm) by the frequency (3) you will know the distance the wave traveled in 1 second: 32.4 x 3 = 97.2 cm. This makes option D the correct one as the distance in 1 second was 97.2 cm.


Now


- Lower mass=Higher acceleration
- Lower Force=Lower Acceleration
Option B has lowest mass and highest force hence its correct
Answer:
The minimum coefficient of friction is 0.27.
Explanation:
To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.
Centripetal force is written as

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

with ω denoting the angular velocity, which we are given. With that, the above becomes:

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

and so 45 rev/min = 4.71 rad/s.

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.
Based on the length of the Ethernet cable and the mass, the tension in the cable can be found to be 80 N.
<h3>How much tension is in the cable?</h3>
The tension in the cable can be found as:
= 4 x mass x length x frequency
Solving for the frequency is:
= 1 / (0.800 / 4)
= 1 / 0.20
= 5.0 Hz
The tension is therefore:
= 4 x 0.20 x 4.00 x 5
= 80N
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Answer:
38.6 J
Explanation:
c = molar heat capacity of silver = 25.35 J/(mol °C)
m = given mass of silver = 9.00 g
M = Molar mass of silver = 108 g
n = Number of moles of silver
Number of moles of silver are given as


n = 0.0833
Q = Energy needed to raise the temperature
ΔT = Change in temperature = 18.3 °C
Energy needed to raise the temperature is given as
Q = n c ΔT
Q = (0.0833) (25.35) (18.3)
Q = 38.6 J