<em>speed</em><em> </em><em>=</em><em> </em><em>wav</em><em>elength</em><em> </em><em>×</em><em> </em><em>freq</em><em>uency</em>
<em>but</em><em> </em><em>the</em><em> </em><em>freq</em><em>uency</em><em> </em><em>was</em><em> </em><em>giv</em><em>en</em><em> </em><em>to</em><em> </em><em>as</em>
<em>1</em><em>0</em><em>k</em><em>h</em><em>z</em><em> </em><em>,</em><em> </em>
<em>whic</em><em>h</em><em> </em><em>mea</em><em>ns</em>
<em></em>
<em>=</em><em> </em><em>1000</em>
<em>so</em><em> </em><em>no</em><em>w</em><em> </em><em>frequ</em><em>ency</em><em> </em><em>=</em><em> </em><em>1000</em><em>h</em><em>z</em>
<em>sub</em><em>stitute</em><em> </em><em>it</em><em> </em><em>in</em><em>to</em><em> the</em><em> equation</em>
<em>speed</em><em> </em><em>=</em><em> </em><em>1000</em><em>×</em><em>2</em>
<em>speed</em><em> </em><em>=</em><em> </em><em>2</em><em>0</em><em>0</em><em>0</em><em>m</em><em>s</em><em>^</em><em>-</em><em>1</em>
Answer:
Approximately 0.00760 m (that's 7.60 mm.)
Explanation:
Refer to the first diagram attached. For a double-slit diffraction, the angle (angular separation) between the m-th maximum and the central maximum satisfies the following equation:
, where
- is the wavelength of the light, and
- is the separation between the two slits.
(Young’s Double Slit Experiment, OpenStaxCollege)
If is much larger than , the value of will be considerably small. The value of could thus be approximately as:
.
For this problem,
- for a first-order maximum.
- ;
- .
Approximate the value of :
.
Separation between the first and central maximum:
Answer:
1. Which limiting factor(s) in this lab simulation are biotic?
The limiting factors in this lab that are biotic are predators and food.
2. Which limiting factor(s) in this lab simulation are abiotic?
The limiting factor in this lab that is abiotic is pollution.
3. Which limiting factor impacted the cricket frog population the most? Use evidence to support your answer.
The limiting factor that impacted the cricket frog population the most is the predators because the population went to 90K and then all the way down to 15K.
4. Which limiting factor impacted the cricket frog population the least? Use evidence to support your answer.
The limiting factor that impacted the cricket frog population the least is pollution because it stayed the same then went down to 35K.
5. Mosquitoes can carry and transmit disease to animals and humans. Explain how the cricket frog plays an important role in limiting the spread of mosquito-borne illnesses like West Nile virus and malaria.
The cricket frogs play an important role in limiting the spread of mosquito borne illnesses because they consume the mosquitos and decrease the population of contaminated mosquitos.
6. Predict the long-term effects of these limiting factors on the cricket frog population in the pond ecosystem.
Eventually the cricket frog population may die out because of all the negative factors. If there are too many mosquitoes, then the frog population can begin to die out. With pollution, the frogs do not have a safe and clean environment to leave in and can get infected, sick, and can die after a while. Crocodiles need to eat, so they are also bringing down the population.
Answer:
(a)7.5 rad/s2
(b)1.83s
(c)0.03 kgm2
(d)At the outer edge
Explanation:
The moments of inertia of the record table can be calculated as:
(a) If he is pulling with a constant linear acceleration of a= 1.2 m/s2, then the constant angular acceleration is
(b)The time it takes for the child to pull a distance of s = 2m given a = 1.2 m/s2
Then the angular speed the turntable would have achieved by that time is
(c) By the law of conservation in angular momentum:
where I1 is the initial moment of the turn table before spaghetti drop, and I2 is after.
(d) For the same force, the child could generate different amount of torque, depending on where he's pressing his thumb. If it's near the the rotational axis, the moment arm is very small, or not at all, making the torque small. If it's at the edge, then the moment arm is large, making greater torque, so less work.
Answer:
t = 5.4 s
Explanation:
from the question we are given :
height (s) = 20 m
mass of paint (Mp) = 4 kg
mass of nails (Mn) = 3 kg
acceleration due to gravity (g) = 9.8 m/s^{2}
- The net force accelerating the can of paint should be equal to the difference in weight of the can of paint and the can of nails.
weight of nails = mass of nails x g = 3 x 9.8 = 29.4 N
weight of paint = mass nails x g = 4 x 9.8 = 39.2 N
net force = 39.2 - 29.4 = 9.8 N
- net force = total mass x acceleration
9.8 = (3 +4) x a
a = 1.4 m/s^{2}
- from S = Ut + 0.5at^{2} we can get the time the carpenter has to catch the nails
where U is the initial velocity and is 0 since the can was initially at
rest
20 = (0 x t) + (0.5 x 1.4 x t^{2})
20 = 0.7 x t^{2}
t^{2} = 28.6
t = 5.4 s