Answer:
The answer is not B or D because they need light co2 and water to make their food so B and D are out. and unless someone is walking during winter to feed it. It will not be A so the answer is C.
Explanation:
Answer:
COMPLETE QUESTION
A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?
Explanation:
Given that,
Extension of spring
x = 0.0208m
Mass attached m = 3.39kg
Additional mass to have a frequency f
Let the additional mass be m
Using Hooke's law
F= kx
Where F = W = mg = 3.39 ×9.81
F = 33.26N
Then,
F = kx
k = F/x
k = 33.26/0.0208
k = 1598.84 N/m
The frequency is given as
f = ½π√k/m
Make m subject of formula
f² = ¼π² •(k/m
4π²f² = k/m
Then, m4π²f² = k
So, m = k/(4π²f²)
So, this is the general formula,
Then let use the frequency above
f = 3Hz
m = 1598.84/(4×π²×3²)
m = 4.5 kg
Answer:
when the mug is heated thus its temperature rises increasing the kinetic energy of the molecules , the oscillations around the rest position in the mug(solid) increases which increase the spaces between molecules and the mug expands. what cause the cracking is that outside of the mug expands before the inside and the mug cracks
Answer:
The speed the bat is gaining on its prey is 0.03m/s
Explanation:
Given;
speed of the bat, v₀ = 3.7 m/s
frequency of the bat, F₀ = 36 kHz
frequency of the source, Fs = 36.79
This is relative motion between a source of the sound and the observer. The phenomenon is known as Doppler effect.
Apply the following equation to determine the speed of the insect which is the source;
![F_0 = F_s[\frac{v+v_0}{v-v_s} ]\\\\\frac{F_0}{F_s} = [\frac{v+v_0}{v-v_s} ]\\\\\frac{36.79}{36} = \frac{340+3.7}{340-v_s}\\\\1.0219 = \frac{343.7}{340-v_s}\\\\ 340-v_s = \frac{343.7}{1.0219}\\\\340-v_s = 336.33\\\\v_s = 340-336.33\\\\v_s = 3.67 \ m/s](https://tex.z-dn.net/?f=F_0%20%3D%20F_s%5B%5Cfrac%7Bv%2Bv_0%7D%7Bv-v_s%7D%20%5D%5C%5C%5C%5C%5Cfrac%7BF_0%7D%7BF_s%7D%20%3D%20%5B%5Cfrac%7Bv%2Bv_0%7D%7Bv-v_s%7D%20%5D%5C%5C%5C%5C%5Cfrac%7B36.79%7D%7B36%7D%20%3D%20%5Cfrac%7B340%2B3.7%7D%7B340-v_s%7D%5C%5C%5C%5C1.0219%20%3D%20%5Cfrac%7B343.7%7D%7B340-v_s%7D%5C%5C%5C%5C%20%20340-v_s%20%3D%20%5Cfrac%7B343.7%7D%7B1.0219%7D%5C%5C%5C%5C340-v_s%20%3D%20336.33%5C%5C%5C%5Cv_s%20%3D%20340-336.33%5C%5C%5C%5Cv_s%20%3D%203.67%20%5C%20m%2Fs)
The speed the bat is gaining on its prey = 3.7m/s - 3.67m/s = 0.03 m/s
Therefore, the speed the bat is gaining on its prey is 0.03m/s
<span>A van is traveling on a road at a speed of 55 km/h relative to a
stationary observer on the side of the road. A girl sitting near the
driver of the van throws a paper airplane to a boy at the back of the
van with a speed of 2 km/h relative to the girl, the boy, and the van.
The speed of the paper airplane, relative to the same stationary observer
on the side of the road, is (55 - 2) = 53 km/h. No rounding is necessary.</span>
