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blagie [28]
4 years ago
6

How does an octopus or a squid uses Newton's third law of motion to escape from enemies

Physics
1 answer:
Sloan [31]4 years ago
4 0
I think everytime they swim away, the water pushes the enemy back, which makes the octopus go faster. Every action has an equal and opposite reaction. Hope this helps.
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a kid on a playground swing makes 6 complete to-and-from swings each 30 seconds (a) the frequency of the swinging is? (b) the pe
Mariana [72]

The frequency is how many per second:

       (6 swings)/(30 sec) = (6/30) swing/sec = 0.2 per sec = 0.2 Hz .

The period is how long each one takes, or seconds per swing.
It's exactly the flip of frequency.
So we could just take the frequency, flip it, and find   1 / 0.2 ,
but let's do it the long way:

      (30 sec) / (6 swings) = (30/6) sec/swing =  5 seconds  .  

      
4 0
4 years ago
Read 2 more answers
Answer please I need help
madam [21]

Answer:

1) F

2) A

3) B

4) E

5) C

6) D

6 0
4 years ago
You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int
uysha [10]
In a RC-circuit, with the capacitor initially uncharged,  when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
Q(t) = Q_0 (1-e^{-t/\tau})
where t is the time, Q_0 = CV is the maximum charge on the capacitor at voltage V, and \tau = RC is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using R=39.1 \Omega and C= 65.7 mF=65.7\cdot 10^{-3}F, the time constant of the circuit is:
\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s

2) To find the charge on the capacitor at time t=1.95 \tau, we must find before the maximum charge on the capacitor, which is
Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C
And then, the charge at time t=1.95 \tau is equal to
Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be Q_0 = 0.59 C. We can see this also from the previous formule, by using t=\infty:
Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C

4 0
3 years ago
B) Smartphone, iPad, and tablet are also a kind of computer.
valentinak56 [21]

Answer:

Yes

Explanation:

7 0
3 years ago
A scene in a movie has a stuntman falling through a floor onto a bed in the room below. The plan is to have the actor fall on hi
tekilochka [14]

Answer:

The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg

Explanation:

Hi there!

Due to conservation of energy, the potential energy (PE) of the mass at a height of 3.32 m will be transformed into elastic potential energy (EPE) when it falls on the mattress:

PE = EPE

m · g · h = 1/2 k · x²

Where:

m = mass.

g = acceleration due to gravity.

h = height.

k = spring constant.

x = compression distance

The maximum compression distance is 0.1289 m, then, the maximum elastic potential energy will be the following:

EPE =1/2 k · x²

EPE = 1/2 · 65144 N/m · (0.1289 m)² = 541.2 J

Then, using the equation of gravitational potential energy:

PE = m · g · h =  541.2 J

m =  541.2 J/ g · h

m = 541.2 kg · m²/s² / (9.8 m/s² · 3.32 m)

m = 16.6 kg

The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg.

6 0
3 years ago
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