Answer:
Tension = 0.012 N
Explanation:
If the black widow spider is hanging vertically motionless from the ceiling above. Then, the weight of the spider must be balancing the tension in the spider web. Therefore,
Tension = Weight
Tension = mg
where,
m = mass of spider = 1.27 g = 0.00127 kg
g = acceleration due to gravity = 9.8 m/s²
Therefore,
Tension = (0.00127 kg)(9.8 m/s²)
<u>Tension = 0.012 N</u>
Answer:
a. The thickness of the wire is 2.5 mm.
b. The wire is 0.25 cm thick.
Explanation:
Number of turns of the wire = 10
The length of total turns = 25 mm
a. The thickness of the wire can be determined by;
thickness of the wire = 
= 
= 2.5 mm
Therefore, the wire is 2.5 mm thick.
b. To determine the thickness of the wire in centimetre;
10 mm = 1 cm
So that,
2.5 mm = x
x = 
= 0.25 cm
The wire is 0.25 cm thick.
Answer: d constint speed
Explanation: im 15 and no the answer
Answer:
1.19 m/s²
Explanation:
The frequency of the wave generated in the string in the first experiment is f = n/2l√T/μ were T = tension in string = mg were m = 1.30 kg weight = 1300 g , μ = mass per unit length of string = 1.01 g/m. l = length of string to pulley = l₀/2 were l₀ = lent of string. Since f is the second harmonic, n = 2, so
f = 2/2(l₀/2)√mg/μ = 2(√mg/μ)/l₀ (1)
Also, for the second experiment, the period of the wave in the string is T = 2π√l₀/g. From (1) l₀ = 2(√mg/μ)/f and from (2) l₀ = T²g/4π²
Equating (1) and (2) we ave
2(√mg/μ)/f = T²g/4π²
Making g subject of the formula
g = 2π√(2√(m/μ)/f)/T
The period T = 316 s/100 = 3.16 s
Substituting the other values into , we have
g = 2π√(2√(1300 g/1.01 g/m)/200 Hz)/3.16
g = 2π√(2 × 35.877/200 Hz)/3.16
g = 2π√(71.753/200 Hz)/3.16
g = 2π√(0.358)/3.16
g = 2π × 0.599/3.16
g = 1.19 m/s²
Answer: Resting Membrane Potential
Explanation:
The <u>resting membrane potential</u> refers to the difference in voltage between the inside and outside of the cell membrane when the cell is at physiological rest. It should be noted that <u>the cell membrane is a selective semipermeable barrier, which only allows the transit through it of certain molecules and prevents the transit of others.
</u>
This selectivity causes an uneven distribution of charged particles (ions), as the membrane only accepts some types of ions.
Now, in the case of neurons, which are electrically excitable nerve cells; the transport of electrical signals is due to these changes in the permeability and asymmetric distribution of ions (mainly sodium and potassium) when the neuron is not excited (at rest).
