Answer: A voltmeter must have a high resistance where as an ammeter must have a low resistance.
Explanation:
A voltmeter is a device which is connected in parallel to the component across which voltage needs to be measured. In a parallel circuit voltage drop is same at the nodes. The parallel connection must not offer easier path for current to divert from the main circuit and travel. Thus, a voltmeter must have high resistance.
On the other hand, an ammeter which is used to measure current in the circuit must have low resistance as it is connected in series. It should not offer resistance as it would reduce the actual current and measurement would be inaccurate.
The potential energy= mass times gravity times height. However, we are missing height. Gravity is a constant that is 9.8 on Earth. We then solve for height by dividing 350 by 10 and then 9.8 to get about 3.5.
TLDR: 3.5
Answer:
a) t = 0.0185 s = 18.5 ms
b) T = 874.8 N
Explanation:
a)
First we find the seed of wave:
v = fλ
where,
v = speed of wave
f = frequency = 810 Hz
λ = wavelength = 0.4 m
Therefore,
v = (810 Hz)(0.4 m)
v = 324 m/s
Now,
v = L/t
where,
L = length of wire = 6 m
t = time taken by wave to travel length of wire
Therefore,
324 m/s = 6 m/t
t = (6 m)/(324 m/s)
<u>t = 0.0185 s = 18.5 ms</u>
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b)
From the formula of fundamental frquency, we know that:
Fundamental Frequency = v/2L = (1/2L)(√T/μ)
v = √(T/μ)
where,
T = tension in string
μ = linear mass density of wire = m/L = 0.05 kg/6 m = 8.33 x 10⁻³ k gm⁻¹
Therefore,
324 m/s = √(T/8.33 x 10⁻³ k gm⁻¹)
(324 m/s)² = T/8.33 x 10⁻³ k gm⁻¹
<u>T = 874.8 N</u>
It’s 49% of its original gig hope this helps!
