Answer:
346.66 Hz
Explanation:
= Length of string which is unfingered = l
= Length of string which is vibrate when fingered = 
= Unfingered frequency = 260 Hz
= Fingered frequency
Frequency is inversely proportional to length

So,

The frequency of the fingered string is 346.66 Hz
Answer:
Δ v = 125 m/s
Explanation:
given,
mass of space craft = 435 Kg
thrust = 0.09 N
time = 1 week
= 7 x 24 x 60 x 60 s
change in speed of craft = ?
Assuming no external force is exerted on the space craft
now,



a = 2.068 x 10⁻⁴ m/s²
using equation of motion
Δ v = a t
Δ v = 2.068 x 10⁻⁴ x 7 x 24 x 60 x 60
Δ v = 125 m/s
The new acceleration is 
Explanation:
We can answer this problem by applying Newton's second law, which states that:

where
F is the net force on an object
m is the mass of the object
a is its acceleration
The equation can be rewritten as

In this problem, the initial acceleration is

Later:
- The net force is tripled: 
- The mass is halved: 
Therefore, the new acceleration is:

which means that the new acceleration is 6 times the original acceleration, therefore

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The total work done on the car is 784Joule.
<h3>What's the acceleration of the car?</h3>
- As per Newton's equation of motion, V= U+at
- U= initial velocity= 0 m/s
V= vinal velocity= 20m/s
t= time = 10s
a= acceleration
=> a= 20/10= 2m/s²
<h3>What's the distance covered by the car in 10 seconds?</h3>
- As per Newton's equation of motion,
V²-U² = 2aS
- S= distance covered by the car
- So, 20²-0=2×2×S=4S
=> 400= 4S
=> S= 400/4= 100m
<h3>What's the work done on the car due to frictional force?</h3>
Work done by frictional force= frictional force × distance
= (0.2×4×9.8)×100
= 784Joule
Thus, we can conclude that the work done on the car is 784Joule.
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