answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Gennadij
17 days ago
7

10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????

How far will this pulse travel in the same time if the tension is doubled?
Physics
1 answer:
serg [2.5K]17 days ago
4 0

Answer: Tension = 47.8N, Δx = 11.5×10^{-6} m.

              Tension = 95.6N, Δx = 15.4×10^{-5} m

Explanation: The speed of a wave on a string under tension can be determined using the following:

|v| = \sqrt{\frac{F_{T}}{\mu} }

F_{T} denotes tension (N)

μ refers to linear density (kg/m)

Calculating the velocity:

|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }

|v| = \sqrt{0.00874 }

|v| = 0.0935 m/s

Distance a pulse traveled in 1.23ms:

\Delta x = |v|.t

\Delta x = 9.35.10^{-2}*1.23.10^{-3}

Δx = 11.5×10^{-6}

With a tension of 47.8N, the distance a pulse will cover is Δx = 11.5×10^{-6}  m.

When tension is doubled:

|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }

|v| = \sqrt{2.0.00874 }

|v| = \sqrt{0.01568}

|v| = 0.1252 m/s

Distance in the same time:

\Delta x = |v|.t

\Delta x = 12.52.10^{-2}*1.23.10^{-3}

\Delta x = 15.4×10^{-5}

With the increased tension, it moves \Delta x = 15.4×10^{-5} m

You might be interested in
A compact car has a maximum acceleration of 4.0 m/s2 when it carries only the driver and has a total mass of 1200 kg . you may w
inna [2205]
<span>Let F represent the maximum thrust produced by the car's motor. Thus, F = ma = 1300 x 3.0 = 3900 N. After adding the load, F stays the same, leading to the equation F = 1700a, which results in a = F/1700 = 3900/1700 = 2.3 m/s².</span>
3 0
1 month ago
Read 2 more answers
A beaker contain 200mL of water<br> What is its volume in cm3 and m3
Sav [2226]
The volumes are 200cm3 and 0.0002m3
7 0
14 days ago
Read 2 more answers
A rope connects boat A to boat B. Boat A starts from rest and accelerates to a speed of 9.5 m/s in a time t = 47 s. The mass of
ValentinkaMS [2425]

Answer: 339.148N

Explanation:

Given data:

Time (t) = 47s

Initial speed (U) = 0m/s

Final speed (V) = 9.5m/s

Mass of B = 540kg

Frictional force on B = 230N

Since both boats are linked, movement of A causes B to move as well.

What is the acceleration of boat A?

Applying the motion formula:

V = u + at

9.5 = 0 + a * 47

a = 9.5 / 47

a = 0.2021 m/s²

To determine the force necessary to accelerate boat B, as both boats experience the same force:

F = Mass * acceleration

F = 540 * 0.2021 = 109.14N

Given that there is a frictional force of 230N acting on boat B, the overall force (Tension) becomes:

Tension = frictional force + applied force = (109.14 + 230)N = 339.148N

7 0
16 days ago
Most calculators operate on 6.0 V. If, instead of using batteries, you obtain 6.0 V from a transformer plugged into 110-V house
kicyunya [2264]
1/0.0545. The transformation ratio of primary coil turns to secondary coil turns is directly proportional to the voltage transformation occurring. With 6.0 V on the secondary side (output) and 110 V on the primary side (input), the voltage ratio is calculated as 6/110 = 0.0545. This means for each turn in the primary coil, there are 0.0545 turns in the secondary coil.
6 0
11 days ago
A uniformly charged spherical droplet of mercury has electric potential Vbig throughout the droplet. The droplet then breaks int
kicyunya [2264]

Answer:

\frac{V_{big}}{V_{small}} = n^{2/3}

Explanation:

Let the charge on the large droplet be denoted as Q.

When the radius of the droplet is R, the electric potential for the larger droplet can be expressed as:

V_{big} = \frac{KQ}{R}

If it splits into n identical droplets, let the charge of each be "q" and their radius be "r".

Applying volume conservation gives us:

\frac{4}{3}\pi R^3 = n(\frac{4}{3}\pi r^3)

r = \frac{R}{n^{1/3}}

Now, the potential for the smaller droplets is given as:

V_{small} = \frac{kq}{r}

V_{small} = \frac{K(Q/n)}{\frac{R}{n^{1/3}}}

V_{small} = \frac{1}{n^{2/3}}\frac{KQ}{R}

\frac{V_{big}}{V_{small}} = n^{2/3}

7 0
17 days ago
Other questions:
  • An archer tests various arrowheads by shooting arrows at a pumpkin that is suspended from a tree branch by a rope, as shown to t
    14·1 answer
  • A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the
    5·2 answers
  • Kayla, a fitness trainer, develops an exercise that involves pulling a heavy crate across a rough surface. The exercise involves
    6·1 answer
  • A satellite revolves around a planet at an altitude equal to the radius of the planet. the force of gravitational interaction be
    14·2 answers
  • A 7.0-kilogram cart, A, and a 3.0-kilogram cart, B, are initially held together at rest on a horizontal, frictionless surface. W
    7·1 answer
  • Estimate the monthly cost of using a 700-W refrigerator that runs for 10 h a day if the cost per kWh is $0.20.
    12·1 answer
  • Vinny is on a motorcycle at rest, 200 m away from a ramp that jumps over a gully. Calculate the minimum constant acceleration Vi
    9·1 answer
  • Alana is skateboarding at 19 km/h and throws a tennis ball at 11 km/h to her friend Oliver who is behind her leaning against a w
    5·1 answer
  • A motorist enters a freeway at 45 km/h and accelerates uniformly to 99 km/h. From the odometer in the car, the motorist knows th
    9·1 answer
  • The distance between two slits is 1.50 *10-5 m. A beam of coherent light of wavelength 600 nm illuminates these slits, and the d
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!