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Xelga [282]
2 years ago
8

A rope, under a tension of 221 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement

of the rope is given by y = (0.10 m)(sin πx/2) sin 12πt, where x = 0 at one end of the rope, x is in meters, and t is in seconds.
What are:
a. the length of the rope.
b. the speed of the waves on the rope
c. the mass of the rope
d. If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation.
Physics
1 answer:
mamaluj [8]2 years ago
3 0

Answer:

sup qwertyasdfghjk

Explanation:

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Which of the following is Not true about the noble gases
Alexandra [31]
Noble gases are not highly reactive
3 0
3 years ago
If half of the weight of a flatbed truck is supported by its two drive wheels, what is the maximum acceleration it can achieve o
Scilla [17]

Answer:

Maximum acceleration will be equal to 3.43m/sec^2

Explanation:

We have given coefficient of kinetic friction \mu _k=0.7

And coefficient of static friction \mu _s=1

Acceleration due to gravity g=9.8m/sec^2

When truck moves maximum force will be equal to F=\mu _kmg

It is given that half of the weight is supported by its drive wheels

So force required =\frac{\mu _kmg}{2}

From newtons law maximum acceleration will be equal to a=\frac{\frac{\mu _kmg}{2}}{m}=\frac{\mu _kg}{2}=\frac{0.7\times 9.8}{2}=3.43m/sec^2

8 0
3 years ago
In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5.0-m-diameter hollow steel cylinder with their b
lesya692 [45]

Answer:2.55 rad/s

Explanation:

Given

Diameter of ride=5 m

radius(r)=2.5 m

Static friction coefficient range=0.60-1

Here Frictional force will balance weight

And limiting  frictional force is provided by Centripetal force

f=\mu N=\mu m\omega ^2\cdot r

weight of object=mg

Equating two

f=mg

\mu m\omega ^2\cdot r=mg

\omega ^2=\frac{g}{\mu r}

\omega =\sqrt{\frac{g}{\mu r}}

\omega =2.55 rad/sec

8 0
3 years ago
The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .
ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is W=1.20 \cdot 10^{-3}J

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: W_e=q\Delta V

where

q is the charge

\Delta V is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

K_f - K_i = W + W_e = W+q\Delta V

and since the charge starts from rest, K_i = 0, so the formula becomes

K_f = W+q\Delta V

In this problem, we have

W=1.20 \cdot 10^{-3}J is the work done by the external force

q=-6.70 \mu C=-6.7\cdot 10^{-6}C is the charge

K_f = 4.72\cdot 10^{-4}J is the final kinetic energy

Solving the formula for \Delta V, we find

\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V

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3 years ago
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