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alisha [4.7K]
3 years ago
5

What is a rubens tube

Physics
1 answer:
Free_Kalibri [48]3 years ago
6 0

Answer:

its an antique physics apparatus for demonstrating acoustic standing waves in a tube.

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A 3.0 kg cart moving due east at 4.0 m/s collides with a 6.0 kg cart moving due west. the carts stick together and come to rest
denpristay [2]
The problem about describes a perfectly inelastic collision. We are tasked to find the initial velocity of an object having a mass of 6 kg moving due west. It is given in the problem that after collision the cart sticks together and it stops. Thus, the final mass is the sum of the two cart and the final velocity is zero. For a perfectly inelastic collision,

m1v1-m2v2=vf(m1+m2)
By Substitution,

3(4)-6(v2)=0
6v2=12
v2=2

Therefor, the initial velocity if a 6 kg cart is 2 m/s
3 0
4 years ago
1. What is energy? What can an object with energy do?
kvv77 [185]

Answer:

Kinetic energy is the energy an object has because of its motion. If we want to accelerate an object, then we must apply a force. Applying a force requires us to do work. After work has been done, energy has been transferred to the object, and the object will be moving with a new constant speed.

3 0
3 years ago
7. Electrical forces are extremely what
Reptile [31]

Answer:

it's extremely strong?

8 0
3 years ago
Read 2 more answers
As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labe
spin [16.1K]

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

6 0
3 years ago
An air compressor does 2.5 × 106 joules of work in raising the pressure of a quantity of air while the temperature of the air re
damaskus [11]

Answer:

2.5\cdot 10^6 J

Explanation:

According to the 1st law of thermodynamics, we have:

\Delta U = Q-W

where

\Delta U is the change in internal energy of the system

Q is the heat absorbed by the system

W is the work  done by the system

We also know the change in internal energy of a system only depends on the change in temperature:

\Delta U \propto \Delta T

Since here the temperature of the air remains constant, the change in internal energy is zero:

\Delta T=0 \rightarrow \Delta U = 0

So the first equation becomes

Q=W

The work done by the system here is

W=2.5\cdot 10^6 J

Therefore, the heat added to the system is

Q=W=2.5\cdot 10^6 J

7 0
4 years ago
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