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IRISSAK [1]
3 years ago
7

A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what d

istance from the left end of the rod should a 0.6kg mass be hung to balance the rod?
a.48 cm
b. 50 cm
c. 45 cm
d. the rod can not be balanced with this mass. e.42 cm
NO LINKS. ​
Physics
1 answer:
Aleks04 [339]3 years ago
8 0
I’m pretty sure it’s b
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A system of two objects has ΔKtot = 6 J and ΔUint = -5 J. Part A How much work is done by interaction forces? Express your answe
Elina [12.6K]

A) +5 J

B) +1 J

Explanation:

A)

The internal forces (interaction forces) acting on a system do not change the mechanical energy (sum of potential and kinetic energy) of the system.

However, these forces are responsible for converting the energy from one form into another; the work done by these forces is equal to the amount of energy converted from one form into the other.

In this problem, we have:

\Delta U=-5 J is the loss in potential energy of the system

\Delta K=+6 J is the gain in kinetic energy of the system

By looking at these numbers, this means that the internal forces have converted 5 J of energy from potential energy into kinetic energy (while the additional +1 J missing is due to external forces, as explained in part B).

Therefore, the work done by internal forces is

W = +5 J

B)

First of all, we calculate the change in mechanical energy of the system.

The mechanical energy of a system is the sum of its kinetic energy (K) and its potential energy (U):

E=K+U

So, the change in mechanical energy is equal to the sum of the changes of kinetic energy and the changes of potential energy:

\Delta E= \Delta K + \Delta U

In this problem:

\Delta K=+6 J

\Delta U=-5 J

So, the change in mechanical energy is:

\Delta E=+6+(-5)=+1 J

According to the work-energy theorem, the work done by external forces on a system is equal to the change in mechanical energy of the system: therefore in this case, the work done by external forces is

W=\Delta E=+1 J

5 0
4 years ago
Which type of surface would most likely be the best reflector of electromagnetic energy?
Len [333]
<span>light colored and smooth surface would most likely be the best reflector of electromagnetic energy.Light, shiny surfaces are the best reflectors of radiation and they will allow the waves to reflect and bounce off rather than absorb. we can consider mirror as the example ,it will only reflect the light energy falling on them and it will not absorb. The darker coloured and rough surfaced substances will definitely absorb some amount of light falling on it. so light coloured smooth or shiny surfaced material would be the best reflector for electromagnetic energy.</span>
5 0
3 years ago
A ladder is slipping down a vertical wall. If the ladder is 10 ft long and the top of it is slipping at the constant rate of 4 ​
Andrei [34K]

Answer:

Speed will be: 1.33ft/s

Explanation:

If ladder is 10 ft long and bottom is 8 ft from the wall then by Pythagoras theorem we can find the height of the wall where ladder touches. (before it started slipping)

10^2 = 8^2 + x^2

thus x^2 = 100 - 64 = 36

Giving x= 6 ft. If the ladder is falling with a speed of 4ft/s it will take 1.5 seconds to cover the 6ft distance.

This shows that the bottom of ladder will travel from 8ft to 10 ft in 1.5 seconds. Thus covering 2 ft in 1.5 seconds making the speed to be:

v = S / t

v = 2 / 1.5

v = 1.33 ft/s

8 0
4 years ago
Without the wheels, a bicycle frame has a mass of 8.29 kg. Each of the wheels can be roughly modeled as a uniform solid disk wit
trapecia [35]

Answer:

69.66 Joule

Explanation:

mass of bicycle frame, mf = 8.29 kg

mass of wheel, mw = 0.820 kg

radius, r = 0.343 m

velocity, v = 3.6 m/s

There are two wheels in the bicycle.

There are two types of kinetic energy of the system one is kinetic energy of rotation and another is rotational kinetic energy.

K = \frac{1}{2}m_{f}v^{2}+ 2\times \frac{1}{2}m_{w}v^{2}+ 2\times \frac{1}{2}I_{w}\omega^{2}

K = \frac{1}{2}m_{f}v^{2}+m_{w}v^{2}+ \frac{1}{2}\times m_{w}v^{2}

K = \frac{1}{2}m_{f}v^{2}+ \frac{3}{2}\times m_{w}v^{2}

K = \frac{1}{2}\times 8.29\times 3.6^{2}+ \frac{3}{2}\times 0.820\times 3.6^{2}

K = 69.66 J

3 0
4 years ago
A piece of indium with a mass of 16.6 g is submerged in 46.3 cm3 of water in a graduated cylinder. The water level increases to
blagie [28]

Answer:

the density of indium is  7.2 g/cm^3

Explanation:

The computation of the density of indium is shown below:

Given that

Mass = 16.6 g

Volume = 48.6 c,^3 - 46.3cm^3 = 2.3 cm^3

Based on the above information

As we know that

Density = mass  ÷ volume

So,

= 16.6g ÷ 2.3 cm^3

= 7.2 g/cm^3

hence, the density of indium is  7.2 g/cm^3

We simply applied the above formula so that the correct value could come

And, the same is to be considered  

8 0
3 years ago
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