All categories

3 years ago

How do u work out weight c and d and what r the answers

Physics

1 answer:

Ket [755]3 years ago

6 0

The whole secret of things that are balanced on a pivot like this is:

The sum of all of the 'moments' is equal on both sides.

The moment of each weight is (the weight) times (its distance from the pivot).
If you add up those for each eight on one side, it has to be equal to the sum
of all the ones on the other side.

<u>2. a).</u>
The moments on the right side are: (4 x 0.15) and (1 x 0.40).
They add up to (0.60 + 0.40) = 1.00

The only moment on the left side is (C x 0.25). Both sides have to be equal.

C x 0.25 = 1.00

Divide each side by 0.25, and you have C = 4 N .

===========================================

<u>2. b).</u>
The only moment on the left side is (5 x 0.40) = 2.00

The moments on the right side are (1 x 0.20) and (D x 0.30)
They add up to (0.3D + 0.2).

Both sides have to be equal. 0.3D + 0.2 = 2.0

Subtract 0.2 from each side: 0.3D = 1.8

Divide each side by 0.3: D = 6 N

You might be interested in

Every winter I fly home to Michigan. It takes 5.2 hours. What is my average speed

Nataliya [291]

Yr average speed would be about 1151.75 really depends

4 0

3 years ago

Distance= 10km due West in 1hour calculate the velocity

Rus_ich [418]

Answer:

Velocity = distance / time

V = 10/1

V = 10km/h

6 0

3 years ago

Read 2 more answers

The center of mass of a 0.30-kg (non-uniform) meter stick is located at its 45-cm mark. What is the magnitude of the torque (in

lbvjy [14]

Answer:

The magnitude of the torque due to gravity if it is supported at the 28-cm mark is 0.5 N-m.

Explanation:

Given that,

Mass of the meter stick, m = 0.3 kg

Center of mass is located at its 45 cm mark.

We need to find the magnitude of the torque due to gravity if it is supported at the 28-cm mark. Torque acting on the object is given by :

$\tau=r\times F\\\\\tau=(45-28)\times 10^{-2}\times 0.3\times 9.79\\\\\tau=0.499\ N-m$

$\tau=0.5\ N-m$

So, the magnitude of the torque due to gravity if it is supported at the 28-cm mark is 0.5 N-m.

6 0

3 years ago

If a voltmeter has a less than ideal resistance, say 1 MΩ, and is used to measure the voltage across a resistor of a comparable

Naddik [55]

Answer:

As the difference between the resistance of voltmeter and the resistance being measured gets reduced the error in the reading of the voltmeter gets increased.

Explanation:

An ideal voltmeter has infinite parallel resistance and because of this it doesn't draw any current from the circuit of measurement which means it will measure the exact voltage across the elements.

But practically speaking, a real voltmeter doesn't has infinite resistance therefore, all the practical voltmeters face loading effect to some extent.

As the difference between the resistance of voltmeter and the resistance being measured gets reduced the error in the reading of the voltmeter gets increased. This is why we want to have a greater value of voltmeter resistance, ideally infinite so that the corresponding error is minimized.

Lets consider the given scenario,

A voltmeter has 1 MΩ parallel resistance and the resistance of of measuring element is 500 kΩ or 0.5 MΩ

lets suppose the supplied voltage is 1 V.

First lets assume that the voltmeter is ideal and it has infinite resistance, so in this case voltmeter will measure a voltage of 1 V across the 0.5 MΩ resistor.

Now consider the loading effect, when we connect the voltmeter across the 0.5 MΩ resistor they both become parallel so the resistance is

R = (1*0.5)/(1+0.5)

R = 0.33 MΩ

As you can see the voltmeter will see a reduced resistance and the corresponding voltage also reduces because resistance and voltage are directly proportional.

Therefore, it is preferred to have a very high parallel resistance of the voltmeter.

8 0

3 years ago

In Figure 10-1, if the force exerted on a 3.0-kg backpack that is initally at rest is 20.0 N and the distance it acts over is 0.

pshichka [43]

I'm going to assume this is over a horizontal distance. You know from Newton's Laws that F=ma --> a = F/m. You also know from your equations of linear motion that v^2=v0^2+2ad. Combining these two equations gives you v^2=v0^2+2(F/m)d. We can plug in the given values to get v^2=0^2+2(20/3)0.25. Solving for v we get v=1.82 m/s!

4 0

3 years ago