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3 years ago

How many significant digits are in the measurement 589.040 newtons

Physics

2 answers:

dimaraw [331]3 years ago

6 0

Answer:

F = 589.040 N

Number of significant digits = 6

Explanation:

As we know that all the measurements in which the value is given with some decimal then in that case all the numbers in the given measurement is always significant digits.

Since here the force measurement gives the value with decimal as

F = 589.040 N.

So here we can say that total digits in this measurement is significant figures.

So correct answer is 6 digits

Neporo4naja [7]3 years ago

5 0

5... 589. 40....even though the zeros don't really count

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2 years ago

Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part

Fantom [35]

a) Angular acceleration: $17.0 rad/s^2$

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

$W=mg$

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

$W_p =mg sin \theta$

where $\theta$ is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

$\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta$

where L is the length of the pencil.

The relationship between torque and angular acceleration $\alpha$ is

$\tau = I \alpha$ (1)

where

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is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for $\alpha$ , we find:

$\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}$

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when $\theta=10^{\circ}$ is

$\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2$

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude $mg$

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

$\tau = Fd$

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

$\tau =mg\frac{L}{2} cos \theta$

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0

3 years ago

The cylinder in the picture is rotating at 500 RPMs. The friction coefficients between the cylinder and block B are static=0.5 a

Tju [1.3M]

Coefficient of static friction = 0.5
Coefficient of Kinetic friction = 0.3
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<h3>The Radius of the System</h3>

Let R be the radius of cylinder

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The normal force

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Since the radius is very little for two block to execute circular motion so system will slide down.

Learn more on coefficient of static friction here;

brainly.com/question/11841776

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