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3 years ago

Please help I’m soooo confused

Physics

2 answers:

hammer [34]3 years ago

6 0

Answer:

Your foot is pushing the ground

alexandr402 [8]3 years ago

6 0

A or Number 1 your foot is pushing off the ground therefore giving the force needed to keep moving

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A piece of steel is 11.5cm long at 22C. It is heated to 1221C, close to its melting point. How long is it, in cm, at the high te

Nataly [62]

Answer:

The length at the final temperature is 11.7 cm.

Explanation:

We need to use the thermal expansion equation:

$\Delta L=\alpha L_{0}\Delta T$

Where:

L(0) is the initial length
ΔT is the differential temperature, final temperature minus initial temperature (T(f)-T(0))
ΔL is the final length minus the initial length (L(f)-L(0))
α is the coefficient of linear expantion of steel (12.5*10⁻⁶ 1/°C)

So, we have:

$L_{f}-L_{0}=\alpha L_{0}(T_{f}-T_{0})$

$L_{f}=L_{0}+\alpha L_{0}(T_{f}-T_{0})$

$L_{f}=0.115+(12.5*10^{-6})(0.115)(1221-22)$

$L_{f}=0.117\: m$

Therefore, the length at the final temperature is 11.7 cm.

I hope it helps you!

7 0

3 years ago

What is an example of a negative incentive for producers

Hatshy [7]

An example of a negative incentive for producers is the sharp increase in production costs. Producers are the one who manage the production costs and even the production budget. Anything that relates the production department is entitled to the management of production producers.

There is what we called positive and negative incentives and both of these can affect consumers and producers. Positive incentives are those situations which will give a certain outcome that will benefit the producers, for example, during the peak season there will be a high demand of products, and this gives the chance of producers to demand a higher price from the consumers, in this situation, there will be a big chance of increase sales. A sharp increase in production costs is a loss for the producers. If there will be an increase in production costs, the budget will be greatly affective and even though it is not a peak season, there’s a big chance also to increase prices which we know, consumers are not fond of.

7 0

3 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

$I=\frac{1}{3}mL^2$

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

¿Qué distancia recorre un móvil que lleva una aceleración de 5m/s durante 10seg?

coldgirl [10]

Answer:

250 m

Explanation:

The car in this problem is moving of uniform accelerated motion, so we can use the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

Assuming the car starts from rest,

u = 0

Also we know that

a = 5 m/s^2 (acceleration of the car)

t = 10 s

Substituting, we find the distance covered:

$s=0+\frac{1}{2}(5)(10)^2=250 m$

3 0

3 years ago

A group of students were investigating the force of gravity. They began by dropping a foam ball from a height of 3 meters into a

kondaur [170]

Answer:B

Explanation:

4 0

3 years ago