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Anika [276]
3 years ago
9

What is the smallest particle of an element that still retains of that element

Physics
1 answer:
Thepotemich [5.8K]3 years ago
7 0

Answer:

atom

Explanation:

the first guy who gave link this is for you:

we cannot use links in brainly

Thankyou.

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Advantages of outsourcing project work may likely include all of the following EXCEPT:
yaroslaw [1]

Answer:

e. Reduced conflict.

Explanation:

Outsourcing is a tool that allows you to hire a provider outside the company for the execution of secondary activities, such as cleaning or mail, or covering other areas of the company, such as financial or accounting systems or the human resources area

- Advantages of Outsourcing

- Cost reduction

- Focus on the main activity

- Transformation of fixed costs into variables

- Reduce risk

- Improve quality

- Productivity increase

- Improve innovation processes

- Greater flexibility

- Access to the latest technologies

- Increase in competitiveness

3 0
3 years ago
At t = 0, a car registers at 30 miles/hr. Forty seconds later, the car’s velocity is now at 50 miles/hr. Assuming constant accel
ratelena [41]
D.

50 mph - 30 mph= 20 mph net velocity
change.
20mph/3600 seconds/hour= .00555 MPS
.0055 miles per second


40 seconds to complete the change
.0055/40= .000138

7 0
2 years ago
Forces acting on a ball as it is being squeezed
goblinko [34]
Squishy pls explain
6 0
3 years ago
When compared to other types of waves, electromagnetic waves differ because they are
expeople1 [14]
They differ because they are transverse wave. That is their direction of travel is perpendicular to its vibrations.
7 0
3 years ago
1. A student lifts a box of books that weighs 185 N. The box is
aksik [14]

1)  148 J

When lifting an object, the work done on the object is equal to its change in gravitational potential energy. Mathematically:

W = \Delta U = (mg) \Delta h

where

mg is the weight of the object

\Delta h is the change in height

For the box in this problem,

mg = 185 N

\Delta h = 0.800 m

Substituting into the equation, we find:

W=(185)(0.800)=148 J

2) (a) 28875 J

The work done by a force applied parallel to the direction of motion of the object is given by

W=Fd

where

F is the magnitude of the force

d is the displacement

In this problem,

F = 825 N is the force applied by the two students together

d = 35 m is the displacement of the car

Substituting,

W=(825)(35)=28875 J

2) (b) 57750 J

As seen previously, the equation that gives the work done by the force is

W=Fd

We see that the work done is proportional to the magnitude of the force: therefore, if the force is doubled, then the work done is also doubled.

The work done previously was

W = 28875 J

Now the force is doubled, so the new work done will be

W' = 2(28875)=57750 J

3) 4.4 J

In this case, the force acting on the ball is the force of gravity, whose magnitude is:

F = mg

where

m = 0.180 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity

Solving the equation,

F=(0.180)(9.8)=1.76 N

Now we find the work done by gravity using the same formula applied before:

W=Fd

where d = 2.5 m is the displacement of the ball. We can apply this version of the formula since the force is parallel to the displacement. Substituting,

W=(1.76)(2.5)=4.4 J

4) 595.2 kg

In this case, we have the work done on the box:

W = 7.0 kJ = 7000 J

And we also know the change in height of the box:

\Delta h = 1.2 m

As we stated in part a), the work done on the box is equal to its change in gravitational potential energy:

W=mg \Delta h

Solving for m, we find

m=\frac{W}{g \Delta h}

And substituting the numerical values, we find the mass of the box:

m=\frac{7000}{(9.8)(1.2)}=595.2 kg

5) They do the same work

In fact, the net work done by each person on the box is equal to the change in gravitational potential energy of the box:

W=mg \Delta h

Where \Delta h is the difference in height between the final position and the initial position of the box.

This means that the work done on the box depends only on its initial and final position, not on the path taken. The two men carry the box along different paths, however the reach at the end the same position, and they started from the same position: this means that the value of \Delta h is the same for both of them, so the work they have done is exactly the same.

5 0
3 years ago
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