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Elena L [17]
3 years ago
11

A bag weighing 20 N is pushed horizontally a distance of 35 m across a

Physics
1 answer:
topjm [15]3 years ago
6 0

Answer:

350J

Explanation:

Given parameters:

Weight of bag  = 20N

Distance moved horizontally  = 35m

Force applied  = 10N

Unknown:

Work done on the bag  = ?

Solution:

Work done is the force applied to move a body through given distance.

  Work done  = Force applied x distance

So;

 Work done  = 10 x 35  = 350J

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Metals in group 1 on the periodic table most commonly form which type of ion
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Group 1 - the alkali metals. The Group 1 elements in the periodic table are known as the alkali metals.

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If you are traveling at 75 km/h how long will it take to travel 32 km?
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While Bob is demonstrating the gravitational force on falling objects to his class, he drops an 1.0 lb bag of feathers from the
BartSMP [9]

In the experiment of free fall bob released a bag of mass 1 lb

so here we can say that initial speed of the bag is Zero

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a = 9.8 m/s^2

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3 0
3 years ago
Solve for work when
BlackZzzverrR [31]

So, <u>the value of the work is approximately 84.65 J</u>.

<h2>Introduction</h2>

Hi ! Here I will help you to discuss the subject about work that caused by force in amount value of angle. Work is affected by the force and displacement.

  • If related to the magnitude of the force, the amount of work will be proportional to the magnitude of the applied force. Thats mean, if the value of the force that applied on it is greater, then the value of the work will be greater.
  • If related to the magnitude of shift, the amount of work will be proportional to the magnitude of shift of object. Thats mean, if the value of the shift on it is greater, then the value of the work will be greater.
<h3>Formula Used</h3>

The work done by a moving object can be expressed in the equation:

If the Angle Is Ignored

\boxed{\sf{\bold{W = F \times s}}}

If the Angle Effect on Work

\boxed{\sf{\bold{W = F \times s \times \cos(\theta)}}}

With the following condition:

  • W = work that done by object (J)
  • F = force that applied (N)
  • s = shift or distance (m)
  • \sf{\theta} = angle of elevation (°)

<h3>Solution</h3>

We know that :

  • F = force that applied = \sf{1.41 \times 10^4} N
  • s = shift or distance = 84.9 m
  • \sf{\theta} = angle of elevation = 45°

What was asked ?

  • W = work that done by object = ... J

Step by step :

\sf{W = F \times s \times \cos(\theta)}

\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \cos(45^o)}

\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \frac{\sqrt{2}}{2}}

\sf{W = 119.709 \times \frac{\sqrt{2}}{2}}

\sf{W = 59.8545 \sqrt{2}}

\boxed{\sf{W \approx 84.65 \: J}}

<h3>Conclusion</h3>

So, the value of the work is approximately 84.65 J.

3 0
1 year ago
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