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erastova [34]
2 years ago
11

24. An elevator is moving vertically up with an acceleration a. The force exerted on the floor by a passenger of mass m is

Physics
1 answer:
Wewaii [24]2 years ago
3 0

<u>Given </u><u>:</u><u>-</u>

  • An elevator is moving vertically up with an acceleration a.

<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>

  • The force exerted on the floor by a passenger of mass m .

<u>Solution</u><u> </u><u>:</u><u>-</u>

As the man is in a accelerated frame that is <u>non </u><u>inertial</u><u> frame</u><u> </u>, we would have to think of a pseudo force .

  • The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .

For the FBD refer to the attachment . From that ,

\implies Weight_{apparent}= mg + ma

<u>Hence</u><u> </u><u>option</u><u> </u><u>d </u><u>is </u><u>correct</u><u> </u><u>choice </u><u>.</u>

<em>I </em><em>hope</em><em> this</em><em> helps</em><em> </em><em>.</em>

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At baseball practice, Mason and Alfredo both picked up the same bat and neither would let go until one of them had it for himsel
Illusion [34]

Answer:

Option B. O because the net force was 5 N in Alfredo's direction

Explanation:

To know the the correct answer to the question given above, we shall determine the net force acting on the bat. This can be obtained as follow:

Force of pull by Mason (Fₘ) = 15 N

Force of pull by Alfredo (Fₐ) = 20 N

Net force (Fₙ) =?

Fₙ = 20 – 15

Fₙ = 5 N in Alfredo's direction

From the calculation made above, we can see that the net force is 5N in Alfredo's direction. This is the reason why Alfredo end up having the bat.

7 0
3 years ago
Which particles contribute to the net charge and how does each change the net charge?
Licemer1 [7]

The two subatomic particles that contribute to the net charge of an ion are electrons and protons.

<h3>What is an atom?</h3>

Atom is the smallest possible amount of matter which still retains its identity as a chemical element, now known to consist of a nucleus surrounded by electrons.

The atom is made up of three components called subatomic particles as follows;

  • Protons
  • Electrons
  • Neutrons

The proton is the positively charged subatomic particle forming part of the nucleus of an atomwhile the electron is the subatomic particle having a negative charge and orbiting the nucleus.

This suggests that the two subatomic particles that contribute to the net charge of an ion are electrons and protons. That is;

Net charge = protons - electrons

Learn more about subatomic particles at:brainly.com/question/13303285

#SPJ1

3 0
2 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!<br><br> Joules could be used to measure
lyudmila [28]

All of the above, work is a measurement of energy transfer, in Joules.

Potential energy = Joules

Kinetic energy = Joules

The key thing here is that anything having to do with just energy or energy transfer is measured in joules.

4 0
3 years ago
How much work was done by a hot air balloon to lift up a 100 Newton to a height of 300 meters?
Monica [59]

So, the work was done by that hot air-balloon is <u>30,000 J or 30 kJ</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. <u>Work is the amount of force exerted to cause an object to move a certain distance from its starting point</u>. In physics, the amount of work will be proportional to the increase in force and increase in displacement. Amount of work can be calculated by this equation :

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

With the following condition :

  • W = work (J)
  • F = force (N)
  • s = shift or displacement (m)

Now, the s (displacement) can be written as ∆h (altitude change) because the object move to vertical line. The formula can also be changed to:

\boxed{\sf{\bold{W = F \times \Delta h}}}

With the following condition :

  • W = work (J)
  • F = force (N)
  • \sf{\Delta h} = change of altitude (m)

If an object has mass, then the object will also be affected by gravity. Always remember that F = m × g. So that :

\sf{W = F \times \Delta h}

\boxed{\sf{\bold{W = m \times g \times \Delta h}}}

With the following condition :

  • W = work (J)
  • m = mass of the object (kg)
  • g = acceleration of the gravity (m/s²)
  • \sf{\Delta h} = change of altitude (m)

<h3>Problem Solving</h3>

We know that :

  • F = force = 100 N
  • \sf{\Delta h} = change of altitude 300 m

What was asked :

  • W = work = ... J

Step by step :

\sf{W = F \times \Delta h}

\sf{W = 100 \times 300}

\boxed{\sf{W = 30,000 \: J = 30 \: kJ}}

<h3>Conclusion</h3>

So, the work was done by that hot air-balloon is 30,000 J or 30 kJ.

<h3>See More :</h3>
  • Work that he had done to lift object brainly.com/question/26341717
  • Converting work to potential energy brainly.com/question/26487284
5 0
3 years ago
A cat is shot horizontally out of a cannon that is on top of a 125 meter tall cliff. If the cat lands 286 m away from the base o
Elden [556K]

The initial velocity of the cat is 56.6 m/s

Explanation:

The motion of the cat is a projectile motion, consisting of:

- a uniform horizontal motion with constant velocity

- a vertical accelerated motion with constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time it takes for the cat to reach the ground. We use the following suvat equation:

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

where we have (choosing downard as positive direction)

s = 125 m is the vertical displacement of the cat

u = 0 is the initial vertical velocity

t is the time taken to reach the ground

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

Solving for t, we find

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(125)}{9.8}}=5.05 s

Now we can analzye the horizontal motion. Since the motion is uniform and the horizontal velocity is constant, it is given by

v_x =\frac{d}{t}

where

d = 286 m is the horizontal distance travelled

t = 5.05 s is the time taken

Solving the equation,

v_x = \frac{286}{5.05}=56.6 m/s

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

3 0
3 years ago
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