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4 years ago

Molecules of which gas would exert the greatest collision force

Physics

2 answers:

mojhsa [17]4 years ago

7 0

Answer: D

Explanation:

AleksAgata [21]4 years ago

3 0

gas S is the answer

I hope this helped you

have a good day :)

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if an object is being pulled by two forces, one 4n to the left, and the other 2 n to the right, what is the net force acting on

Naddika [18.5K]

Answer:

2 N to the left

Explanation:

6 0

3 years ago

The weight of a luggage is 69.3 N on the moon. Find its weight on the Earth.

kiruha [24]

Answer:

415.8N

Explanation:

Given:

weight of a luggage = 69.3 N

But

✓(weight on the Earth) is equivalent to 6 times weight if the moon.

✓Then, weight on the Earth) =( 69.3N × 6)

✓weight on the Earth) = 415.8N

7 0

3 years ago

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arr

Nitella [24]

Answer:

1. Part A: 648N
2. Part B: 324J
3. Part C: 1,296N

Explanation:

1. Part A:

The magnitude of the force is calculated using the Hook's law:

$|F|=k\Delta x$

You know $\Delta x=0.250m$ but you do not have $k$ .

You can calculate it using the equation for the work-energy for a spring.

The work done to compress the springs a distance $\Delta x$ is:

$Work=\Delta PE=(1/2)k(\Delta x)^2$

Where $\Delta PE$ is the change in the elastic potential energy of the "spring".

Here you have two springs, but you can work as if they were one spring.

You know the work (81.0J) and the length the "spring" was compressed (0.250m). Thus, just substitute and solve for k:

$81.0J=(1/2)k(0.250m)^2\\\\k=2,592N/m$

In reallity, the constant of each spring is half of that, but it is not relevant for the calculations and you are safe by assuming that it is just one spring with that constant.

Now calculate the magnitude of the force:

$|F|=k\Delta x=2,592N/m\times 0.250m=648N$

2. Part B. How much additional work must you do to move the platform a distance 0.250 m farther?

The additional work will be the extra elastic potential energy that the springs earn.

You already know the elastic potential energy when Δx = 0.250m; now you must calculate the elastic potential energy when Δx = 0.250m + 0.250m = 0.500m.

$\Delta E=(1/2)2,592n/m\times(0.500m)^2=324J$

Therefore, you must do 324J of additional work to move the plattarform a distance 0.250 m farther.

3. Part C

What maximum force must you apply to move the platform to the position in Part B?

The maximum force is when the springs are compressed the maximum and that is 0.500m

Therefore, use Hook's law again, but now the compression length is Δx = 0.500m

$|F|=k\Delta x=2,592N/m\times 0.500m =1,296N$

8 0

3 years ago

WRONG ANSWERS WILL BE REPORTED

Dmitry_Shevchenko [17]

The answer to number 9 is C, decreased muscle tension. I believe number 10 is D, prepares an organism to respond to stress.

7 0

3 years ago

How much work is done when a hoist lifts a 290-kg rock to a height of 7 m?

Zepler [3.9K]

Given: Mass m = 290 Kg; Height h = 7 m

Required: Work = ?

Formula: Work = Force x distance but, Force F = mg

W= fd

W = mgh

W= (290 Kg)(9.8 m/s²)(7 m)

W = 19,894 Kg.m²/s²

W = 19,894 J

5 0

4 years ago