Captain Kiddo is trapped inside a submarine. In order to escape, she first needs to open a heavy revolving door. a. Would be easier for her to apply all her force close to the axle of the door or as far away as possible from it?
b. To finally get the hatch open she needs to grab hold of the wheel with her hands on opposite sides and rotate the wheel by exerting a force toward the top of the door on one side while exerting the same force towards the bottom of the door on the opposite side. Is there a net force applied on the wheel? Is there a net torque applied? Explain your answers

Answer:

61 degrees, I just did the test.

Explanation:

Answer: 61 degrees

Explanation:

I just did the question and got it right

Answer:

4.93 m

Explanation:

According to the question, the computation of the height is shown below:

But before that first we need to find out the speed which is shown below:

As we know that

= 9.92 m/s

Now

98.4064 = 19.96 × height

So, the height is 4.93 m

We simply applied the above formulas so that the height i.e H could arrive

Final answer:

The height of the water slide is 5.04 meters.

Explanation:

The problem described in this question involves a water slide, where swimmers start from rest at the top and leave the slide traveling horizontally. To determine the height of the slide, we can use the equations of motion in the horizontal direction. The horizontal displacement (x) is given as 5.00 m and the time (t) is given as 0.504 s. Assuming no friction or air resistance, we can use the equation x = v*t, where v is the horizontal velocity. Rearranging the equation, we can solve for v, which is equal to x/t. Substituting the given values, we have v = 5.00 m / 0.504 s = 9.92 m/s. The horizontal velocity (v) is constant throughout the motion, so we can use the equation v = sqrt(2*g*H), where g is the acceleration due to gravity (9.8 m/s^2) and H is the height of the slide. Rearranging the equation, we can solve for H, which is equal to v^2 / (2*g). Substituting the known values, we have H = (9.92 m/s)^2 / (2*9.8 m/s^2) = 5.04 m.

Answer:

liquids

Explanation

Answer:

Heat and temperature are related but very different.

Explanation:

Heat: The total energy of molecular motion in a substance

Temperature: A measure of the average energy of molecular motion in a substance

For further help:

Examples

Heat                                                             Temperature

-Heat is a form of energy that can             -The degree of hotness and

transfer from hot body to cold body           coldness of the body

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-Heat flows from hot body to cold             -It rises when heated and falls down

body                                                              when an object is cooled down

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-Total kinetic energy and potential            -Temp. is the average kinetic

energy obtained by molecules in                energy of molecules in a

an object                                                        substance

Answer:

t = 39.60 s

Explanation:

Let's take a careful look at this interesting exercise.

In the first case the two motors apply the force in the same direction

            F = m a₀          

           a₀ = F / m

with this acceleration it takes t = 28s to travel a distance, starting from rest

           x = v₀ t + ½ a t²

           x = ½ a₀ t²

           t² = 2x / a₀

           28² = 2x /a₀          (1)

in a second case the two motors apply perpendicular forces

we can analyze this situation as two independent movements, one in each direction

in the direction of axis a, there is a motor so its force is F/2

the acceleration on this axis is

          a = F/2m

          a = a₀ / 2

so if we use the distance equation

             x = v₀ t + ½ a t²

as part of rest v₀ = 0

             x = ½ (a₀ / 2) t²

let's clear the time

             t² = (2x / a₀)  2

we substitute the let of equation 1

             t² = 28² 2

             t = 28 √2

             t = 39.60 s

Answer:

1. 1 s
2. 19.6 m
3. 2 s
4. 0.8 m/s^2
5. 28 m/s
6. 79 m/s
7. 0.37 s
8. 26 m/s
9. 242 m/s
10. 19,930 m

Explanation:

In physics, many of the relationships between speed, distance, and acceleration are tied up in the equations for potential and kinetic energy. For an object of mass M* at height h in a gravity field with acceleration g, the potential energy is

  PE = Mgh

At velocity v, the kinetic energy of the object is ...

  KE = 1/2Mv^2

When an object is dropped or launched from rest, the height and velocity are related by the fact that kinetic energy gets translated to potential energy, or vice versa. This gives rise to ...

  PE = KE

  Mgh = (1/2)Mv^2

The mass (M) can be factored out of this, so we have ...

  2gh = v^2

This can be solved for height:

  h = v^2/(2g) . . . . [eq1]

or for velocity:

  v = √(2gh) . . . . [eq2]

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When acceleration is constant, as assumed here, the velocity changes linearly (to/from 0). So, over the time of travel, the average velocity is half the final velocity. That is,

  t = 2h/v

Depending on whether you start with h or with v, this resolves to two more equations:

  t = 2(v^2/(2g))/v = v/g . . . . [eq3]

  t = 2h/(√(2gh)) = √(4h^2/(2gh)) = √(2h/g) . . . . [eq4]

The last of these can be rearranged to give distance as a function of time:

  h = gt^2/2 . . . . [eq5]

or acceleration as a function of time and distance:

  g = 2h/t^2 . . . . [eq6]

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These 6 equations can be used to solve the problems posed. Just "plug and chug." For problems in Earth's gravity, we use g=9.8 m/s^2. (You may want to keep these equations handy. Be aware of the assumptions they make.)

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* M is used for mass in these equations so as not to get confused with m, which is used for meters.

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1) Use [eq4]: t = √(2·6 m/(9.8 m/s^2)) ≈ 1.107 s ≈ 1 s

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2) Use [eq5]: h = (9.8 m/s^2)(2 s)^2/2 = 19.6 m

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3) Use [eq4]: t = √(25 m/(4.9 m/s^2)) ≈ 2.259 s ≈ 2 s

__

4) Use [eq6]: g = 2(10 m)/(5 s)^2 = 0.8 m/s^2

__

5) Use [eq2]: v = √(2·9.8 m/s^2·40 m) = 28 m/s

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6) Use [eq2]: v = √(2·9.8 m/s^2·321 m) ≈ 79.32 m/s ≈ 79 m/s

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7) Using equation [eq3], we will find the time until Tina reaches her maximum height. Her actual off-the-ground total time is double this value. Using [eq3]: t = v/g = (1.8 m/s)/(9.8 m/s^2) = 9/49 s. Tina is in the air for double this time:

  2(9/49 s) ≈ 0.37 s

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8) Use [eq2]: v = √(2·9.8 m/s^2·33.5 m) ≈ 25.624 m/s ≈ 26 m/s

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9) Use [eq2]: v = √(2·9.8·3000) m/s ≈ 242.49 m/s ≈ 242 m/s

(Note: the terminal velocity in air is a lot lower than this for an object like a house.)

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10) Use [eq1]: h = (625 m/s)^2/(2·9.8 m/s^2) ≈ 19,930 m

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Additional comment

Since all these questions make use of the same equation development, I have elected to answer them. Your questions are more likely to be answered if you restrict your posts to 3 or fewer questions each.