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Anvisha [2.4K]
3 years ago
6

Suppose you find yourself in your friend's third floor apartment building when you smell smoke coming from outside the door. you

open the door an see nothing but flames protruding from the hallway. Realizing that your only means of escape is through the window of the apartment, your freind agrees to donate many pairs of nylons which you tie together to form a makeshift rope. this nylon rope(no pun intended) can withstand a maximum tension of 500 N without breaking. you have a mass of 70Kg on this particular day.
a) why can't you slide down the rope at a near constant speeed?
b) What is the smallest prssible acceleration you must achieve in order to slide down the rope without it breaking? ans:2.9m/s square
Physics
1 answer:
Alborosie3 years ago
8 0

(a) Because the tension in the rope would be larger than the maximum value allowed

When you are sliding down the slope, there are two forces acting on you:

- Your weight, W=mg, acting downward, where

m = 70 kg is your mass

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

- The tension in the rope, T, acting upward

Therefore, the equation of motion is

T-mg=ma

where a is the acceleration.

If you want to slide at constant speed, then the acceleration must be zero:

a = 0

And so the equation becomes

T-mg=0

This means that the tension in the rope must be:

T=mg=(70 kg)(9.8 m/s^2)=686 N

Which is larger than the maximum tension allowed in the rope (500 N), so the rope will break.

(b) 2.66 m/s^2 downward

Again, we must refer to the equation of the forces:

T-mg=ma

In this case, we want the tension in the rope to be the maximum allowed value,

T = 500 N

the other data are

m = 70 kg is your mass

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

Substituting into the equation, we can find the corresponding value of acceleration:

a=\frac{T-mg}{m}=\frac{500-(70)(9.8)}{70}=-2.66 m/s^2

where the negative sign means the acceleration is downward.

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A meter stick with a mass of 0.167 kg is pivoted about one end so it can rotate without friction about a horizontal axis. The me
SSSSS [86.1K]

Answer:

Part a)

change in potential energy is given as

\Delta U = 0.82 J

Part B)

angular speed of the rod is given as

\omega = 5.42 rad/s

Part c)

Linear speed of the end of the rod is given as

v = 5.42 m/s

Part d)

when a particle falls from rest to distance d = 1 m

v = 4.42 m/s

Explanation:

Part A)

As we know that the gravitational potential energy change is given as

\Delta U = mgH

\Delta U = 0.167(9.81)(0.5)

\Delta U = 0.82 J

Part B)

As we know that change in gravitational energy is equal to gain in kinetic energy

so we have

\Delta U = \frac{1}{2}I\omega^2

0.82 = \frac{1}{2}(\frac{mL^2}{3})\omega^2

\omega^2 = \frac{6 \times 0.82}{0.167 (1)^2}

\omega = 5.42 rad/s

Part c)

Linear speed of the end of the rod is given as

v = L\omega

v = 5.42 m/s

Part d)

when a particle falls from rest to distance d = 1 m

so we will have

v = \sqrt{2gL}

v = \sqrt{2(9.81)(1)}

v = 4.42 m/s

4 0
3 years ago
Now suppose that the Earth had the same mass and radius as it currently does, but all of the mass was concentrated into a thin (
KiRa [710]

Answer:

The force of gravity at the shell will be extremely great on me due to the huge mass collapsed into the small radius.

<em>At the center of the shell, the gravitational forces all around should cancel out, giving me a feeling of weightlessness; which will be a lesser force compared to that felt while standing on the shell.</em>

<em></em>

Explanation:

For the collapsed earth:

mass = 5.972 × 10^24 kg

radius = 1 ft

according to Newton's gravitation law, the force of gravity due to two body with mass is given as

Fg = GMm/R^{2}

Where Fg is the gravitational force between the two bodies.

G is the gravitational constant

M is the mass of the earth

m is my own mass

R is the distance between me and the center of the earths in each case

For the case where I stand on the shell:

radius R will be 1 ft

Fg = GMm/1^{2}

Fg = GMm

For the case where I stand stand inside the shell, lets say I'm positioned at the center of the shell. The force of gravity due to my mass will be balanced out by all other masses around due to the shell of the hollow earth. This cancelling will produce a weightless feeling on me.

3 0
3 years ago
Different structural forms of the same element are called
julia-pushkina [17]

Answer:

allotropes

Explanation:

4 0
3 years ago
A carnival game consists of a two masses on a curved frictionless track, as pictured below. The player pushes the larger object
Harman [31]

Answer:

v₁₀ = 1.90 m / s

Explanation:

In this exercise we are given the maximum height data, with energy we can know how fast the body came out

Final mechanical energy, maximum height

    Em_{f} = U = m g h

Initial mechanical energy, in the lower part of the track

    Em₀ = K = ½ m v²

    Em=   Em_{f}

    ½ m v² = m g h

    v = √ 2gh

Now we can use the moment to find the speed with which objects collide

The large object has a mass M = 5.41 kg a velocity starts v₁₀, the small object has a mass m = 1.68 kg an initial velocity of zero v₂₀ = 0 and  final velocity v

Initial before the crash

    p₀ = M v₁₀ + 0

Final after the crash

      p_{f} = M v1f + m v

   p₀ =   p_{f}

   M v₁₀ = M v_{1f}+ m v

As the shock is elastic the kinetic energy is conserved

     K₀ = K_{f}

    ½ M v₁₀² = ½ M v_{1f}² + ½ m v²

Let's write the system of equations

    M v₁₀ = M  v_{1f} + m v

    M v1₁₀² = M v_{1f}² + m v²

We cleared v1f in the first we replaced in the second

   v_{1f} = (M v₁₀ - mv) / M

    M v₁₀² = M (M v₁₀ - mv)² / M² + m v²

    M v₁₀² = 1 / M (M² v₁₀² - 2mM v v₁₀ + m² v²) +m v²

     v₁₀² (M - M) + 2 m v v₁₀ - v² (m2 + m) / M = 0

     2 m v₁₀ - v (m + 1) m/ M = 0

     v₁₀ = v (m +1) / (2M)

Let's substitute the value of v

     v1₁₀= √ (2gh) (m +1) / (2M)

Let's calculate

    v₁₀ = √ (2 9.8 3) (1+ 1.68) / (2  5.41)

    V₁₀ = 7.668 (2.68) / 10.82

   v₁₀ = 1.90 m / s

5 0
3 years ago
What is the science principle that explains magnetic fields? I need this fast, please!
bekas [8.4K]

Answer:

I literally just learned this last week and if I remember correctly it is Faraday's Law of Induction.

Explanation: Hope this helps also I hope you have/had an amazing day today<3

3 0
1 year ago
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