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dangina [55]
4 years ago
10

A horse 1 m tall running towards a tree at a constant velocity of 20m/s,

Physics
1 answer:
kykrilka [37]4 years ago
7 0

The horse's position on the ground at time <em>t</em> is

<em>x</em> = (20 m/s) <em>t</em>

The baboon's height from the ground at time <em>t</em> is

<em>y</em> = 3 m - 1/2 <em>g</em> <em>t</em>²

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The baboon falls and lands on the horse, so that the two animals meet when the baboon's height is 2 m from the ground, which happens after

2 m = 3 m - 1/2 <em>g</em> <em>t</em>²

1/2 <em>g</em> <em>t</em>² = 1 m

<em>t</em>² = (2 m) / (9.80 m/s²)

<em>t</em> ≈ 0.452 s

In this time, the horse reaches the tree, so its distance from it is

(20 m/s) * (0.452 s) ≈ 9.04 m

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A 74 kg firefighter slides, from rest, 4.9 m down a vertical pole. (a) If the firefighter holds onto the pole lightly, so that t
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Answer:

Her speed is 9.8 meter per second

Explanation:

Newton's second law states that acceleration (a) is related with force (F) by:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

Here the only force acting on the firefighter is the weight F=mg so (1) is:

mg=ma

Solving for a:

a=g

Now with the acceleration we can use the Galileo's kinematic equation:

Vf^{2}=Vo^{2}+2a\varDelta x (2)

With Vf the final velocity, Vo the initial velocity and Δx the displacement, because the firefighter stars from rest Vo=0 so (2) is:

Vf^{2}=2a\varDelta x

Solving for Vf

Vf=\sqrt{2g\varDelta x}=\sqrt{2(9.81)(4.9)}

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6 0
3 years ago
a dog pulls on a pillow with a force of 8.4 N at an angle of 31 degrees above the horizontal. what is the x component of this fo
Natasha_Volkova [10]
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What has more energy a 2 kg mass at 10 m/s or a 2 kg mass 5 m above the ground
Alexxandr [17]

Answer:

Both have same energy

Explanation:

Given:

Sample 1

Mass m = 2 kg

Velocity v = 10 m/s

Sample 2

Mass m = 2 kg

Height h = 5 m

Assume g = 10 m/s²

Find:

What has more energy

Computation:

In sample 1

Ke = 1/2(m)(v)²

Ke = 1/2(2)(10)²

Ke = 100 joule

In sample 2

Pe = mgh

Pe = (2)(10)(5)

Pe = 100  joules

Both have same energy

7 0
3 years ago
A heavy rope, 60 ft long, weighs 0.5 lb/ft and hangs over the edge of a building 130 ft high. (Let x be the distance in feet bel
Nataliya [291]

Answer:

Riemann sum

W = lim n→∞ Σ 0.5xᵢΔx (with the summation done from i = 1 to n)

Integral = W = ∫⁶⁰₀ 0.5x dx

Workdone in pulling the entire rope to the top of the building = 900 lb.ft

Riemann sum for pulling half the length of the rope to the top of the building

W = lim n→∞ Σ 0.5xᵢΔx (but the sum is from i = 1 to n/2)

Integral = W = ∫⁶⁰₃₀ 0.5x dx

Work done in pulling half the rope to the top of the building = 675 lb.ft

Step-by-step explanation:

Using Riemann sum which is an estimation of area under a curve

The portion of the rope below the top of the building from x to (x+Δx) ft is Δx.

The weight of rope in that part would be 0.5Δx.

Then workdone in lifting this portion through a length xᵢ ft would be 0.5xᵢΔx

So, the Riemann sum for this total work done would be

W = lim n→∞ Σ 0.5xᵢΔx (with the summation done from i = 1 to n)

The Riemann sum can easily be translated to integral form.

In integral form, with the rope being 60 ft long, we have

W = ∫⁶⁰₀ 0.5x dx

W = [0.25x²]⁶⁰₀ = 0.25 (60²) = 900 lb.ft

b) When half the rope is pulled to the top of the building, 60 ft is pulled up until the length remaining is 30 ft

Just like in (a)

But the Riemann sum will now be from the start of the curve, to it's middle

Still W = lim n→∞ Σ 0.5xᵢΔx (but the sum is from i = 1 to n/2)

W = ∫⁶⁰₃₀ 0.5x dx

W = [0.25x²]⁶⁰₃₀ = 0.25 (60² - 30²) = 675 lb.ft

Hope this Helps!!!

7 0
3 years ago
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