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m_a_m_a [10]
3 years ago
13

Once​ Kate's kite reaches a height of 52 ft ​(above her​ hands), it rises no higher but drifts due east in a wind blowing 6 ft d

ivided by s. How fast is the string running through​ Kate's hands at the moment that she has released 105 ft of​ string?
Physics
1 answer:
Andru [333]3 years ago
7 0

Answer:

5.213ft

Explanation:

Z² = x² + y²

x = √(z² - y²)

y = 52ft, dx = 6ft, z = 105ft, dz = ?

d(z² = x² + y²)

2zdz = 2xdx

dz = xdx/z

But x = √(z² - y²)

dz = √(z² - y²)/z * dx

dz = [√(105² - 52²)/105] * 6

dz = √(8521)/ 17.5

dz = 5.213ft

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While walking between gates at an airport, you notice a child running along a moving walkway. Estimating that the child runs at
Mashcka [7]

Answer:

The speed of the moving walkway is 1.50 m/s

Explanation:

The position of the child can be calculated using the following equation:

x = x0 + v · t

Where :

x = position of the child at time t.

v = velocity of the child.

t = time.

When the child runs in the same direction as the walkway, the velocity of the child will be its  velocity relative to the walkway plus the velocity of the walkway. Then, if we place the origin of the frame of reference at the start of the walkway:

x = x0 + v · t

25 m = 0 m + (2.8 m/s + v) · t₁

Where v is the velocity of the walkway

On its way back, the velocity of the child relative to the walkway is in the opposite direction to the velocity of the walkway. Then:

x = x0 + v · t

0 m = 25 m + (-2.8 m/s + v) · t₂

We also know that t₁ + t₂ = 25 s

Then: t₁ = 25 - t₂

So, we can write the following system of equations:

25 m = (2.8 m/s + v) · (25 s - t₂)

-25 m = (-2.8 m/s + v) · t₂

Let´s take the second equation and solve it for t₂

-25 m / (-2.8 m/s + v) = t₂

Now, let´s replace t₂ in the first equation:

25 m = (2.8 m/s + v) · (25 s + 25 m / (-2.8 m/s + v))

Let´s sum the fraction: 25 s + 25 m / (-2.8 m/s + v)

25 m = (2.8 m/s + v) · (25 s ·(-2.8 m/s + v) + 25 m) / (-2.8 m/s + v)

multiply by (-2.8 m/s + v) both sides of the equation:

25 m(-2.8 m/s + v) = (2.8 m/s + v) · (-70 m + 25 s · v + 25 m)

Apply distributive property:

-70 m²/s +25 m·v = -196 m²/s +70 m·v +70 m²/s -70 m·v +25 s ·v² + 25 m v

56 m²/s = 25 s · v²

56 m²/s / 25 s = v²

v = 1.50 m/s

The speed of the moving walkway is 1.50 m/s

7 0
3 years ago
A briefcase sits stationary in an elevator. The mass of the briefcase if 4.5 kg. The elevator then begins accelerating upwards a
Korvikt [17]

Answer:

D. 48.985 N

Explanation:

Newton's second law states that:

\sum F = ma

which means that the net force acting on an object is equal to the product between the object's mass and its acceleration.

The equation of the forces for the briefcase in the elevator therefore is given by:

N-mg=ma

where

N is the normal reaction exerted on the briefcase

(mg) is the weight of the briefcase, with

m = 4.5 kg being its mass

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

a = 1.10 m/s^2 is the acceleration

Here we chose upward as positive direction.

Solving for N, we find the normal force:

N=mg+ma=m(g+a)=(4.5)(9.81+1.10)=49.095 N

So the closest answer is

D. 48.985 N

3 0
3 years ago
Leta stetter hollingworth conducted pioneering work on __________. a. identity development in ethnic minorities b. cognitive pro
Elena L [17]

D.  Leta stetter hollingworth conducted pioneering work on <u>adolescent development and gifted children</u>.

<h3>Who is Leta stetter hollingworth?</h3>

Leta stetter hollingworth is an early feminist and active member of the Women's Suffrage Party.

Leta Stetter Hollingworth is best known for her landmark contributions to the psychology of women and to education of the gifted. That is adolescent development and gifted children.

Thus, we can conclude that Leta stetter hollingworth conducted pioneering work on <u>adolescent development and gifted children</u>.

Learn more about Leta stetter hollingworth here: brainly.com/question/2680369

#SPJ1

4 0
1 year ago
Potential energy is the ability to:
Ymorist [56]

Answer:

potential energy is the ability to do work

8 0
4 years ago
A 0.017-kg acorn falls from a position in an oak tree that is 18.5 meters above the ground.
Marianna [84]

Answer:

Part a)

Final speed of the corn is 19.05 m/s

Part b)

Kinetic energy of the corn is 3.1 J

Explanation:

Part a)

As we know that the initial position of the corn is

h = 18.5 m

now we also know that it will fall from rest and moving under constant acceleration so we will have

v_f^2 - v_i^2 = 2 a d

v_f^2 - 0 = 2(9.81)(18.5)

v_f = 19.05 m/s

Part b)

Kinetic energy of the corn is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.017)(19.05)^2

K = 3.1 J

4 0
4 years ago
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