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Angelina_Jolie [31]
3 years ago
6

A twenty-acre park was created to give residents a place to hike, bike, and enjoy other recreational activities. Kathy was hired

to design a map that would show residents how they could easily access different areas of the park. Sean was hired to monitor the number of visitors they have per month and analyze the park’s popularity for funding possibilities. Joan was hired to offer assistance and teach residents about the plants and animals they may encounter while at the park.
Which best describes the tasks Kathy, Sean, and Joan perform?

Kathy and Joan perform tasks common to the Engineering and Technology pathway and Sean performs tasks common to the Science and Math pathway.
Kathy performs tasks common to the Engineering and Technology pathway and Sean and Joan perform tasks common to the Science and Math pathway.
Sean performs tasks common to the Engineering and Technology pathway and Kathy and Joan perform tasks common to the Science and Math pathway.
Sean and Joan perform tasks common to the Engineering and Technology pathway and Kathy performs tasks common to the Science and Math pathway.
Physics
1 answer:
Aleksandr-060686 [28]3 years ago
6 0
Tell that person above me to translate that
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5. Susan exerted 400 newtons of force while pushing on a huge boulder. The boulder moved 0 meters. Calculate work.
Stells [14]

Answer:

0 J

Explanation:

Work is the product of force and distance in the direction of force.

The formula is ; Work = Force * Distance

Given that :

Force = 400 N

Distance = 0 meters

Work = 400 * 0 = 0 J

No work was done because the boulder did not move.

4 0
3 years ago
The diagram shows a skydiver at different points of her jump. At what point would her
8_murik_8 [283]
I’m pretty sure it’s B! :)
6 0
3 years ago
How can we tell when forces are acting on an object (science)
iren [92.7K]

If an object's speed changes, or if it changes the direction it's moving in,
then there must be forces acting on it. There is no other way for any of
these things to happen.

Once in a while, there may be <em><u>a group</u></em> of forces (two or more) acting on
an object, and the group of forces may turn out to be "balanced".  When
that happens, the object's speed will remain constant, and ... if the speed
is not zero ... it will continue moving in a straight line.  In that case, it's not
possible to tell by looking at it whether there are any forces acting on it. 


4 0
3 years ago
Maggie completed a 10000-m race at an average speed of 160
Gala2k [10]

Answer: 200m/min

Explanation:

Divide 10000m by 160m/min, you will get the answer 62.5. You then subtract 12.5 from 62.5 to understand what you will need your answer for the other person’s speed will be. 10000m divided by 50min is 200m/min.

3 0
4 years ago
A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w
amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

(c) the fraction of the initial energy lost is 0.71

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁  + m₂u₂ = v(m₁  +  m₂)

10 x 3   + 25 x 0   = v(10  +  25)

30  = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\

the final kinetic energy of the package;

K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J

the fraction of the initial energy lost;

= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71

7 0
3 years ago
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