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alex41 [277]
3 years ago
11

Two children of mass 27 kg and 51 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the c

hildren are separated by a distance of 4 m, at what distance from the pivot point is the small child sitting in order to maintain the balance?
Physics
1 answer:
Flauer [41]3 years ago
8 0

Answer:

2.62 m

Explanation:

Let the small child sit at a distance x from the pivot.

The distance of big child from the pivot is 4 - x .

By using the concept of moments.

Clockwise moments = anticlockwise moments

27 x = 51 ( 4 - x )

27 x = 204 - 51 x

78 x = 204

x = 2.62 m

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A 221 g cart starts from rest and rolls in the right direction (positive) down an incline. The incline is at a height of 5 cm. A
Julli [10]

Answer:

1) p₀ = 0.219 kg m / s, p = 0, 2)  Δp = -0.219 kg m / s, 3) 100%

Explanation:

For the first part, which is speed just before the crash, we can use energy conservation

Initial. Highest point

            Em₀ = U = mg y

Final. Low point just before the crash

           Emf = K = ½ m v²

          Em₀ = Emf

          m g y = ½ m v²

           v = √ 2 g y

Let's calculate

           v = √ (2 9.8 0.05)

           v = 0.99 m / s

1) the moment before the crash is

           p₀ = m v

           p₀ = 0.221 0.99

           p₀ = 0.219 kg m / s

After the collision, the car's speed is zero, so its moment is zero.

           p = 0

2) change of momentum

           Δp = p - p₀

            Δp = 0- 0.219

            Δp = -0.219 kg m / s

3) the reason is

     Δp / p = 1

In percentage form it is 100%

3 0
3 years ago
Please I need help with these 2 questions. Thank you.
Lemur [1.5K]
First one is D and Second one is B
7 0
3 years ago
The unit of length mist suitable for measuring the thickness of a cell phone is a_________? ​
lbvjy [14]

Answer:

Explanation: Width

4 0
3 years ago
6
Arturiano [62]

Explanation:

A light bulb changes electrical energy into <em>heat energy and light energy .</em>

4 0
3 years ago
(a) How much work is required to lift a 35-kg object from the ground 3.0 m into the air? (b) How much gravitational potential en
V125BC [204]

Answer:

(a) work required to lift the object is 1029 J

(b) the gravitational potential energy gained by this object is 1029 J

Explanation:

Given;

mass of the object, m = 35 kg

height through which the object was lifted, h = 3 m

(a) work required to lift the object

W = F x d

W = (mg) x h

W = 35 x 9.8 x 3

W = 1029 J

(b) the gravitational potential energy gained by this object is calculated as;

ΔP.E = Pf - Pi

where;

Pi is the initial gravitational potential energy, at initial height (hi = 0)

ΔP.E = (35 x 9.8 x 3) - (35 x 9.8 x 0)

ΔP.E = 1029 J

7 0
2 years ago
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