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irinina [24]
2 years ago
13

Two objects are placed on a scale so that it balances. One object weighs 5 N and is placed 0.5 m from the fulcrum of the scale.

The other object is placed 1 m from the fulcrum. How heavy is the other object?
Physics
1 answer:
insens350 [35]2 years ago
5 0

The weight of the other object placed 1 m from the fulcrum that balances the other object is 2.5 N.

<h3>Principle of moment</h3>

The weight of the second object is determined by applying principle of moment as shown below;

take moment about the pivot,

sum of clockwise moment = sum of anticlockwise moment

F₁r₁ = F₂r₂

5 x 0.5 = F₂ x 1

2.5 N = F₂

Thus, the weight of the other object placed 1 m from the fulcrum that balances the other object is 2.5 N.

Learn more about moment here: brainly.com/question/6278006

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Answer:

decreases 1/4 of the original value

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An air balloon is moving upward at a constant speed of 3 m/s. Suddenly a passenger realizes that she left her camera on the grou
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Answer:t=0.3253 s

Explanation:

Given

speed of balloon is u=3\ m/s

speed of camera u_1=20\ m/s

Initial separation between camera and balloon is d_o=5\ m

Suppose after t sec of  throw camera reach balloon then,

distance travel by balloon is

s=ut

s=3\times t

and distance travel by camera to reach balloon is

s_1=ut+\frac{1}{2}at^2

s_1=20\times t-\frac{1}{2}gt^2

Now

\Rightarrow s_1=5+s

\Rightarrow 20\times t-\frac{1}{2}gt^2 =5+3t

\Rightarrow 5t^2-17t+5=0

\Rightarrow t=\dfrac{17\pm \sqrt{17^2-4(5)(5)}}{2\times 5}

\Rightarrow t=\dfrac{17\pm 13.747}{10}

\Rightarrow t=0.3253\ s\ \text{and}\ t=3.07\ s

There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .

(b)When passenger catches the camera time is  t=0.3253\ s

velocity is given by

v=u+at

v=20-10\times 0.3253

v=16.747\ m/s

and position of camera is same as of balloon so

Position is =5+3\times 0.3253

=5.975\approx 6\ m

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2 years ago
A 12-kg dog jumps up in the air to catch a ball. The dog's center of mass is normally 0.20 m above the ground, and he is 0.50 m
ad-work [718]

Explanation:

Given Data:

mass of dog = 12 Kg

dog's center of mass = 0.20m

length of dog = 0.50m

height of dog's jump = ?

Solution:

Work done of gravitational force = Gain in Potential energy

2.1 × mgΔh = mg (h - 0.1)

2.1 × (0.3 - 0.1) = (h - 0.1)

h = 0.52 m

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3 years ago
A football punter accelerates a .55 kg football
Ronch [10]

Answer:

17.6 N

Explanation:

The force exerted by the punter on the football is equal to the rate of change of momentum of the football:

F=\frac{\Delta p}{\Delta t}

where

\Delta p is the change in momentum of the football

\Delta t=0.25 s is the time elapsed

The change in momentum can be written as

\Delta p = m(v-u)

where

m = 0.55 kg is the mass of the football

u = 0 is the initial  velocity (the ball starts from rest)

v = 8.0 m/s is the final velocity

Combining the two equations and substituting the values, we  find the force exerted on the ball:

F=\frac{m(v-u)}{\Delta t}=\frac{(0.55)(8.0-0)}{0.25}=17.6 N

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3 years ago
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