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max2010maxim [7]
2 years ago
14

A ladybug rests on the bottom of a tin can that is being whirled horizontally on the end of a string. Since the ladybug, like th

e can, moves in a circle, there must be a force on it. What exerts this force?
a. Gravityb. The stringc. Your handd. The can
Physics
1 answer:
Lorico [155]2 years ago
7 0

Answer:uigu

Explanation:nvhv

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3. A 75kg man sits at one end of a uniform seesaw pivoted at its center, and his 24kg son sits at the
bulgar [2K]

Answer:

The wife have to sit at 0.46 L from the middle point of the seesaw.

Explanation:

We need to make a sketch of the seesaw and the loads acting over it.

And by the studying of the Newton's law we can find the equation useful to find the distance of the mother sitting on the seesaw with respect to the center ot the pivot point.

A logical intuition will give us the idea that the mother will be on the side of her son to make the balance.

The maximum momentum with respect to the pivot point (0) will be:

M=75 *\frac{L}{2}

Where L/2 is the half of the distance of the seesaw

Therefore the other loads ( mom + son) must be create a momentum equal to the maximum momentum.

7 0
2 years ago
According to the author, Picasso’s artistic achievements were in large part the result of his:
Harrizon [31]

Answer:

Picasso’s artistic achievements were in large part the result of his contribution to help bring the Nazi' devastating casual bombing and Spanish civil war in Guernica to the world's attention through his paintings.

4 0
2 years ago
A large container, 120 cm deep is filled with water. If a small hole is punched in its side 77.0 cm from the top, at what initia
prisoha [69]

Answer:

The water will flow at a speed of 3,884 m/s

Explanation:

Torricelli's equation

v = \sqrt{2gh}

*v = liquid velocity at the exit of the hole

g = gravity acceleration

h = distance from the surface of the liquid to the center of the hole.

v = \sqrt{2*9,8m/s^2*0,77m} = 3,884 m/s

6 0
2 years ago
A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.
fredd [130]

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

F=\dfrac{mv^2}{r}-mg

F=m(\dfrac{v^2}{r}-g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)

F = 18 N

(b) The net tension in the string when the ball is at the bottom of the circle is given by :

F=\dfrac{mv^2}{r}+mg

F=m(\dfrac{v^2}{r}+g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}

T=\dfrac{m}{r}(g(r-h)+v^2)

Since, v^2=u^2-2gh

T=\dfrac{m}{r}(u^2-3gh+gr)

Hence, this is the required solution.

6 0
2 years ago
Suppose your friend claims to have discovered a mysterious force in nature that acts on all particles in some region of space. H
kirill [66]

Answer:

             U = 1 / r²

Explanation:

In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related

             

         F = - dU / dr

this derivative is a gradient, that is, a directional derivative, so we must have

          dU = - F. dr

the esxresion for strength is

         F = B / r³

let's replace

          ∫ dU = - ∫ B / r³  dr

in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product

let's evaluate the integrals

            U - Uo = -B (- / 2r² + 1 / 2r₀²)

To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)

             U = B / 2r²

we substitute the value of B = 2

             U = 1 / r²

5 0
2 years ago
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