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3 years ago

Examples of applied force

Physics

2 answers:

Vladimir [108]3 years ago

4 0

Answer:

Push - The most common form of force is a push through physical contact (like a lawnmower or shopping cart)

Pull - You can apply a force by directly pulling on an object (like pulling a wagon)

Explanation:

dem82 [27]3 years ago

3 0

-- you sit on a chair

-- gravity pulls something down

-- water pushes a log up to float

-- air holds the skin of a balloon out, in a ball-shape

-- my dog bites a bone

-- air holds the tires of my car out, in donut shapes

-- a book presses on a table

-- my shoes press down on the floor

-- my feet press down on the bathroom scale

-- the bathroom scale presses up on my feet

-- gravity pulls the Earth up toward me

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The charged particles in the beams that Thomson studied came from atoms. As these particles moved away from their original atoms

aalyn [17]

The question to the above information is;

What is the best use of an atomic model to explain the charge of the particles in Thomson's beams?

Answer;

An atom's smaller negative particles are at a distance from the central positive particles, so the negative particles are easier to remove.

Explanation;

-Atoms are comprised of a nucleus consisting of protons (red) and neutrons (blue). The number of orbiting electrons is the same as the number of protons and is termed the "atomic number" of the element.

J.J. Thomson discovered the electron. Atoms are neutral overall, therefore in Thomson’s ‘plum pudding model’:

atoms are spheres of positive charge
electrons are dotted around inside

5 0

3 years ago

Read 2 more answers

How high does a rocket have to go above Earth's surface before its weight is half of what it is on Earth?

Contact [7]

Answer:

$h=1.6\times 10^6\ m$

Explanation:

As we know that the acceleration due to gravity decreases with height.

At certain height it will get to the half of its value on the surface of the earth.

As we know that the weight on the surface of the earth is given as:

$w=m.g$

where:

m = mass of the object

g = acceleration due to gravity of the substance

Since mass of the substance is constant so the variation is weight is possible only due to change in the acceleration due to gravity.

<u>We know that the variation of the acceleration due gravity with height is given as:</u>

$g_{_h}=g\times (1-\frac{2h}{R} )$

where:

$g_{_h}=$ value to acceleration due to gravity at height h

g = acceleration due to gravity at the earth's surface

h = height of the object

R = radius of the earth = $6400\ km$

according to question the weight becomes half, so,:

$4.9=9.8\times (1-\frac{2h}{6400\times10^3} )$

$h=1.6\times 10^6\ m$ is the height a rocket has to go above Earth's surface before its weight is half of what it is on Earth.

4 0

3 years ago

The Millennium Falcon was constructed 43.6 m long. a) How fast is it traveling past Luke when he measures its length to be 30.1

Zinaida [17]

Answer:

a)0.5564c

b)43.6 m

Explanation:

Given proper length of falcrum L₀= 43.6 m

improper length L=30.1 m (when viewed from moving frame)

we know that $L=L_{0}\sqrt{1-\frac{v^2}{c^2}}$

$30.1=43.6\sqrt{1-\frac{v^2}{c^2}$

⇒ $\{1-\frac{v^2}{c^2}}$ = $\frac{30.1}{43.6}$

${1-\frac{v^2}{c^2}}=0.6904$

$\frac{v^2}{c^2}=0.3096$

$v^{2}=3096c^{2}$

v=0.5564c

this is the required speed of falcon when it passes luke

b). Since Han solo is on the Falcon its reference frame will be falcon itself hence there wont be any change in the length of Falcon that its length will be

43.6 m

8 0

3 years ago

A 22.0 kg bucket of concrete is connected over a very light frictionless pulley to a 375 N box on the roof of a building as show

Veronika [31]

Answer:

vf = 3.27[m/s]

Explanation:

In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.

With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.

With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.

Then we equalize the two stress equations and we can clear the acceleration.

a = 3.58 [m/s^2]

As we know that the bucket descends 1.5 [m], this same distance is traveled by the box, as they are connected by the same rope.

$x = \frac{1}{2} *a*t^{2}\\1.5 = \frac{1}{2}*(3.58) *t^{2} \\t = 0.91 [s]$