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3 years ago

Examples of the six simple machines

Physics

2 answers:

Rudiy273 years ago

6 0

<span>1. lever

2. wedge

3. pulley

4. screw

5. wheel and axle

6. inclined planes.</span>

lakkis [162]3 years ago

5 0

Car, computer,drill,air plane,television, vending machine

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A cat chases a mouse across a 0.65 m high

lora16 [44]

Answer:

V = 7.5 m/s

Explanation:

Projectile motion is a two dimensional motion experienced by an object or particle that is subjected near the surface of the Earth and moves along a curved path under the influence of gravity only. The path followed by projectile motion is called projectile path.

As the cat followed the projectile path, and we know that in projectile motion the horizontal component of the velocity remains constant throughout its motion. So there is no acceleration along horizontal path.

Given data :

Height = y = 0.65 m

Horizontal distance = x = 2.7 m

Gravitational acceleration = g = 9.81 m/s²

Velocity when cat slid off the table = V = ?

As the cat slides off horizontally, So there is no vertical component of velocity or vertical component of velocity is zero.

Using 2nd equation of motion

y = 0.5 gt²

0.65 = 0.5*9.8*t²

t² = 0.1326

t = 0.36 s

Velocity when cat slid off the table = V = x / t

V = 2.7 / 0.36

V = 7.5 m/s

7 0

3 years ago

When a force is exerted on an object, work is done only if the object

Kazeer [188]

Is moving, i'm almost positive

4 0

3 years ago

Mercury has an average disease to the sun of 0.39 AU. In two or more complete sentences, explain how to calculate the orbital pe

alukav5142 [94]

Answer: 88 Earth days

Explanation:

According to the Kepler Third Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

<em />

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit:

$T^{2}=a^{3}$ (1)

If we assume the orbit is circular and apply Newton's law of motion and the Universal Law of Gravity we have:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (2)

Where $M$ is the mass of the massive object and $G$ is the universal gravitation constant. If we assume $M$ constant and larger enough to consider $G$ really small, we can write a general form of this law:

$MT^{2}=a^{3}$ (3)

Where $T$ is in units of Earth years, $a$ is in AU (<u>1 Astronomical Unit is the average distane between the Earth and the Sun)</u> and $M$ is the mass of the central object in units of the mass of the Sun.

This means when we are making calculations with planets in our solar system $M=1$ .

Hnece, in the case of Mercury:

$(1)T^{2}=(0.39 AU)^{3}$ (4)

Isolating $T$ :

$T=\sqrt{(0.39 AU)^{3}}$ (5)

$T=0.243 Earth-years \frac{365 days}{1 Earth-year}=88.6 days \approx 88 days$ (6)

This means the period of Mercury is 88 days.

7 0

3 years ago

A uniform rod is 2. 0 m long and has mass 15 kg. What is most nearly the rod's mass moment of inertia?

trapecia [35]

The rod's mass moment of inertia is 5kgm².

<h3>Moment of Inertia:</h3>

The "sum of the product of mass" of each particle with the "square of its distance from the axis of rotation" is the formula for the moment of inertia.

The Parallel axis Theorem can be used to compute the moment of inertia about the end of the rod directly or to derive it from the center of mass expression. I = kg m². We can use the equation for I of a cylinder around its end if the thickness is not insignificant.

If we look at the rod we can assume that it is uniform. Therefore the linear density will remain constant and we have;

or = M / L = dm / dl

dm = (M / L) dl

$I = \int\limits^M_0 {r^2} \, dm$