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Murrr4er [49]
2 years ago
11

Which has more inertia - a 2,750 gram object or a 2,500 gram object?

Physics
2 answers:
noname [10]2 years ago
5 0

Question:

Which has more inertia - a 2,750 gram object or a 2,500 gram object?

Answer:

2,750 has more interia..

k0ka [10]2 years ago
3 0

Answer:

a 2,750 gram object has more inertia.

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Describe the earth's rotation on it axis
Lisa [10]
<span>Earth's rotation is the rotation of the planet Earth around its own axis. The Earth rotates from the west towards east. As viewed from North Star or polestar Polaris, the Earth turns counter-clockwise.</span>
3 0
3 years ago
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A 15-kg block at rest on a horizontal frictionless surface is attached to a very light ideal spring of force constant 450 N/m. T
den301095 [7]

Answer:

0.266 m

Explanation:

Assuming the lump of patty is 3 Kg then applying the principal of conservation of linear momentum,

P= mv where p is momentum, m is mass and v is the speed of an object. In this case

m_pv_p=v_c(m_p+m_b) where sunscripts p and b represent putty and block respectively, c is common velocity.

Substituting the given values then

3*8=v(15+3)

V=24/18=1.33 m/s

The resultant kinetic energy is transferred to spring hence we apply the law of conservation of energy

0.5(m_p+m_b)v_c^{2}=0.5kx^{2} where k is spring constant and x is the compression of spring. Substituting the given values then

(3+15)*1.33^{2}=450*x^{2}\\x\approx 0.266 m

7 0
3 years ago
A toaster draws 8 A of current with a voltage of 120 V. Which is the power used by the toaster?
dexar [7]
U=120 \text{ V}\\&#10;I=8\text{ A}\\&#10;P=U\cdot I\\\\&#10;P=120\cdot8=960 \text{ W}&#10;
7 0
3 years ago
Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1
hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

\sum F=ma

We know that the force between two bodies due to gravity is given by the following equation:

F_{g} = G\frac{m_{1}m_{2}}{r^{2}}

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

F_{g} =G\frac{Mm}{r^{2}}

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

F_{c}=ma_{c}

where the centripetal acceleration is given by:

a_{c}=\omega ^{2}r

where

\omega = \frac{2\pi}{T}

Where T is the period, and \omega is the angular speed of the planet, so:

a_{c} = ( \frac{2\pi}{T})^{2}r

or:

a_{c}=\frac{4\pi^{2}r}{T^{2}}

so:

F_{c}=m(\frac{4\pi^{2}r}{T^{2}})

so now we can do the sum of forces:

\sum F=ma

F_{g}=ma_{c}

G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})

in this case we can get rid of the mass of the planet, so we get:

G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})

we can now solve this for T^{2} so we get:

T^{2} = \frac{4\pi ^{2}r^{3}}{GM}

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

T_{1}^{2}>T_{2}^{2}

So let's see what's going on there, we'll call:

M_{1}= mass of Star 1

M_{2}= mass of Star 2

So:

\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}

we can get rid of all the constants so we end up with:

\frac{1}{M_{1}}>\frac{1}{M_{2}}

and let's flip the inequality, so we get:

M_{2}>M_{1}

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

6 0
3 years ago
You wad up a piece of paper and throw it into the wastebasket. How far will
astraxan [27]

Answer:

Since the paper is wadded up tight, and if there's any

air resistance left we assume there isn't any, it might

just as well be a stone that's tossed.  This is just a

stripped down projectile situation.

You said "an angle of 36 degrees", but you didn't say relative

to what.  I'll assume that it's 36 degrees above horizontal, and

now I'll proceed to answer the question with the information that

I just gave myself.

-- The vertical component of the velocity is  1.4 sin(36)

                                                                        = 0.823 m/s up.

-- The projectile rises for (0.823/9.8) second, runs out of gas,

and then falls for another (0.823/9.8) second to its original height.

So it's in the air for

                                  2 (0.823/9.8) = 0.168 second

                                                            (not very long at all)

-- The horizontal component of the velocity is  1.4 cos(36)

                                                                           = 1.133 m/s  

                                                             and it doesn't change.

-- During the 0.168 second that it's in the air,

the wad travels horizontally

                                              (0.168 s) x (1.133 m/s)

                                          =            0.19 meter

                                              (19 cm, ~ 7.5 inches)

If you find my mistake on this one, please please tell me.  

As of now, it looks like with that velocity at that angle, your

paper wad only makes it 7.5 inches from your hand into the can.

Explanation:

6 0
3 years ago
Read 2 more answers
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