What is the name for a star and planets held together by gravity? A. solar system B. galaxy C. black hole D. supernova

You have three objects of varying shapes and sizes: Object 1 is a rectangular block of tin. Object 2 is a cube of aluminum. Obje

Over [174]

Answer: a. m = 7.7 kg

b. V = 435.52 in³

c. m = 1927 kg

d. V = 335.37 cm³

e. m = 3 kg

Explanation: Density is the ratio of mass per volume, i.e., it's the measure of an object's compactness. Its representation is the greek letter ρ.

The formula for density is

$\rho=\frac{m}{V}$

Density's unit in SI is kg/m³, but it can assume lots of other units.

Some unit transformations necessary for the resolution of the question:

1 L = 1 dm³ = 1000 cm³

1 in³ = 16.3871 cm³

1 g = 0.001 kg

a. V = 1.34 L = 1340 cm³

$\rho=\frac{m}{V}$

$m=\rho.V$

m = 5.75 * 1340

m = 7705 g => 7.705 kg

Mass of object 1 with volume 1.34L is 7.7 kg.

b. A cube's volume is calculated as V = side³

V = 7.58³

V = 435.52 in³

Volume of object 2 is 435.52 in³.

c. Using 1 in³ = 16.3871 cm³ to change units:

V = 435.52 * 16.3871

V = 713689.4 cm³

Then, mass will be

$m=\rho.V$

m = 2.7 * 713689.4

m = 1926961.4 g => 1927 kg

Mass of object 2 is 1927 kg.

d. Volume of a sphere is calculated as $V=\frac{4}{3}.\pi.r^{3}$

Diameter is twice the radius, then r = 4.31 cm.

Volume is

$V=\frac{4}{3}.\pi.(4.31)^{3}$

V = 335.37 cm³

Volume of object 3 is 335.37 cm³.

e. $m=\rho.V$

m = 8.96 * 335.37

m = 3004.91 g => 3 kg

Mass of object 3 is 3 kg.

4 0

3 years ago

A thin rod of length 1.4 m and mass 140 g is suspended freely from one end. It is pulled to one side and then allowed to swing l

m_a_m_a [10]

Answer:

a The kinetic energy is $KE = 0.0543 J$

b The height of the center of mass above that position is $h = 1.372 \ m$

Explanation:

From the question we are told that

The length of the rod is $L = 1.4m$

The mass of the rod $m = 140 = \frac{140}{1000} = 0.140 \ kg$

The angular speed at the lowest point is $w = 1.09 \ rad/s$

Generally moment of inertia of the rod about an axis that passes through its one end is

$I = \frac{mL^2}{3}$

Substituting values

$I = \frac{(0.140) (1.4)^2}{3}$

$I = 0.0915 \ kg \cdot m^2$

Generally the kinetic energy rod is mathematically represented as

$KE = \frac{1}{2} Iw^2$

$KE = \frac{1}{2} (0.0915) (1.09)^2$

$KE = 0.0543 J$

From the law of conservation of energy

The kinetic energy of the rod during motion = The potential energy of the rod at the highest point

Therefore

$KE = PE = mgh$

$0.0543 = mgh$

$h = \frac{0.0543}{9.8 * 0.140}$

$h = 1.372 \ m$

5 0

3 years ago

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A head-on, elastic collision between two particles with equal initial speed v leaves the more massive particle (mass m1) at rest

ZanzabumX [31]

1/3 The key thing to remember about an elastic collision is that it preserves both momentum and kinetic energy. For this problem I will assume the more massive particle has a mass of 1 and that the initial velocities are 1 and -1. The ratio of the masses will be represented by the less massive particle and will have the value "r" The equation for kinetic energy is E = 1/2MV^2. So the energy for the system prior to collision is 0.5r(-1)^2 + 0.5(1)^2 = 0.5r + 0.5 The energy after the collision is 0.5rv^2 Setting the two equations equal to each other 0.5r + 0.5 = 0.5rv^2 r + 1 = rv^2 (r + 1)/r = v^2 sqrt((r + 1)/r) = v The momentum prior to collision is -1r + 1 Momentum after collision is rv Setting the equations equal to each other rv = -1r + 1 rv +1r = 1 r(v+1) = 1 Now we have 2 equations with 2 unknowns. sqrt((r + 1)/r) = v r(v+1) = 1 Substitute the value v in the 2nd equation with sqrt((r+1)/r) and solve for r. r(sqrt((r + 1)/r)+1) = 1 r*sqrt((r + 1)/r) + r = 1 r*sqrt(1+1/r) + r = 1 r*sqrt(1+1/r) = 1 - r r^2*(1+1/r) = 1 - 2r + r^2 r^2 + r = 1 - 2r + r^2 r = 1 - 2r 3r = 1 r = 1/3 So the less massive particle is 1/3 the mass of the more massive particle.

8 0

3 years ago

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People in finance work mostly with: Insurance Products Information Money

zloy xaker [14]

The answer is money. Hope this helps.

6 0

3 years ago

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Please answer in your own words and I will mark brainlist, Compare all matters from the view of passing light through them (name

prisoha [69]

Answer:

Light passes through the gas

Light passes through the pure water

Light passes through some solids

Explanation:

In gasses, there are many spaces between the molecules. These spaces allow light to pass through them without any interruption.

In pure water, there are some spaces between particles. these particles allow some light rays to pass theough, some to move through the common boundary and reflec5 some of them.

in solids, some allow light to pass through as they are transparent or translucent

4 0

3 years ago