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valentinak56 [21]
3 years ago
13

If the mass of a material is 111 grams and the volume of the material is 23 cm3, what would the density of the material be? g/cm

3
Physics
1 answer:
Butoxors [25]3 years ago
4 0
Density can be kg/m^3 or g/cm3
In g/cm3 density =mass /volume =111g/23cm3
=4.826g/cm3.

In kg/m3,density=mass/volume. converting mass in grams to kg, 1000g=1kg,111g=0.111kg.
cm3 to m3, 1cm3=10^-6m3, 23cm3=0.000023m3
density=0.111kg/0.000023m3 or 2.3*10^-5=4,826.1kg/m3.

the other is a long process.
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If a substance is found to be reactive flammable soluble and explosive what observation is also a physical property
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Answer:

  • <u><em>soluble</em></u>

Explanation:

Chemical properties only manifest when a chemical reaction occurs. Being reactive, flammable and explosive are chemical properties, because they involve chemical reactions: the substances are changed; the chemical bonds of some substances, called reactants, are broken, and the chemical bonds are created, forming other substances, called products.

Solubility is a<em> physical property</em> because during dissolution no new substances are formed. You can prove it when the solvent evaporates leaving behind the same original substance.

The the observation that the substance is <em>soluble</em> is describing a <em>physical property.</em>

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3 years ago
Which situation best describes the act of reducing? paying bills online instead of sending paper through the mail donating gentl
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3 years ago
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When a 5.0kg cart undergoes a 2.2m/s increase in speed, what is the impulse of the cart
Karolina [17]

Answer:

11.0 kg m/s

Explanation:

The impulse exerted on the cart is equal to its change in momentum:

I=\Delta p=m\Delta v

where

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At one instant, the center of mass of a system of two particles is located on the x-axis at 2.0 cm and has a velocity of (5.0 m/
Nata [24]

Answer:

Explanation:

Given that,

At one instant,

Center of mass is at 2m

Xcm = 2m

And velocity =5•i m/s

One of the particle is at the origin

M1=? X1 =0

The other has a mass M2=0.1kg

And it is at rest at position X2= 8m

a. Center of mass is given as

Xcm = (M1•X1 + M2•X2) / (M1+M2)

2 = (M1×0 + 0.1×8) /(M1 + 0.1)

2 = (0+ 0.8) /(M1 + 0.1)

Cross multiply

2(M1+0.1) = 0.8

2M1 + 0.2 =0.8

2M1 = 0.8-0.2

2M1 = 0.6

M1 = 0.6/2

M1 = 0.3kg

b. Total momentum, this is an inelastic collision and it momentum after collision is given as

P= (M1+M2)V

P = (0.3+0.1)×5•i

P = 0.4 × 5•i

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c. Velocity of particle at origin

Using conversation of momentum

Momentum before collision is equal to momentum after collision

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We are told that M2 is initially at rest, then, V2=0

So, P(before) = 0.3V1

We already got P(after) = 2 •i kgm/s in part b of the question

Then,

P(before) = P(after)

0.3V1 = 2 •i

V1 = 2/0.3 •i

V1 = 6 ⅔ •i m/s

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