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3 years ago

A sample of gold has a mass of 38.6 grams and a volume of 2 cm3 what is the density of gold

Physics

2 answers:

Contact [7]3 years ago

7 0

Answer:

Density, $d=19.3\ g/cm^3$

Explanation:

Mass of the sample of gold, m = 38.6 grams

Volume of gold sample, $V=2\ cm^3$

The density of gold sample is given by mass per unit volume. Mathematically, it is given by :

$d=\dfrac{m}{V}$

$d=\dfrac{38.6\ g}{2\ cm^3}$

$d=19.3\ g/cm^3$

So, the density of the sample of gold is $19.3\ g/cm^3$ . Hence, this is the required solution.

Rudik [331]3 years ago

3 0

Density = Mass ÷ Volume

Your mass = 38.6 grams
Your volume = 2 cm3

38.6 ÷ 2 = 19.3 grams per cm3

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Answer:

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2 years ago

he triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00 1

meriva

Answer:

Moment of inertia = 0.3862kg-m²

Explanation:

2.00x10³

2.80cm

145 rad

r = r⊥ x F

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r⊥ is the distance between the applied force and axis

Force exerted = 2.00x10³

r⊥ = 2.8cm = 0.028m

Alpha = 145rad/s²

r = 0.028m x 2.00x10³

r = 56.0N-m

To get the moment of inertia

56.0N-m² = (145rad/s²) x I

The I would be:

I = (56.0N-m²)/(145rad/s²)

I = 56/145

= 0.3862Kg-m²

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2 years ago

Water with a volume flow rate of 0.001 m3/s, flows inside a horizontal pipe with diameter of 1.2 m. If the pipe length is 10m an

galben [10]

Answer:

$\triangle P=1.95*10^{-4}$

Explanation:

Mass $m=0.001$

Diameter $d=1.2m$

Length $l=10m$

Generally the equation for Volume flow rate is mathematically given by

$Q=AV$

$V=\frac{Q}{\pi/4D^2}$

$V=\frac{0.001}{\pi/4(1.2)^2}$

$V=8.84*10^{-4}$

Generally the equation for Friction factor is mathematically given by

$F=\frac{64}{Re}$

Where Re

Re=Reynolds Number

$Re=\frac{pVD}{\mu}$

$Re=\frac{1000*8.84*10^{-4}*1.2}{1.002*10^{-3}}$

$Re=1040$

Therefore

$F=\frac{64}{Re}$

$F=\frac{64}{1040}$

$F=0.06$

Generally the equation for Friction factor is mathematically given by

$Head loss=\frac{fLv^2}{2dg}$

$H=\frac{0.06*10*(8.9*10^-4)^2}{2*1.2*9.81}$

$H=19.9*10^{-9}$

Where

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3 years ago

Two carts, one of mass 2m and one of mass m, approach each other with the same speed, v. When the carts collide, they hook toget

Nikitich [7]

Answer:

<em>Second option</em>

Explanation:

<u>Linear Momentum</u>

The linear momentum of an object of mass m and speed v is

P=mv

If two or more objects are interacting in the same axis, the total momentum is

$P_t=m_1v_1+m_2v_2+...$

Where the speeds must be signed according to a fixed reference

The images show a cart of mass 2m moves to the left with speed v since our reference is positive to the right

$P_1=-2mv$

The second cart of mass m goes to the right at a speed v

$P_2=mv$

The total momentum before the impact is

$P_t=-2mv+mv=-mv$

The total momentum after the collision is negative, both carts will join and go to the left side

The first option shows both carts with the same momentum before the collision and therefore, zero momentum after. It's not correct as we have already proven

The third option shows the 2m cart has a positive greater momentum than the other one. We have proven the 2m car has negative momentum. This option is not correct either

The fourth option shows the two carts keep separated after the collision, which contradicts the condition of the question regarding "they hook together".

The second option is the correct one because the mass $m_2$ has a negative momentum and then the sum of both masses keeps being negative