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3 years ago

6

A locomotive engine pulls a train with a constant force comment on whether its acceleration will increase or decrease if more co

aches are added to the train

1 answer:

julia-pushkina [17]3 years ago

7 0

Explanation:

the acceleration will be unchanged according to newton second law of motion

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A uniform copper sphere and a uniform aluminum sphere have the same mass.What is the ratio of the radius of the aluminum sphere

neonofarm [45]

Answer:

$\frac{r_{Al}}{r_{Zn} }= 1.625 : 1$

Explanation:

$density = \frac{mass}{volume}$

ρ = $\frac{m}{v}$

making mass subject of formula we have

m=ρv

from the question aluminium and zinc have same mass hence we can say;

$m_{Al}=m_{Zn}$

which means; $ρ_{Al} V_{Al}=ρ_{Zn} V_{Zn}$

$\frac{ρ_{Al} }{ρ_{Zn} }= =\frac{V_{Zn} }{V_{Al} }$

$\frac{ρ_{Al} }{ρ_{Zn} }= =\frac{4\pi r_{Zn} ^{3} }{4\pi r_{Al}^{3} }$

$\frac{ρ_{Al} }{ρ_{Zn} }= =\frac{r_{Zn} ^{3} }{r_{Al}^{3} }$

$\frac{ρ_{Al} }{ρ_{Zn} }= =(\frac{r_{Zn}}{r_{Al} })^{3}$

$\frac{r_{Zn}}{r_{Al} }= \sqrt{\frac{ρ_{Al} }{ρ_{Zn} }}$

${ρ_{Al}= 2700{/tex] [tex]{ρ_{Zn}= 7130{/tex][tex]\frac{r_{Zn}}{r_{Al} }=\sqrt{\frac{2700 }{7130} }}$

$\frac{r_{Zn}}{r_{Al} }= 0.6154$

$\frac{r_{Al}}{r_{Zn} }=\frac{1}{0.6154}$

$\frac{r_{Al}}{r_{Zn} }= 1.625 : 1$

4 0

3 years ago

6) K2O Use IUPAC nomenclature rules to properly identify this compound. A) potassium oxide B) dipotassium oxide C) potassium (I)

tino4ka555 [31]

NO,,The answer is A) potassium oxide

4 0

3 years ago

Impedance plethysmography is a technique that can be used to detect thrombosis--the presence of clots in a blood vessel--by meas

likoan [24]

Answer:

a) The resistance of the calf between the electrodes is $85\Omega$

b) The average resistivity of this part of the leg is $57.81m\Omega/cm^{3}$

Explanation:

Hi

a) Using Ohm's law $V=IR$ , solving for $R$ , we obtain $R=\frac{V}{I}=\frac{17mV}{200uA}=85\Omega$

b) The volume of the calf is like a cylinder, so $Vol=\pi r^{2}h$ , with $h=13cm$ and $r=d/2=6cm$ , therefore $Vol=\pi (6cm)^{2}(13cm)=468\pi cm^{3}$ . Then we can use $R_{av}=\frac{R}{Vol} =\frac{85\Omega}{468\pi cm^{3}} =0.05781\Omega /cm^{3}=57.81m\Omega/cm^{3}$ , this is the average resistivity of this part of the leg.

4 0

4 years ago

A bullet of mass 11.1 g is fired into an initially stationary block and comes to rest in the block. The block, of mass 1.01 kg,

Kryger [21]

Answer:

a) The initial speed of the bullet is 488 m/s

b) The loss of kinetic energy is 1.3 × 10³ J.

Explanation:

Hi there!

To solve this problem we have to use the conservation of momentum:

initial momentum of the bullet + initial momentum of the block =

final momentum of the block-bullet system

The momentum of an object is calculated as follows:

p = m · v

Where:

p = momentum

m = mass of the object.

v = velocity.

Then, in our system:

p₁₁ = initial momentum of the bullet.

p₂₁ = initial momentum of the block.

p₃₂ = final momentum of the block-bullet system.

p₁₁ + p₂₁ = p₃₂

The initial momentum of the bullet will be:

p₁₁ = m · v

p₁₁ = 0.0111 kg · v

The initial momentum of the block will be:

p₂₁ = 1.01 kg · 0 m/s = 0 kg · m/s

The final momentum of the block-bullet system will be:

p₃₂ = (1.01 kg + 0.0111 kg) · 5.30 m/s

Then, by conservation of the momentum:

initial momentum of the bullet = momentum of the block-bullet system

0.0111 kg · v = (1.01 kg + 0.0111 kg) · 5.30 m/s

v = ((1.01 kg + 0.0111 kg) · 5.30 m/s)/ 0.0111 kg

v = 488 m/s

The initial speed of the bullet is 488 m/s

b) The initial kinetic energy (KE) of the system is the kinetic energy of the bullet because the block is at rest:

KE = 1/2 · m · v²

KE = 1/2 · 0.0111 kg · (488 m/s)²

KE = 1.32 × 10³ J

The final kinetic energy of the system will be the kinetic energy of the block-bullet system:

KE = 1/2 · (1.01 kg + 0.0111 kg) · (5.30 m/s)²

KE = 14.3 J

The loss of kinetic energy will be:

initial kinetic energy - final kinetic energy

1.32 × 10³ J - 14.3 J = 1.3 × 10³ J

The loss of kinetic energy is 1.3 × 10³ J.

8 0

3 years ago

A 16 Ω resistor and a 6 Ω resistor are connected in series to an ideal 6 V battery.

GrogVix [38]

Answer:

Explanation:

a) series resistors carry the same current

A = V/Re = 6/(16 + 6) = 0.2727272... ≈ 27 mA

b) V = V₀(R/Re) = 6(16/(16 + 6)) = 4.363636 ≈ 4.4 V

c) V = V₀(R/Re) = 6(6/(16 + 6)) = 1.636363 ≈ 1.6 V

or V = 6 - 4.4 = 1.6 V

3 0

3 years ago

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