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3 years ago

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The record blue whale shown below had a mass of 1.90 × 105 kg. Assuming that its average density was 0.900 g/cm3, as has been me

asured for other blue whales, what was the volume of the whale in cubic meters (m3)?

1 answer:

Agata [3.3K]3 years ago

8 0

Answer:

replace 920 with 900

Explanation:

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My Notes An electron is released from rest on the axis of a uniform positively charged ring, 0.500 m from the ring's center. If

dsp73

Answer:

Velocity of the electron = v = 1.2\times 10^8\ m/s.

Explanation:

Given,

Mass of the electron = $m_e\ =\ 9\times 10^{-31}\ kg$
Charge on the electron = $q_e\ =\ 1.62\times 10^{-19}\ C$
Charge density of the ring = $\rho\ =\ +1.00\times 10^{-6}\ C/m$
Radius of the ring = R = 0.70 m
Distance between the electron ant the center or the ring = x = 0.5 m

Now total charge on the ring = $Q\ =\ \rho\times 2\pi R$

Potential energy due to the charged ring to the point on the x-axis is

$P.E.\ =\ \dfrac{KQq_e}{\sqrt{R^2\ +\ x^2}}\\$

Let v be the velocity of the electron at the center of the ring.

Total kinetic energy of the electron = $\dfrac{1}{2}m_ev^2\\$

Now, From the conservation of energy,

the total potential energy of the electron at initially is converted to the total kinetic energy of the electron at the center of the ring,

$\therefore P.E.\ =\ K.E.\\\Rightarrow \dfrac{KQ}{\sqrt{R^2\ +\ x^2}}\ =\ \dfrac{1}{2}m_ev^2\\\Rightarrow v\ =\ sqrt{\dfrac{2 KQq_e}{m_e\sqrt{R^2\ +\ x^2}}}\\\Rightarrow v\ =\ \sqrt{\dfrac{2Kq_e\rho \times 2\pi R}{m_e\sqrt{R^2\ +\ x^2}}}\\\Rightarrow v\ =\ \sqrt{\dfrac{2\times 9\times 10^9\times 1.0\times 10^{-6}\times 2\times 3.14\times 0.7\times 1.6\times 10^{-19}}{9\times 10^{-31}\times \sqrt{0.7^2\ +\ 0.5^2}}}\\\Rightarrow v\ =\ 1.2\times 10^8\ m/s.$

Hence the velocity of the electron on the center of the charged ring is $1.2\times 10^8\ m/s.$

5 0

3 years ago

A 15-kN tensile load will be applied to a 50-m length of steel wire with E = 200 GPa. Determine the smallest diameter wire that

sasho [114]

Answer:

d=13.81 mm

Explanation:

Given that

P = 15 KN ,L = 50 m

E= 200 GPa

ΔL = 25 mm

σ = 150 MPa

Lets take d=Diameter

There are we have two criteria to find out the diameter of the wire

Case I :

According to Stress ,σ = 150 MPa

P = σ A

$A=\dfrac{P}{\sigma}$

$d=\sqrt{\dfrac{4P}{\pi \sigma}}$

By putting the values

$d=\sqrt{\dfrac{4\times 15000}{\pi \times 150}}$

d= 11.28 mm

Case II:

According to elongation ,ΔL = 25 mm

$\Delta L=\dfrac{PL}{AE}$

$A=\dfrac{PL}{E\Delta L}$

$A=\dfrac{4PL}{\pi E\Delta L}$

$d=\sqrt{\dfrac{4\times 15000\times 50000}{\pi \times 200\times 1000\times 25}}$

d=13.81 mm

Therefore the answer will be 13.81 mm .Because it satisfy both the conditions.

3 0

3 years ago

During which phase of a four stroke engine are waste gases removed?

Fed [463]

At the phase of D. Exhaust Stroke, the waste gases are removed during a four stroke engine. It is one of the phases of the four stroke engine. The cylinder from the fuel ignited during the compression step and removed from the cylinder.

4 0

3 years ago

A rifle fires a bullet at a target. The speed of the bullet is 600m/s. The target is located 400m away. How long does it take fo

Alex

0.67s

Explanation:

Given parameters:

Speed of bullet = 600m/s

Distance of target = 400m

Unknown:

Time taken for bullet to reach target = ?

Solution:

Speed is a physical quantity that expresses the rate of change of distance with time;

Speed = $\frac{distance}{time taken}$

Since time is unknown, we make it the subject of the expression;

time = $\frac{distance }{speed}$ = $\frac{400}{600}$

time = 0.67s

Learn more:

Speed brainly.com/question/10048445

#learnwithBrainly

5 0

3 years ago

Graph the following data tables on different graphs.

Anna [14]

Answer:

Sjjsjsjsjsjsjsjwjwjw

8 0

3 years ago