Một chất điểm DĐĐH có phương trình x=Acos(omega t). Tìm thời gian chất điểm qua vtcb theo chiều dương lần thứ 2?

The electric flux through a square-shaped area of side 5 cm near a large charged sheet is found to be 3 × 10−5 N · m2 /C when th

Zanzabum

Answer:

$\sigma=2.124\times 10^{-13}C/m^{2}$

Explanation:

Given:

Electric Flux = $3\times10^{-5}N.m^2/C$

Side of sheet = 5cm

Area of the square sheet, A= 5×5 = 25cm²=25×10⁻⁴ m²

Now

the electric flux (Φ) is given as:

$\phi =EA$

where, E = Electric field

or

$E=\frac{\phi}{A}$

substituting the values in the above equation, we get

$E=\frac{3\times10^{-5}N.m^2/C}{25\times 10^{-4}}$

$E=0.012N/C$

Now the charge density (σ) on a sheet is given as:

$\sigma=2\epsilon_oE$

where, $\epsilon_o$ = Permittivity of the free space = 8.85×10⁻¹²

substituting the values in the above equation, we get

$\sigma=2\times 8.85\times10^{-12}\times 0.012$

$\sigma=2.124\times 10^{-13}C/m^{2}$

5 0

2 years ago

In 1865, jules verne proposed sending men to the moon by firing a space capsule from a 220-m-long cannon with final speed of 10.

Aneli [31]

I'm going to assume that the 10.97 km/s is the velocity at the end of the cannon.

vi = 0
d = 220m
vf = 10.97 km/s = 10970 m/s
a = ???

vf^2 = vi^2 + 2*a*d
10970^2 = 0 + 2*a*220
10970^2 / 440 = a You are right this is a pretty big number.
2.73 * 10^5 = a

5 0

3 years ago

How is a pareto chart different from a standard vertical bar graph?

Alona [7]

Answer:

A Pareto chart, named after an Italian economist, combines a bar chart with a line graph. The bar chart is different from a histogram in more than one way. For example, the vertical bars need not touch one another as per a histogram

Explanation:

5 0

3 years ago

You, Archimedes, suspect that the king’s crown is not solid gold but is instead gold-plated lead. To test your theory, you weigh

hjlf

Answer:

a) 16675.75 Kg/m³ b) 77.6%

Explanation:

the weight of the crown = 60 N, density of gold = 19300 Kg/m^3, density of lead = 11340 kg/m^3, density of water = 1000kg/m^3 and acceleration due to gravity = 9.8 m/s^2

upthrust on the crown = weight in air - weight when fully submerged in water = 60 - 56.4 = 3.6 N

mass of water displaced = 3.6 / 9.8 since weight = mass × g

mass of water displaced = 0.367 Kg

density of water = mass / volume

1000 = 0.367 / volume

cross multiply and find volume

volume of the crown = 0.367 / 1000 = 0.000367 m³ since the crown will displace water of equal volume according to Archimedes principle

Let V1 represent the volume of Gold and let V2 represent the volume of lead

Total volume of the crown = V1 + V2

also

density of gold = mass of gold / V1 and density of lead = mass of lead / V2

19300 = mass of gold in the crown / V1 and 11340 = mass of lead in the crown / V2

19300 V1 = mass of gold and 11340 V2 = mass of lead

add the two together

19300 V1 + 11340 V2 = weigth of the crown / 9.8

19300 V1 + 11340 V2 = 6.12 also

V1 + V2 = 0.000367

make V1 subject of the formula in equation 2

V1 = 0.000367 - V2

substitute for V1 in equation 1

19300 (0.000367 - V2) + 11340 V2 = 6.12

open the bracket

7.083 - 19300 V2 + 11340 V2 = 6.12

rearrange the equation

-7960 V2 = 6.12 - 7.083

-7960 V2 = -0.963

V2 = -0.963 / -7960 = 0.000121 (volume of lead in the crown)

substitute V2 into equation 2

V1 + 0.000121 = 0.000367m³

V1 = 0.000367 - 0.000121 = 0.000246m³ (volume of gold in the crown)

so mass of gold in the crown = 19300 × 0.000246 = 4.748 kg

and mass of lead = 11340 × 0.000121 = 1.372 kg

average density of the crown = (mass of gold + mass of lead) / total volume = 6.12 / 0.000367 = 16675.75 kg/ m³

b) percentage make of gold = mass of gold / total mass × 100 = 77.6 % approx

4 0

2 years ago

A wheelbarrow is a good example of a second-class lever. True or False

olganol [36]

Answer:

true

Explanation:

a wheelbarrow has its load situated between the fulcrum and the force the wheel Barrow is 2nd class because of its resistance between the force and the axis

3 0

2 years ago